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Exercise as it relates to Disease/Is physical activity during pregnancy influential on maternal weight and obstetric outcomes?

This page is a critique on research article The impact of physical activity during pregnancy on maternal weight and obstetric outcomes. Mizgier M, Mruczyk K, Jarząbek-Bielecka G, Jeszka J.Ginekologia polska.

What is the background to this research?
Regular physical activity in all phases of life, including pregnancy, promotes health benefits for the mother and reduces the risk of complications related to foetal development and chronic diseases in children later in life. However, the majority of the woman of childbearing age are overweight or obese. Inactivity in pregnancy and excessive weight gain have been recognised as an independent probability for maternal obesity and related pregnancy complications. Maternal weight pre and post-pregnancy affect mothers and children's health by significantly impacting the in utero environment on fetal development and the child's health throughout their lifespan. Underweight women tend to have low birth weight babies as opposed to obese women who have larger gestation (macrosomia) babies. A low-birth-weight infant is defined as less than 2,500g. Macrosomia, defined as birth weight greater than 4,000g, increases the risk of complications at delivery such as caesarean birth, vacuum extraction, episiotomy, acute haemorrhage, immediate adverse health effects, gestational diabetes, pregnancy-induced hypertension and an increased risk of obesity in childhood.

The US institute of medicine (IOM) published gestational weight gain recommendations for women based on pre-pregnancy body mass index (BMI) in the guidelines are as follows.

Women who exhibited recommended weight gain are at minor risk of complications throughout pregnancy and delivery. Mothers with excess weight gain relative to IOM recommendations more often have neonates that score below the APGAR'5 scale and are more inclined to experience complications. In previous studies, the risk of preterm birth, larger gestation babies (macrosomia), diabetes, preeclampsia, eclampsia and caesarean section increases with increased BMI.

Where is the research from?
This study was conducted by researches of the Institute of Dietetics, Department of Hygiene and Human Nutrition, Poznań University of Life Sciences; at the Gynaecology and Obstetrics Training Hospital, Poznań University of Medical Sciences. The study project was approved by the Independent Ethics Committee at the Karol Marcinkowski University of Medical Sciences, Poznań.

What kind of research was this?
This research was a quantitative assessment designed to assess the intervention of physical activity program of pregnant women throughout their second and third trimester, determining the effect on variables birth weight, appropriate maternal weight gain, the method of delivery and week of gestation. The data was measured through questionnaires and recorded with the ActiGraph GT3X monitor. The evidence provided can be considered a strong article as there isn’t much research comparing short and long term moderate physical activity in pregnant woman.

What did the research involve?
The method consisted of 57 pregnant women with uncomplicated single pregnancies participating in a 10 to 18-week physical activity program through their second and third trimesters. The woman were divided into two groups, short and long moderate physical activity, in comparison by their age, height, weight and body mass index pre and post-pregnancy. ActiGraph GT3X monitor monitored physical activity. Individual consultation with a dietitian was implemented every three weeks to obtain healthy nutrition throughout pregnancy. Data on delivery consisted of body weight, method of delivery, week of birth, gestation birth weight, baby length on study completion.

Exercise program

The physical activity program implemented was designed around the guidelines of the American Congress of obstetricians and gynaecologists. The program consisted of safe pregnancy activities of individualised pregnancy workouts, swimming, walking, cycling and yoga for 30 mins every day or most days of the week. In addition, applicants took part in a training session to assess their current health and fitness status and were advised on adjustments to their level of physical activity before commencement. Participant's physical activity was recorded with the ActiGraph GT3X monitor. In addition, the participants wore the devices on a flexible waistline belt due in daily activities.

Women were advised to not participate in physical activity if they experienced dizziness, headaches, chest pains, muscle weakness or oedema of the calves or any other abnormal signs.

The study's limitations were the small number of participants than anticipated inhibiting the association of physical activity and the pregnancy results of the investigation as well as participants lack of motivation to exercise can also limit findings. Furthermore, addressing the fetal physiological response to maternal physical activity such as heart rate changes would have been an interesting concept. Research has recognised that fetal heart rate increases by 10-30 beats per minute over the baseline of exercise exertion in minimum to moderate exercise.

What were the basic results?
Overall, the researchers concluded that there was a slight variance in weight gain. The women in the shorter duration of physical activity had six times greater excess weight gain than the normal range. The longer duration physical activity showed lower weight gain under the standard recommendation. The shorter duration physical activity group had twice as many cases of macrosomia and vacuum extraction occurring in the group. There were three cases of preterm births and three times as many low birth weight cases in the comparison group. No significance was found in terms of low birth weight or gestation at birth.

Unfortunately, due to the low participant rate, there wasn’t a possibility to show significance in the relationship between the duration of physical activity and macrosomia and obstetric outcomes. Therefore, further investigation is required.

What conclusions can we take from this research?
From the aspect of pregnant women, recent publications are increasing their focus on the impact of physical activity in pregnancy, whether beneficial or adverse to the pregnancy outcome. The research, to a degree, enhances the benefits of increased duration of moderate physical activity as it improves physical fitness and prevents excessive weight gain. However, it reported the shorter intensity and duration of physical activity decreased due to the increased weight changes in the third trimester, resulting in less than 21 minutes of physical activity. The recommendation for exercise is 30 minutes per day. Unfortunately, most pregnant women don’t adhere to regular exercise; therefore, they don’t meet the US institute of medicine standard for the minimum amount of exercise to stay healthy per day. The study also confirmed that physical activity positively affects obstetric outcomes as the number of vaginal births was comparable in both groups.

Further research on the influence of optimal type, frequency and intensity of physical activity programs and the affects on obstetric outcomes need further observation in larger population intervention.

Practical advice
Physical activity guidelines recommend at least 150 minutes of moderate-intensity activity spread out through the week. Women who regularly engaged in vigorous-intensity aerobic exercise or were physically active before conception can continue these activities under the guidance of a doctor, obstetrician - gynaecologist or obstetric care provider. Women with or without a sedentary lifestyle should be encouraged to participate in physical activity. When starting a new exercise program during pregnancy, the Australian guidelines suggest beginning at a light to a moderate exercise program with constant consultation with various professionals to adjust the program accordingly. Guidelines also stipulate if to cease exercise if a woman feels dizzy, abdominal pain, pain or oedema of the calf, vaginal bleeding/fluid, pelvic pain excessive fatigue.

Further information/resources

 * Pregnancy and exercise
 * Health channel pregnancy and physical activity
 * Active pregnancy
 * Pregnancy,birth,baby


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World of Dinosaurs/Allosaurus

Basics
Allosaurus means "other lizard"

Allosuarus fragilis is the state fossil of Utah

Allosaurus lived during the Jurassic Period

Anatomy
Allosaurus is a lanky meat-eating dinosaur.

Look for these features:

- furcula (wish bone, which is found in the theropod clade)

- smooth ankle and smooth hips

- ankle raised off the ground (found in derived theropods)

Phylogeny
Allosaurus is a meat-eating beasty, a theropod dinosaur.

All theropods are dinosaurs.

All dinosaurs are amniotes and all dinosaurs are tetrapods.

Biomechanics
Allosaurus have jaws that can offer a stronger bite than an alligator, but not quite as powerful as a Tyrannosaurus rex.

Fossil Record
Visit the paleobiology database, at paleodb.org

Allosaurus fossils are relatively common in Utah, including many scattered (disarticulated) skeletons at the Cleveland Lloyd Quarry in Utah.

Ecology
Allosaurus is a meat-eating dinosaur which co-occured with Stegosaurus.

Ferns, shrubs, and trees would have been common, but flowering plants, fruit trees, or grass were NOT around yet!

Biogeography
Rocks from the Jurassic period indicate hotter, wetter conditions than we experience on Earth today.

Rocks do not show signs of ice caps near the poles, of traveling ice-bergs, or other signs of "ice house" conditions.

Allosaurus would have roamed deserts and river systems. The super-continent Pangea was splitting up throughout the Jurassic Period, adding volcanic activity, particularly along the west coast, and warm, seasonally-variable habitats.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: World of Dinosaurs/Allosaurus


text: Category:Book:World of Dinosaurs

title: Category:Book:World of Dinosaurs


text:
User talk:KRitterbush

You are copying content from https://uudino.fandom.com/wiki/Clades_of_Interest. Just because something is a wiki doesn't mean that the content is free, or reusable, at Wikibooks. The site needs to have a license that is compatible. As far as I can tell, the content at fandom is not licensed in a way that enables reuse here. Unless you can provide a link to a GFDL or CC-BY-SA compatible license we'll have to delete the pages you've added. QuiteUnusual (discuss • contribs) 11:26, 24 August 2021 (UTC)

 * Thanks for getting in touch, as I'm new to this interface. I created all the content that is posted at the uudino.fandom site - I wrote it and did all the illustrations and photographs. Some of the content is a link-out to other useful features (sketchfab 3D models posted by a museum in Japan), so those items I will ensure here are linked out properly. I prefer to publish my content as CC-BY-NC (non commercial). But the fandom site is completely overrun with advertisements and I can no longer use it. I am a paleontology professor at the University of Utah and I created the UUDino wiki to replace the textbook we'd been using. I made the wiki during the first COVID summer so it was slap-dash, but I worked very hard on all the illustrations. (You can see my name on the CC-BY-NC over at the fandom.) I'm happy to bring that content over to wikimedia, wikicommons, and wiki books. I understand that people can now use the content for commecial purposes, but it's worth it to move to a more mature interface. I will spend this semester creating a useful book here at wikibooks, along with at least two helpers, who are undergrads at the University of Utah employed to help me with the course. I appreciate any advice and help as I try to learn and follow the new forms and format here. Also, I can dis-able or remove the fandom pages as I move content over here. Thank you! KRitterbush (discuss • contribs) 13:23, 24 August 2021 (UTC)

title: User talk:KRitterbush


text:
User:KRitterbush

Kathleen Ritterbush (she/her) is a paleontologist and an Assistant Professor at the University of Utah in Salt Lake City.

She researches ecosystems in deep time, with emphasis on mass extinctions, state shifts, and biomechanics.

View her academic faculty page here.

View her Google Scholar publication listings here.

title: User:KRitterbush


text:
World of Dinosaurs/Animals to Know

We chose a specific list of animals for focus in this book. We will build a page specific to this book for each animal listed below. While we update our custom pages, links may visit the related wikipedia page.

Dimetrodon

Crocodylus

Pteranodon

Magpie

Confuciusornis

Archaeopteryx

Utahraptor

Hagryphus (Oviraptor)

Nothrychus (Therizinosaur, aka "slicing lizard")

Tyrannosaurus rex

Allosaurus fragilis

/Spinosaurus/

Ceoelophysus

Plateosaurus

Eoraptor

Barasaurus

Diplodicus

Camerasuaurs

Brachiosaurus

Fruitadens

Akainacephalus

Scutelosaurus

Stegosaurus

Iguanadon

Parasaurolophus

Psittacasaurus (aka "parrot lizard")

title: World of Dinosaurs/Animals to Know


text:
World of Dinosaurs/Animals to Know/Crocodylus

Basics
Crocodylus is the latin name for the crocodile.

Anatomy
Crocodiles have short legs with webbed feet.

They have very powerful bite forces, but the muscles to open their jaws are very weak.

Phylogeny
Crocodiles are alive today, and are related to a great variety of animals that lived during the time of large dinosaurs.

Biomechanics
Salt water crocodiles have a bite force of [https://www.nationalgeographic.com/animals/article/120315-crocodiles-bite-force-erickson-science-plos-one-strongest#:~:text=The%20%22winners%22%E2%80%94saltwater%20crocodiles,1%2C000%20psi%20(4%2C450%20newtons). 3,700 pounds per square inch, or 16,460 Newtons.]

Fossil Record
Ancient crocodilians have been found as far back as the Jurassic Period.

The oldest true crocodile fossil is 95 million years old and was found in Portugal.

Ecology
Crocodiles are semiaquatic animals.

They tend to live in fresh water bodies, though some can live in salt water.

They are carnivorous and can eat a wide variety of species.

Crocodiles in Africa are perhaps the most famous due to the abundance of nature documentaries on them.

Biogeography
They live in fresh water bodies, though some species live in salt water.


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World of Dinosaurs/Animals to Know/Dimetrodon

Basics
Dimetrodon means "Two Measures of Teeth".

It gained this name because it has teeth of varying sizes.

It lived during the Permian Period.

Anatomy
Dimetrodon is a four-legged egg-laying animal that lived before the time of dinosaurs.

They are most famous for their sail that is made out of spines extending from the vertebrae.

Phylogeny
Dimetrodons are synapsids.

This means they are more closely related to mammals than any living reptile.

Biomechanics
Dimetrodons probably used their sails to regulate their temperature.

Fossil Record
Fossils are found in Oklahoma, Ohio, Texas, Utah, New Mexico, and Arizona.

It is also found in Germany.

Ecology
Dimetrodon lived near aquatic environments where fish would have thrived.

Biogeography
Dimetrodon probably inhabited areas surrounding deltas.


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Talk:World of Dinosaurs/How We Organize Life

I am the explicit creator of all the content I'm posting to this wikibook, although of course other people will be able to augment it. I previously wrote and posted the content to fandom as a way to reach my university students during the first summer of COVID. Now I am trying to move the material over to here. The illustrations and writing is mine. The illustrations I first published to zenodo: 10.5281/zenodo.3958352KRitterbush (discuss • contribs) 15:03, 24 August 2021 (UTC)


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World of Dinosaurs/Clades of Interest

This text will focus on a specific list of clades that help us organize life.

Bilaterians
We propose that the last common ancestor of a cat and a crab had bilateral symmetry. If you draw a line down the middle of a cat, both sides match. Same with a crab.

 * The clade that includes the last common ancestor, and all of its descendants, we call bilaterians.
 * bi = two, lateral = along
 * Do ALL animals in this group still show bilateral symmetry? No.
 * Feel your heart in your chest. It's not exactly in the middle, is it?
 * A garden snail has a coiled shell, and some of its organs are a bit twisty. That's ok. It's still a bilaterian.

Amniotes
We propose that the last common ancestor of a cat and of a lizard had the ability to bear young on land, by protecting the egg from drying out.

 * The clade that includes the last common ancestor of the cat and the lizard, and all of its descendants, we call amniotes.
 * The amnion is a layer in an egg that protects the developing fetus from drying up.
 * Frogs and other animals that are not in this clade need to lay eggs near or in water, so they don't dry out.
 * Sea turtles and lizards and chickens can lay eggs on land. Ever notice the thick stretchy layer inside the crunchy part of a chicken egg? That's the amnion.
 * Mammals, including dogs and cats and humans, have young that develop internally. The amniotic sac is the stretchy layer protecting the egg as it grows. Ever heard of a pregnant woman having her "water break"? This is when the amniotic sac is ready to, well, hatch!

Tetrapods
We propose that the last common ancestor of a cat and a frog had a pelvis.

 * We will call that last common ancestor, and ALL of its descendants, tetrapods.
 * Literally translated, tetrapods means "four feet".
 * But do all animals included in this clade STILL have four FEET? Nope.

What's the deal with tetrapods?

 * A pelvis forms a bridge between the legs and spine.

 * Similarly, shoulder blades and collar bones form a bridge between the arms and spine.

 * Each of these (the pelvis and the shoulder stuff) is easily considered as a package: a girdle.
 * The pelvic girdle manages the legs.
 * The pectoral girdle (think about lifting weights to get swole and have huge pecs) manages the arms, or fore-limbs, or front legs, or wings, or whatever you've got up there.

Fish already have a pectoral girdle, which is how their pectoral fins are linked to the spine.

By adding the pelvis, and with it the hind limbs, this new lineage of animals can generally be imagined as animals with FOUR LIMBS.

Tetrapods have been around for a LONG time, based both of fossils and DNA observations.

 * Because different subclades within tetrapods have been evoloving for such a long time, a lot of them are now very DERIVED.
 * Fossils of basal tetrapods include Acanthostega, and Ichthyostega.
 * Many derived lineages of tetrapods do not currently walk on four legs.
 * Birds
 * Humans
 * Whales
 * Snakes

Next time you see a rattlesnake while hiking, you probably don't want to shout, "Hey! Watch out for that tetrapod!" because this will be confusing to the average person. But you can always give the snake a respectful nod and say, "Nice rattle, there, fellow tetrapod. Look at us, being so derived."

Amniotes
We interpret that the last common ancestor of a cat and a bird had an amnion. We refer to this clade as amniotes.

 * An amnion allows an animal to reproduce away from water, by protecting the embryo in a special layer within an egg (i.e., lizard, platypus, chicken) or womb (cat, rat, etc.).

Synapsids
"With a window". The traits that help us diagnose membership in this clade are often hidden on derived, living animals, so let's dig into the puzzle.

Today, we can easily observe that lots of mammals have a mouth full of differently-shaped teeth. Consider a human and a wolf.

 * Humans have teeth that can cut, crush, and grind.
 * Wolves have teeth that can stab and crush.

Taking advantage of these teeth to do interesting things (bite into food, crack it open, chew it up) requires muscles to move the jaws, and cheeks to keep the food from falling out of the mouth!

 * The skulls of humans and wolves have a large arch of bone (on each side of the skull) that allows the muscles to pass between the skull and lower jaw.
 * The arch of bone also provides attachment points for muscular cheeks.
 * In biology, we call this the "zygomatic arch" which is fun to say!
 * We can look at this arch and consider it as a window for the passage of chewing muscles.

Now, let us consider the fossil animal Dimetrodon.

 * Dimetrodon means "two size teeth", named for the differently sizes teeth in it's mouth.
 * Dimetrodon has a skull with a large round hole placed behind each eye socket.
 * Paleontologists interpret that this hole allowed for muscle attachments for chewing.
 * Our current interpretation is that Dimetrodon is a synapsid - "with a window" - and that it is a basal member of our clade.

The clade synapsid contains the last common ancestor of a monkey and a Dimetrodon, and all of its descendants.

Casually, in this class, we will ignore synapsids.

 * So you'll often see us group Dimetrodon with some familiar mammal, to really drive the point home.

 * For example, see our Dinosaur Time Tree. We leave synapsids alone in their little lineage, and show a basal Dimetrodon back in the Paleozoic Era, and a fuzzy cat in the Cenozoic era.
 * Were synapsids around during the Mesozoic? Of course!
 * Will we learn much about them? Not in this class!

Diapsids
"Two windows". Above, we considered synapsids, which have a single archway in the skull to allow chewing muscles.

Diapsids have TWO archways in the skull behind each eye socket.

 * We interpret that one set of holes aided in muscles to work the jaw.
 * We interpret that the set of holes on TOP of the skull aided in temperature regulation. Let's explore!

In 2019, a team of scientists published peer-reviewed observations of heat-exchanging organs in the skulls of crocodiles and birds.

 * Living crocodiles have holes in the top of their skull.
 * So does the living tetrapod Tuatara.
 * Derived birds don't show these holes in the top of the skull. But many feathered dinosaurs related to birds DO show the holes.

Dr. Ritterbush agrees with the interpretation that the skull-top-holes were used to house heat-regulating organs. Dr. Ritterbush likes to call these holes the HotPockets.

AND, special bonus, there is another useful trait that we can observe in living tetrapods that we group together as diapsids.

Corneous beta protein is a tough material that builds outer body coatings.

 * Corneous beta protein is found in living lizards, alligators, turtles, and birds.
 * Corneous beta protein is composed of tiny protein strings woven together.
 * Corneous beta protein allows construction of strong, flexible materials, such as feathers.
 * It is NOT found in mammals or other synapsids. Mammals make keratin, a different set of proteins.

Observations clarifying this situation are fairly new, such that they haven't made it into mainstream teaching practices or textbooks.

 * Beta proteins were long recognized in microscopes, but were just named beta because they looked different.
 * Recent observations include the microscopic structure of the protein,
 * the number of different beta proteins made by different living animal groups,
 * and the placement of the genes that produce the protein on the DNA patterns of these living animals.

The result is a compelling interpretation that the last common ancestor of birds and lizards had corneous beta proteins.

Diapsids are a case study in the connections between paleontology and biology.

 * Biologists can observe the detailed chemical structures and DNA patterns of living animals.
 * Paleontologists can observe the shapes, sizes, structures, and microscopic appearance of extinct animals.
 * Interpreted relationships between living and dead animals are always changing, because biologists and paleontologists are continually making new observations.

Very important note!

Diapsids and synapsids are SISTER clades, both within the larger clade of amniotes. Diapsids are tetrapod amniotes. Synapsids are tetrapod amniotes. We don't have compelling evidence that either skull configuration is more basal or derived. We typically show them diverging within the amniote clade, and we mark the traits (two windows; one window) on each separate diverging branch.

Archosaurs
"Old lizard". The last common ancestor of a bird and an alligator, and all of its descendants, are a clade we call archosaurs.

Archosaurs are diapsids.

 * Basal archosaurs show a large lump on the rear face of their thigh bone (femur).
 * The lump allows stronger attachment for a muscle that runs between the tail and the thigh (caudal-femoral muscle).
 * The fancy name for this is the "fourth trochanter",
 * but we won't use that term in this class
 * because it implies we are going to learn all the other trochanters,
 * which we definitely will not!

The size of this muscle attachment spot varies in different archosaurs.

 * On fossils, it's easiest to see in crocodiles and dinosaurs.
 * In live animals, it's easier to see in crocodiles.
 * In modern birds:
 * the tail is so reduced and stumpy that,
 * although the muscle is still there
 * and still attaches to the femur,
 * the big lump is not so prominent or easy to see.

Fossils reveal an amazing variety of archosaurs, some of which were enormous and probably terrifying. That's super fun to explore, but we aren't making time in this class.

Dinosaurs
"Scary lizard". The clade dinosaur includes the last common ancestor of a bird and a Triceratops, and all of its descendants.

In this class, we will track two features that dinosaurs acquired.

Find detailed dinosaur cladograms and descriptions over at Dinosaur Clades of Interest.

Dinosaurs have smooth ankles.

 * Crocodile ankles are twisty, which is smooth in the water but sloppy on land.
 * Crocodile ankles concentrate bending motion by oblique shearing of two bones: the calcaneum and the astragulus.
 * In dinosaurs, the astragalus is big and sticks tightly to the shin bone (the tibia).
 * In dinosaurs, the calcaneum is tiny, and just sort of rides alongside the astragulus.
 * I like to remember them this way:
 * The astragalus looks like a big A with a pulley on the bottom.
 * The calcaneum? Barely know 'em!
 * The result in dinosaurs is a smooth hinge ankle that flaps forward and back, like a dog door.
 * Along with dinosaurs, smooth ankles are also found in pterosaurs.
 * Pterosaurs ("wide lizards") are flying archosaurs, like the pterodactyl ("wide finger") and pteranodon ("wide toothless")
 * The clade that includes the last common ancestor of a bird and a pteranodon, and all of its descendents, is called Ornithodira ("bird neck").

Dinosaurs have smooth hips.

 * Crocodile hips are twisty, which is subtle in the water but hilariously awkward on land.
 * Remember that tail-to-thigh muscle that's so prominent in crocodiles?
 * On land they have to really pump that muscle to yank the leg back and prep for a new step.
 * In the water, crocodiles can glide and sneak and creep with subtle adjustments, and their legs are kind of at an angle for little adjustments.
 * On land, crocodiles stagger around with big awkward steps.
 * Crocodile thigh bones have a top notch that fits into the pelvis at a funny angle.
 * Dinosaur hips allow the leg to swing like a clock pendulum. [click link for video of reconstruction!]
 * The dinosaur thigh bone has a top with a sharp 90 degree handle on top, like an upside down golf club.
 * The dinosaur pelvis also has a nice round socket hole for the femur to slide into.
 * The result is a nice smooth swinging motion allowed by the leg,
 * and an upright stance allowed by the whole skeleton.

Check out more clades within the dinosaurs!

Pterosaurs are NOT dinosaurs because they did not inherit the smooth hips.

Birds
The last common ancestor of a crow and an ostrich, and all of its descendants, is the clade of birds.

Birds are a clade within dinosaurs.

A key trait that appears in birds, but not other dinosaurs, is the stumpy tail bone, the pygostyle.

Lots of features we see on a chicken are shared by other dinosaurs:

 * Smooth ankles and smooth hips
 * Upright posture, with the body supported above the ankles.
 * A stout tibia and reduced fibula (shin bones)
 * Long forelimbs that do not walk on the ground.
 * Reduced numbers of fingers and toes.
 * Feathers

Other features we see on a chicken are NOT shared by other dinosaurs:

 * The pygostyle, a stumpy tail bone.
 * The pelvis and sacral spine are so fused together that it's hard to tell what you're even looking at.
 * Compared to their body size, most birds have HUGE eyes and HUGE brains!
 * The adult bird beak has no teeth.
 * The finger bones are fused into a creepy superfinger at the end of the wing.
 * Think about this next time you eat buffalo wings.
 * Or don't. It's disconcerting.

title: World of Dinosaurs/Clades of Interest


text:
Talk:Applications of ICT in Libraries/Locating Information

Url check
Keeping this in mind that this is a Featured Book, the authors of this work need to renew the urls as they are quite dated. For example, this url: http://www.royal.gov.uk/ is giving problems with 504 Error and even the Wayback Machine archived links of that resource are problematic. In the case of troublesome urls, alternate information resources need to be used instead. AshLin (discuss • contribs) 15:47, 24 August 2021 (UTC)


text: Web App Development with Google Apps Script/templates


text:
Stardew Valley/Dumpster Diving

Dumpster diving is a good way to obtain items early on, since it offers more consistent locations than foraging with a wider item variety. There are eight different garbage cans around town that occasionally will give items when clicked on.

Items gained while dumpster diving tend to be linked to the buildings near where they are located. For example, food items can often be found at the trash can of the Stardrop Saloon on lucky days. On unlucky days, typically only trash can be found at most locations. Some items in the trash can only be found after completing certain achievements.

Every character, except Linus, does not like dumpster diving, and their opinion of the player will be reduced if they see them looking in the trash, voicing their displeasure.

One rare item you can get is a trash can lid, a wearable hat, with the description A trash can lid, "upcycled" into a hat.

title: Stardew Valley/Dumpster Diving


text:
Skaneateles Conservation Area/Invasive species/Ficaria verna

Ficaria verna (lesser celandine)
Lesser celandine (Ficaria verna) is a Eurasian groundcover that is considered to be invasive where it occurs in forested floodplains in North America. Its sale is now prohibited in New York State.

There have been no known observations of Ficaria verna at the Skaneateles Conservation Area (SCA), but there is a good-sized infestation at nearby Baltimore Woods Nature Center, near the Mildred Faust Wildflower Garden, where it was likely either planted intentionally or was introduced as a contaminant with another plant.

Given the fact that yard waste is commonly disposed of and recycled at the Skaneateles Transfer Station at the center of the SCA, there seems to be a chance that Ficaria verna could be found at the SCA.

This species was originally included in the genus Ranunculus as R. ficaria L. but is currently placed in the separate genus Ficaria to maintain monophyletic genera within the Ranunculaceae.

The name celandine is used for the the related invasive species Chelidonium majus

Invasiveness ranking for Ficaria verna (lesser celandine)
Lesser celandine (Ficaria verna, assessed as Ranunculus ficaria in 2008) received and invasiveness score of 77/90 = 85.56%. and is now prohibited by New York State law.

1. Ecological impact (23/30)
1.1. Impact on Natural Ecosystem Processes and System-Wide Parameters: Unknown (U)

1.2. Impact on Natural Community Structure: Influences structure in one layer (3/10)

 * Increases density of the herbaceous layer.

1.3. Impact on Natural Community Composition: Causes major alteration in community composition (10/10)

 * May substantially increases herb layer, extirpating many native species, especially spring ephemerals.

1.4. Impact on other species or species groups: Severe impact on other species or species groups (10/10)

 * Large infestations of this plant eliminate spring ephemeral communities in woodlands, including native plants and presumably all species dependent on these plants.

2. Biological characteristics and dispersal ability (25/25)
2.1. Mode and rate of reproduction: Abundant reproduction with vegetative asexual spread documented (4/4)

 * Extensive, highly aggressive vegetative reproduction.

2.2 Innate potential for long-distance dispersal:

2.3. Potential to be spread by human activities:

2.4. Characteristics that increase competitive advantage:

2.5. Growth vigor:

2.6. Germination/Regeneration:

2.7. Other species in the genus invasive in New York or elsewhere:

3. Ecological amplitude and distribution (25/25)
3.1. Density of stands in natural areas in the northeastern USA and eastern Canada:

3.2. Number of habitats the species may invade:

3.3. Role of disturbance in establishment:

3.4. Climate in native range:

3.5. Current introduced distribution in the northeastern USA and eastern Canada:

3.6. Current introduced distribution of the species in natural areas in the 8 NY PRISMs:

4. Difficulty of control (10/10)
4.1. Seed banks:

4.2. Vegetative regeneration:

4.3. Level of effort required:


text:
Template:Associated Wikimedia/doc

This template is designed primarily for use on bookshelves and Wikibooks; all project links are optional. For best results (as regards the aesthetics of the template output), provide values to parameters for at least two different sister projects. Extant link box templates for individual sister projects can be found in the See also section below.

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In this example the topic is set as an unnamed parameter immediately following the template name. This works equally well as using the named Topic parameter shown above, or using the 1 parameter to hold the value; let personal preference dictate your usage.

TemplateData
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title: Template:Associated Wikimedia/doc


text: User talk:71.217.176.107

title: User talk:71.217.176.107


text:
Exercise as it relates to Disease/Lowering the odds of sarcopenia through physical activity in older adults?

This is a critical analysis of the article from the Aging Clinical and Experimental Research Journal, "Physical activity and sarcopenia in older adults" - Nathan F. Meier & Duck-chul Lee (2020)

What is the background of the research?
Physical activity is an essential part of being healthy, however for those that experience sarcopenia it's not as easy to engage in exercise. According to WHO, physical activity is defined as any bodily movement produced by skeletal muscles that requires energy expenditure. Regular physical activity reduces the risk of mortality from several diseases and has a beneficial effect on bone, muscle and joint health. People who experience sarcopenia have a reduction in muscle mass, strength and function as they age. This leaves them weak, frail, and more susceptible to falls, functional decline, loss of independence and mortality. This is why many sarcopenic patients are inactive and reluctant to participate in physical activity as they may fear falling during exercise. However physical activity can be used as an intervention and is extremely beneficial in reducing sarcopenia. Engaging in physical activity may slow, prevent or reverse muscle loss. Exercise training, particularly strength and resistance, is vital and should be prescribed to those that experience sarcopenia.

Where is the research from?
This article was published in the Aging Clinical and Experimental Research Journal by Springer Verlag. It offers a range of multidisciplinary research on the progressing field of gerontology and geriatrics.

The primary author of this paper, Nathan F. Meir is currently an assistant professor at the Concordia University Irvine, USA. He's published 21 articles, many of which are about physical activity reducing and preventing disease. Nathan only finished his Ph.D in 2018, and doesn't have a long history of research papers, with many of his writings only published in the past 5 years. The secondary author, Duck-chul Lee is from Iowa State University, USA. He currently has 215 publications, with his main areas of research being public health, epidemiology, and cardiology. Both authors bring knowledge and understanding of health issues to this research paper.

What kind of research was this?
This study is a cross-sectional design, which looks at the exposure of an intervention on the outcome of a disease. This study looks at the exposure of physical activity, to cause a positive outcome on older adults who experience sarcopenia. However because outcome and exposure are both measured at the same time it is difficult to establish a causal relationship. This means that any associations that are found in the study may be difficult to interpret. Cross-sectional studies can provide key information for further in-depth research studies and provide the basis for creating a hypothesis. This method was appropriate for establishing the basis of how exercise can impact on sarcopenia, but further research is needed to determine if a causal relationship exists.

What did the research involve?
Between 2015-2016, 304 adults were recruited from a mid-west university town in USA, aged 65 years and older. Individuals who had cancer or other severe conditions that make exercise difficult were excluded from the study. The individuals participated in a series of laboratory tests over 2 weeks. Previous studies on sarcopenia predominately used self-reporting data, however, using this type of measure tends to cause over-reporting and recall bias. This study aimed to limit these issues by using measurement devices to accurately capture data. These included:

 * Daily steps using an accelerometer-based pedometer - published cut-points for older adults, >2500-4999 (moderate) and >5000 (high)
 * Muscular strength tests using a one repetition maximum chest and leg press.
 * Full body scan using DXA model to approximate muscle mass in sarcopenia patients.
 * 400m walk test for cardiorespiratory fitness.
 * Handgrip strength using a digital dynamometer.

This methodology wasn't the best approach. Physical activity is a very broad topic requiring multiple variables to understand the whole picture. This method shows that cardiorespiratory fitness, strength, and speed all need to be analysed, which makes it hard to determine cause and effect, and even more difficult with no control group in the study.

What were the basic results?
The results presented in this study had significant outcomes. The prevalence of the outcomes was recorded using an odd ratio (OR):

 * OR=1 Exposure does not affect odds of outcome
 * OR> Exposure associated with higher odds of outcome
 * OR<1 Exposure associated with lower odds of outcome

It was shown that participants with a high step count had lower odds (0.58) of sarcopenia. Cardiorespiratory fitness lowered the odds across all three categories, and was significantly (0.001) associated with lower prevalence of sarcopenia and higher muscle strength. Strength was associated with lower odds of sarcopenia, low muscle mass, and low handgrip strength. Having greater strength was significantly (<0.001) associated with higher muscle mass, and lower prevalence of sarcopenia.

What conclusions can we take from this research?
Muscle strength is extremely important for elderly people, especially those experiencing sarcopenia, as it is a critical component of being able to achieve daily tasks e.g.walking. This study has shown that an association exists between older adults who participate in physical activity, including cardiorespiratory fitness and strength, and lower odds of experiencing sarcopenia. Studies by Hyuntae Park et al, and Michal Steff et al, both show similar results that physical activity increases muscle mass and reduces the odds of experiencing sarcopenia.

Although there was a large research sample, the majority were relatively healthy which may have caused a healthy responder bias. There was also no consideration for ethnic diversity, making the sample hard to generalize to the population. As mentioned previously, physical activity can't be assessed through a single variable, multiple variables are required to understand physical activity as a whole. This means causal links cannot be made from this research, instead only associations and potential relationships that may exist. Further research is required to determine a relationship between the variables.

Practical Advice
Results from this study and several others suggest that older adults should participate in some form of physical activity each day to minimise the risk factors that are associated with sarcopenia. It is advised that the physical activity should include activities that increase strength, cardiorespiratory fitness, and light aerobic activities (e.g walking). It's shown that older adults who engage in physical activity lower their odds of experiencing sarcopenia. However because older adults are at greater risk of falls and other health issues during exercise, it is recommended to seek medical advice before undertaking any form of exercise. Exercise should be prescribed by an allied health professional, and it is advised that informed consent and an ESSA Adult pre screening is undertaken before any physical activity is conducted, to protect both the patient and the practitioner.

Further information/resources
If you would like to know or learn more about this topic, below are some additional resources:

 * Finding a Exercise Professional: https://www.essa.org.au/Public/find-aep.aspx
 * Fact sheet: https://aimss.org.au/wp-content/uploads/2018/07/Sarcopenia-Day-factsheet-for-Patients.pdf
 * Basic Exercises for Sarcopenia: https://www.iowaclinic.com/physical-therapy/how-to-combat-sarcopenia/


text:
Pokémon Crystal/Development/Sandbox Draft

This page was used for staging changes.


text:
Pokémon Crystal/Introduction

About the game
Pokémon Crystal is an enhanced remake of Pokémon Gold and Pokémon Silver for the Nintendo GameBoy Color. This game features greater player customization, more animations, and additional content.

It later received a re-release for the Nintendo 3DS virtual console, with slight modifications to disable certain virtual console functions and to enable trading between 3DS systems and to the then current line up of Pokémon games.

Differences
If you started playing Pokémon before playing this game, understand that there are a few major differences between this generation and others.
 * Fairy typing for Pokémon did not exist this generation. Many Pokémon that become Fairy type in generation six onwards are simply Normal type in this game.
 * This generation introduced Dark and Steel type Pokémon, adding important counters to many types, especially Psychic type.

title: Pokémon Crystal/Introduction


text:
Quantum Chemistry/

This is an introductory course in quantum chemistry. Example Problems

 * /Example 1/
 * /Example 2/
 * /Example 3/
 * /Example 4/
 * /Example 5/
 * /Example 6/
 * /Example 7/
 * /Example 8/
 * /Example 9/
 * /Example 10/
 * /Example 11/
 * /Example 12/
 * /Example 13/
 * /Example 14/
 * /Example 15/
 * /Example 16/
 * /Example 17/
 * /Example 18/
 * /Example 19/

title: Quantum Chemistry/


text:
Pokémon Crystal/Options

Starting the game givers the player two choices, "NEW GAME" and "OPTION".

"OPTION" contains a number of graphical and quality of life features. It is highly recommended that the player makes some adjustments here.

The Option menu includes:
 * Text Speed: Controls how quickly text is displayed.
 * Battle Scene: Turns battle animations on and off.
 * Battle Style: Lets the player elect to choose whether or not they want to change Pokémon after fainting in battle.
 * Shift: This is the default option and allows the player to swap Pokémon more often in battle, giving the player an edge.
 * Set: Somewhat more difficult, which may or may not be desirable. It makes the game more fair to the computer player and can streamline the battle experience.
 * The player may choose between Mono and Stereo audio.
 * Mono is the default choice, and is what should be used if playing on an original Game Boy Color, or on a Game Boy Player equipped GameCube on a Mono speaker TV.
 * Stereo is ideal for playing on a Game Boy Color with headphones, a GameCube with GameBoy Player and Stereo speaker Television, or a Nintendo 3DS.
 * Print settings only applies to gamers using the Game Boy printer accessory.
 * Menu account: Displays a small tooltip for each option when the in-game menu is open.
 * Frame: Allows the player to select one of eight decorative window frames for notifications and text boxes.

When ready choose cancel (Despite the name, your settings will be saved.) or press B to return to the menu and select New Game.

title: Pokémon Crystal/Options


text:
Quantum Chemistry//Example 1

Write a question and its solution that quantitatively demonstrates the Heisenberg uncertainty principle for a quantum harmonic oscillator in the ground state (n=0).

title: Quantum Chemistry//Example 1


text:
Quantum Chemistry/Example 1

Find ⟨x⟩, ⟨x2⟩, ⟨px⟩ and ⟨px2⟩ for a quantum harmonic oscillator in the ground state, then determine the uncertainty on the position and momentum. Is the product of the uncertainty on position and momentum consistent with the Heisenberg's Uncertainty Principle?

$$

The wavefunction of a quantum harmonic oscillator in the ground state is:

$$\qquad \qquad \Psi_0(x)= \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{4} e^{\frac{-\alpha x^2}{2}}$$          Using this wavefunction the average position and the average of the square of the position can be calculated.

The average position:

$$ \qquad \qquad \qquad \qquad \langle x \rangle = \int_{-\infty}^{\infty} \Psi_0(x)^* \cdot x\cdot \Psi_0(x) \ dx $$

$$\qquad \qquad \qquad \qquad \quad \ = \int_{-\infty}^{\infty} \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{4} e^\frac{-\alpha x^2}{2} \cdot x\cdot \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{4} e^\frac{-\alpha x^2}{2} \ dx$$

$$\qquad \qquad \qquad \qquad \quad \ = \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{2} \int_{-\infty}^{\infty} x \cdot e^\frac{-2\alpha x^2}{2} \ dx$$

$$\qquad \qquad \qquad \qquad \quad \ = \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{2} \int_{-\infty}^{\infty} x e^{-\alpha x^2} \ dx$$

$$

$$\qquad \qquad \qquad \qquad \quad \ = \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{2} \biggl[\frac{-1}{2\alpha} \cdot e^{-\alpha x^2}\biggl]_{-\infty}^{\infty} $$

$$\qquad \qquad \qquad \qquad \quad \ = \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{2} \biggl[\frac{-1}{2\alpha} (0)-\frac{1}{2\alpha} (0) \biggl] $$

$$ \qquad \qquad \qquad \qquad \langle x \rangle = 0 $$

The average square of the position:

$$ \qquad \qquad \qquad \qquad\qquad\qquad\qquad \langle x^2 \rangle = \int_{-\infty}^{\infty} \Psi_0(x)^* \cdot x^2\cdot \Psi_0(x) \ dx $$

$$\qquad \qquad \qquad \qquad\qquad\qquad\qquad \qquad = \int_{-\infty}^{\infty} \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{4} e^\frac{-\alpha x^2}{2} \cdot x^2 \cdot \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{4} e^\frac{-\alpha x^2}{2} \ dx$$

$$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad = \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{2} \int_{-\infty}^{\infty} x^2 e^\frac{-2\alpha x^2}{2} \ dx$$

$$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad = \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{2} \int_{-\infty}^{\infty} x^2 e^{-\alpha x^2} \ dx$$

$$

$$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad = \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{2} \Biggl[ \frac{\sqrt{\pi} \text{erf}(\sqrt{\alpha}x)}{4 \alpha^\frac{3}{2}} \ - \ \frac{x e^{-\alpha x^2}}{2\alpha} \Biggl] _{-\infty}^{\infty}

$$

$$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad = \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{2} \Biggl[ \frac{\sqrt{\pi}}{4}\frac{\text{erf}(\sqrt{\alpha} \ \infty )} {\alpha^\frac{3}{2}} \ - \ \lim_{x \to \infty}\frac{x e^{-\alpha x^2}}{2\alpha} \Biggl] \ - \ \Biggl[ \frac{\sqrt{\pi}}{4}\frac{\text{erf}(\sqrt{\alpha} -\infty )} {\alpha^\frac{3}{2}} \ - \ \lim_{x \to -\infty}\frac{x e^{-\alpha x^2}}{2\alpha} \Biggl] $$$$

$$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad = \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{2} \ \frac{\sqrt{\pi}}{4} \ \frac{1}{\alpha^\frac{3}{2}} \ \biggl[ 1 - (-1) \biggl] $$

$$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad =\frac{\sqrt{\alpha}}{\sqrt{\pi}} \ \frac{\sqrt{\pi}}{4} \ \frac{1}{\alpha^\frac{3}{2}} \ \bigl[2 \bigl] $$

$$\qquad \qquad \qquad \qquad\qquad\qquad\qquad \langle x^2 \rangle = \frac{1}{2\alpha} $$

The uncertainty on the position:

$$ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \delta_x= \sqrt{\langle x^2 \rangle-\langle x \rangle ^2} $$

$$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ \ \ = \sqrt{\frac{1}{2\alpha}-0}

$$

$$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \delta_x= \sqrt{\frac{1}{2\alpha}}

$$

The average momentum:

$$ \qquad \qquad \qquad \qquad \qquad \qquad \langle p_x \rangle = \int_{-\infty}^{\infty} \Psi_0(x)^* \cdot \widehat{p}_x \cdot \Psi_0(x) \ dx \qquad \qquad \qquad \qquad \qquad \qquad \text{where} \ \widehat{p}_x = -i\hbar \Bigl(\frac{\partial}{\partial x} \Bigr) $$

$$\qquad \qquad \qquad \qquad\qquad\qquad\qquad = \int_{-\infty}^{\infty} \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{4} e^\frac{-\alpha x^2}{2} \cdot -i\hbar \Biggl(\frac{\partial}{\partial x} \Biggr) \cdot \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{4} e^\frac{-\alpha x^2}{2} \ dx $$

$$\qquad \qquad \qquad \qquad\qquad\qquad\qquad = \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{2} -i\hbar \int_{-\infty}^{\infty} e^\frac{-\alpha x^2}{2} \Biggl(\frac{\partial}{\partial x} \Biggr) \ e^\frac{-\alpha x^2}{2} \ dx $$

$$\qquad \qquad \qquad \qquad\qquad\qquad\qquad = \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{2} -i\hbar \int_{-\infty}^{\infty} e^\frac{-\alpha x^2}{2} \biggl(-\alpha x \ e^\frac{-\alpha x^2}{2} \biggl) \ dx $$

$$\qquad \qquad \qquad \qquad\qquad\qquad\qquad = -\alpha \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{2} -i\hbar \int_{-\infty}^{\infty} e^\frac{-\alpha x^2}{2} \cdot x \ e^\frac{-\alpha x^2}{2} \ dx $$

$$\qquad \qquad \qquad \qquad\qquad\qquad\qquad = -\alpha \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{2} -i\hbar \int_{-\infty}^{\infty} x \ e^ {-\alpha x^2} \ dx $$   $$

$$\qquad \qquad \qquad \qquad\qquad\qquad\qquad = \frac{-\alpha^{\frac{3}{2}}}{\sqrt{\pi}} -i\hbar \

\biggl[ \frac{1}{2 \alpha} \ e^ {-2\alpha x^2} \biggl]_{-\infty}^{\infty} $$

$$ \qquad \qquad \qquad \qquad\qquad\qquad\qquad = \frac{-\alpha^{\frac{3}{2}}}{\sqrt{\pi}} -i\hbar

\biggl[\frac{1}{2\alpha} (0)-\frac{1}{2\alpha} (0) \biggl] $$

$$ \qquad \qquad \qquad \qquad \qquad \qquad \langle p_x \rangle = 0 $$

The average square of the momentum:

$$ \qquad \qquad \qquad \qquad \qquad \qquad \langle p^2_x \rangle = \int_{-\infty}^{\infty} \Psi_0(x)^* \cdot \widehat{p}^2_x \cdot \Psi_0(x) \ dx $$

$$\qquad \qquad\qquad\qquad\qquad \qquad\qquad = \int_{-\infty}^{\infty} \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{4} e^\frac{-\alpha x^2}{2} \cdot \Biggl(-i\hbar \frac{\partial}{\partial x} \Biggr) \Biggl(-i\hbar \frac{\partial}{\partial x} \Biggr) \cdot \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{4} e^\frac{-\alpha x^2}{2} \ dx $$

$$\qquad \qquad\qquad\qquad\qquad \qquad\qquad = \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{2} i^2\hbar^2 \int_{-\infty}^{\infty} e^\frac{-\alpha x^2}{2} \Biggl(\frac{\partial}{\partial x} \Biggr)\Biggl(\frac{\partial}{\partial x} \Biggr) \ e^\frac{-\alpha x^2}{2} \ dx $$

$$\qquad \qquad\qquad\qquad\qquad \qquad\qquad = \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{2} (-1)\hbar^2 \int_{-\infty}^{\infty} e^\frac{-\alpha x^2}{2} \Biggl(\frac{\partial}{\partial x} \Biggr) \ \biggl(-\alpha xe^\frac{-\alpha x^2}{2}\biggl) \ dx $$

$$\qquad \qquad\qquad\qquad\qquad \qquad\qquad = \alpha \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{2} \hbar^2 \int_{-\infty}^{\infty} e^\frac{-\alpha x^2}{2} \biggl( e^\frac{-\alpha x^2}{2}+\alpha x^2e^\frac{-2\alpha x^2}{2}\biggl) \ dx $$

$$\qquad\qquad\qquad\qquad\qquad \qquad\qquad = \frac{\alpha^\frac{3}{2}}{\sqrt{\pi}} \ \hbar^2 \int_{-\infty}^{\infty} e^{-\alpha x^2} \ dx \int_{-\infty}^{\infty} e^\frac{-\alpha x^2}{2} \ \alpha x^2e^\frac{-2\alpha x^2}{2} \ dx $$     $$

$$\qquad \qquad\qquad\qquad\qquad \qquad\qquad = \frac{\alpha^\frac{3}{2}}{\sqrt{\pi}} \ \hbar^2 \biggl(\sqrt{\frac{\pi}{\alpha}} + \alpha \int_{-\infty}^{\infty} x^2e^{-\alpha x^2} \ dx \biggl)

$$ $$

$$\qquad \qquad\qquad\qquad\qquad \qquad\qquad = \frac{\alpha^\frac{3}{2}}{\sqrt{\pi}} \ \hbar^2 \Biggl( \sqrt{\frac{\pi}{\alpha}} + \alpha \biggl[\frac{\sqrt{\pi} \ \text{erf}(\sqrt\alpha \ x) } {4\alpha^{\frac{3}{2}}} - \  \frac{xe^{-\alpha x^2}}{2\alpha} \biggl]_{-\infty}^{\infty} \Biggl) $$

$$\qquad \qquad\qquad\qquad\qquad \qquad\qquad = \frac{\alpha^\frac{3}{2}}{\sqrt{\pi}} \ \hbar^2 \Biggl( \sqrt{\frac{\pi}{\alpha}} + \alpha \Biggl[\frac{\sqrt{\pi}} {4\alpha^{\frac{3}{2}}} \ \text{erf}(\sqrt\alpha \infty) \lim_{x \to \infty} - \ \frac{xe^{-\alpha x^2}}{2\alpha} \Biggl] \ - \ \Biggl[\frac{\sqrt{\pi}} {4\alpha^{\frac{3}{2}}} \ \text{erf}(\sqrt\alpha -\infty) \lim_{x \to -\infty} - \ \frac{xe^{-\alpha x^2}}{2\alpha} \Biggl] \Biggl) $$

$$

$$\qquad \qquad\qquad\qquad\qquad \qquad\qquad = \frac{\alpha^\frac{3}{2}}{\sqrt{\pi}} \ \hbar^2 \Biggl( \sqrt{\frac{\pi}{\alpha}} + \alpha \Biggl[\frac{\sqrt{\pi}} {4\alpha^{\frac{3}{2}}} \ (1) - \ (0) \Biggl] \ - \ \Biggl[\frac{\sqrt{\pi}} {4\alpha^{\frac{3}{2}}} \ (-1) - \ (0) \Biggl] \Biggl) $$

$$\qquad \qquad\qquad\qquad\qquad \qquad\qquad = \frac{\alpha^\frac{3}{2}}{\sqrt{\pi}} \ \hbar^2 \Biggl[ \sqrt{\frac{\pi}{\alpha}} + \alpha \biggl(\frac{2\sqrt{\pi}} {4\alpha^{\frac{3}{2}}} \  \biggl) \Biggl] $$

$$\qquad \qquad\qquad\qquad\qquad \qquad\qquad = \frac{\alpha}{\sqrt{\pi}} \ \hbar^2 \Biggl[ \sqrt{\pi} + \frac{1}{2} \sqrt{\pi} \Biggl] $$

$$ \qquad \qquad \qquad \qquad \qquad \qquad \langle p^2_x \rangle = \frac{1}{2} \alpha \hbar^2 $$

The uncertainty on the momentum:

$$ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \delta p_x= \sqrt{\langle p^2_x \rangle-\langle p_x \rangle ^2} $$

$$ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ = \sqrt{\frac{a \hbar^2}{2}-0 ^2}

$$

$$ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \delta p_x= \sqrt{\frac{a^2}{2}} \ \hbar

$$

The product of the uncertainty on the position and the uncertainty on the momentum is:

$$ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \delta_x \cdot \delta p_x= \frac{1}{\sqrt{2\alpha}} \cdot \sqrt{\frac{\alpha}{2}}\ \hbar

$$

$$ \qquad \qquad \qquad \qquad \qquad \qquad \qquad\qquad\qquad \ = \frac{1}{2} \ \hbar

$$

This is equal to $$\frac{\hbar}{2}$$, therefore, a quantum harmonic oscillator in the ground state is consistent with the Heisenberg Uncertainty Principle.

title: Quantum Chemistry/Example 1


text:
Quantum Chemistry/Example 2

Question
Starting with the wave equation of a 1D box:

$$ \Psi\left( x \right) = \sqrt{\frac{2}{L}} \sin\left( \frac{n\pi}{L}x \right) $$

The classical probability of a particle existing in the region of the box [0, 1/L] is,

$$ P\left( 1/L \right)_{classical} = \frac{1}{L} $$

Where $$ L $$ is the length of the 1D box, $$ n $$ is the principle quantum number, and $$ x $$ is the position of the particle in a 1D box.

Derive an equation that calculates the probability of the particle being in the [0, 1/L] for the quantum mechanical particle in a 1D box. Is this result compatible with the correspondence principle?

title: Quantum Chemistry/Example 2


text:
Quantum Chemistry/Example 3

Question 3
Calculate the number of nodes for a particle in a 1D box when n=2 and n=3 when the length of the box is L=5, and give the x-intercept of the node(s). $$

Part 1
nnodes=n-1 nnodes=2-1=1 We're looking for one x-intercept. $$ 0 \div \sqrt{\frac{2}{5}} = \sqrt{\frac{2}{5}}\sin\left( \frac{2\pi x}{5} \right)\div \sqrt{\frac{2}{5}}$$ $$ 0= \sqrt{\frac{2}{5}}\sin\left( \frac{2\pi x}{5} \right) $$ Use the form $$\sin(kx)$$ where $$ k= \left( \frac{2\pi}{5} \right) $$ $$ Period= \left( \frac{2\pi}{k} \right)$$ $$ Period= \left( \frac{2\pi}{\frac{2\pi}{5}} \right)=5$$ $$ \left( \frac{Period}{2} \right) $$ is the x-intercept x-intercepts are 0, $$\left( \frac{5}{2} \right)$$, 5 0 and 5 are the edges of the box and $$ \left( \frac{5}{2} \right) $$ is the only node.

Part 2
nnodes=n-1 nnodes=3-1=2 We're looking for two x-intercepts. $$ 0 \div \sqrt{\frac{3}{5}} = \sqrt{\frac{3}{5}}\sin\left( \frac{3\pi x}{5} \right)\div \sqrt{\frac{3}{5}}$$ $$ 0= \sqrt{\frac{3}{5}}\sin\left( \frac{3\pi x}{5} \right) $$ $$ k= \left( \frac{3\pi}{5} \right) $$ $$ Period= \left( \frac{3\pi}{\frac{3\pi}{5}} \right)= \frac{10}{3}$$ x-intercepts are $$\left( \frac{5n}{3} \right)$$ x-intercepts = 0, $$\left( \frac{5}{3} \right),\left( \frac{10}{3} \right),5 $$ 0 and 5 are the edges of the box, $$ \left( \frac{5}{3} \right) $$ and $$ \left( \frac{10}{3} \right) $$ are the nodes.

title: Quantum Chemistry/Example 3


text:
Quantum Chemistry/Example 4

Question
An electron in a 1D box emits a photon as the electron transitions to a lower energy level. If the length of the 1D box is equal to 1.0 cm, and the quantum number transition is $$n_{{3}\rightarrow{2}}$$, what is the electromagnetic radiation frequency of the emitted photon?

Solution:

The energy level of a particle in a 1D box at a specific quantum number ($${E_n}$$) is,

$$E_n = \frac{h^2} {8mL^2} {n^2}$$

Where $$h$$ is equal to Planck's constant (6.62607015 x 10-34 J$$\cdot$$s), $$n$$ is equal to the quantum number ($$n$$ = 1, 2, 3, ...), $$m$$ is equal to the mass of the particle, and $$L$$ is equal to the length of the 1D box. For an electron, the mass is equal to 9.10938356 x 10-31 kg.

Since $${E_n}\propto{n^2}$$ (assuming the mass and box length are constant), the energy level increases by a factor of 4 as the quantum number increases by a factor of 2. Therefore, if a particle in a 1D box undergoes an energy level transition, there is a difference between the initial and final quantum number energy levels. The energy level difference ($$\Delta {E}$$) of a particle in a 1D box that has undergone an energy level transition is,

$$\Delta {E} = {E_{n_f}} - {E_{n_i}}$$

$$= \left[\frac{h^2} {8mL^2} {n^2_f}\right] - \left[\frac{h^2} {8mL^2} {n^2_i}\right]$$

$$\Delta {E} = \frac{h^2} {8mL^2} {({n^2_f} - {n^2_i}})$$

Where $$n_f$$ is equal to the final quantum number, and $$n_i$$ is equal to the initial quantum number.

If $$n_f > n_i$$, $$\Delta {E}$$ is a positive value; photon absorbed.

If $$n_f < n_i$$, $$\Delta {E}$$ is a negative value; photon emitted.

Therefore, the energy level difference of an electron in a 1D box with a length of 1.0 cm, which has undergone a $$n_{{3}\rightarrow{2}}$$ transition is,

$$\Delta {E} = \left[\frac{(6.6261 \times 10^{-34} \text{ J}\text{ s})^2} {8(9.1094 \times 10^{-31}\text{ kg})(1.0 \times 10^{-2}\text{ m})^2}\right] {({2}^2 - {3}^2}) = \left[\frac{(6.6261 \times 10^{-34} \text{ m}^2\text{ kg}\text{ s}^{-1})^2} {8(9.1094 \times 10^{-31}\text{ kg})(1.0 \times 10^{-2}\text{ m})^2}\right] {({2}^2 - {3}^2})$$

$$\Delta {E} = -3.01 \times 10^{-33} \text{ J}$$

The energy level difference for the electron which underwent a $$n_{{3}\rightarrow{2}}$$ transition is equal to -3.01 × 10-33 J. Since $$n_f < n_i$$, 3.01 × 10-33 J was emitted from the electron. If the electron underwent a $$n_{{2}\rightarrow{3}}$$ transition, the electron would absorb the same amount of energy that was emitted from the $$n_{{3}\rightarrow{2}}$$ transition which was 3.01 × 10-33 J.

$$\Delta {E} = \left[\frac{(6.6261 \times 10^{-34} \text{ m}^2\text{ kg}\text{ s}^{-1})^2} {8(9.1094 \times 10^{-31}\text{ kg})(1.0 \times 10^{-2}\text{ m})^2}\right] {({3}^2 - {2}^2}) = 3.01 \times 10^{-33} \text{ J}$$

Therefore,

$$E_{\text{photon}_{n_{2\rightarrow3}}} = E_{\text{photon}_{n_{3\rightarrow2}}}$$

The energy of a photon has a specific frequency of electromagnetic (EM) radiation, and the energy is directly proportional to the frequency. The energy of the photon is equal to,

$$E = h\cdot{f}$$

Where $$h$$ is equal to Planck's constant (6.62607015 x 10-34 J$$\cdot$$s), and $$f$$ is equal to EM radiation frequency.

Rearranging this equation allows for the calculation of the photon EM radiation frequency,

$$E = h\cdot{f}$$

$$f = \frac{E} {h}$$

The calculated photon energy was equal to 3.01 × 10-33 J, therefore the EM radiation frequency of the emitted photon from the $$n_{{3}\rightarrow{2}}$$ transition of an electron in a 1D box with a length of 1.0 cm is,

$$f = \frac{3.01 \times 10^{-33} \text{ J}} {6.6261 \times 10^{-34} \text{ J}\text{ s}}$$

$$f = 4.54 \text{s}^{-1} = 4.54 \text{Hz}$$

title: Quantum Chemistry/Example 4


text:
Quantum Chemistry/Example 5

For a particle (assume the particle is an electron with 1 quantum number.) in a 1D box of length 5 cm, the equation of energy levels of a particle in a 1D box is,

$$E_n = \frac{h^2n^2}{8mL^2} $$

a. If the length of the 1D box is increased to 10 cm, what is the change in the energy level of this particle in the box?

b. If the length of the 1D box is decreased to 2 cm, what is the change in the energy level of this particle in the box?

c. Explain the effect of length changes on the energy levels of particles in a 1D box.

Solutions
The energy levels of a particle in 1D box:

$$E_n = \frac{h^2n^2}{8mL^2} $$

where h is Planck's constant equal to $$6.626 \times 10^{-34}\text{ J s}$$

Because this particle is an electron with 1 quantum number:

m is the mass of this particle equal to $$9.109 \times 10^{-31}\text{ kg}$$

n is the quantum number of the particle equal to 1

L is the length of this 1D box equal to $$5 \text{ cm}$$

for this question,

$$\Delta{E} = E_{nf} - E_{ni}$$

$$= \left[\frac{h^2n^2} {8m} \frac{1}{L_f^2}\right] - \left[\frac{h^2n^2} {8m} \frac{1}{L_i^2}\right]$$

$$= \left[\frac{h^2n^2} {8m}\right] \times \left[\frac{1} {L_f^2} - \frac{1}{L_i^2}\right]$$

a. for part a, the initial length of this 1D box: $$L_i = 5 \text{ cm} = 5 \times 10^{-2} \text{ m}$$, the final length of this 1D box: $$L_f = 10 \text{ cm} = 10 \times 10^{-2} \text{ m}$$

$$\Delta{E} = \left[\frac{h^2n^2} {8m}\right] \times \left[\frac{1} {L_f^2} - \frac{1}{L_i^2}\right]$$

$$= \left[\frac{(6.626 \times 10^{-34}\text{ J s})^2 (1)^2 } {(8)(9.109 \times 10^{-31}\text{ kg})}\right] \left[\frac{1} {(10 \times 10^{-2} \text{m})^2} - \frac{1} {(5 \times 10^{-2} \text{m})^2} \right]$$

$$= -1.807 \times 10^{-35} \text{ J}$$

b. for part b, the initial length of this 1D box: $$L_i = 5 \text{ cm} = 5 \times 10^{-2} \text{ m}$$, the final length of this 1D box: $$L_f = 2 \text{ cm} = 2 \times 10^{-2} \text{ m}$$

$$\Delta{E} = \left[\frac{h^2n^2} {8m}\right] \times \left[\frac{1} {L_f^2} - \frac{1}{L_i^2}\right]$$

$$= \left[\frac{(6.626 \times 10^{-34}\text{ J s})^2 (1)^2 } {(8)(9.109 \times 10^{-31}\text{ kg})}\right] \left[\frac{1} {(2 \times 10^{-2} \text{m})^2} - \frac{1} {(5 \times 10^{-2} \text{m})^2} \right]$$

$$= 1.265 \times 10^{-34} \text{ J}$$

c. Based on the calculations in part a and b, the energy level of the particles in the 1D box is decreased by the increase of this 1D box length, and the energy level of the particles in the 1D box is increased by the decrease of this 1D box length. Therefore, the energy level of the particles in the 1D box is negatively related to the length of this 1D box.

title: Quantum Chemistry/Example 5


text:
Quantum Chemistry/Example 6

Using the following equation calculate the ground state energies of H, He+ and Li2+, $$E=- \left( \frac{Z^{2}m_ee^4}{32\pi^2\epsilon_0^2\hbar^2} \right) \frac{1}{n^2}$$, then draw a conclusion on the effect of increasing atomic number on the energy of the ground state.

Solution:

Using the Schrödinger equation (ĤΨ=EΨ) the wave functions of the hydrogen atom have been solved. From the Schrödinger equation, the energy of the quantized states of the hydrogen atom can also be determine, as energy is the eigenvalue of the Schrödinger equation. The equation for energy for each energy level in the hydrogen atom is as follows,

$$

In the equation stated above Z is the nuclear charge, me (9.10938356 x 10-31 kg) is the mass of the electron, e (1.60217662 x 10-19 C) is the elementary charge, ε0 (8.8541878128 x 10-12 Fm-1 ) is the vacuum permittivity, ħ is the reduced Planck's constant and n is the principal quantum number.

The wave functions of the hydrogen atom can be extended to any atom with only one electron. Any atom with only one electron will have the electron in the same energy level as the hydrogen atom as electrons must be placed in the lower energy levels before higher ones. The only thing that will differ between hydrogen and other atoms with one electron is the attraction between the nucleus and the electron. As other atoms such as He+ and Li2+ have a greater number of protons in their nucleus resulting in a greater Coulombic attraction of the electron to the nucleus affecting the orbital size. As a result the energy of the ground state will differ between one electron systems with different nuclear charges.

This can be demonstrated by calculating the ground state energies for H, He+ and Li2+ and seeing how the energies differ. Starting with calculating the ground state of hydrogen where Z=1 and n=1,

$$E=- \left( \frac{Z^{2}m_ee^4}{32\pi^2\epsilon_0^2\hbar^2} \right) \frac{1}{n^2}$$

$$ E=- \left ( \frac {(1)^{2} (9.10938356 \times10^ {-31} \text{kg}) (1.60217662 \times10^{-19} \text{C})^{4}} {32\pi^2 (8.8541878128 \times10^{-12} \text{Fm})^{2}(1.054571817 \times10^{-34} \text{Js})^{2}} \right) \frac{1}{1^2}$$

$$ E=-2.17987225 \times10^{-18} \text {J}$$

$$ 1 \text {eV} = 1.602176634 \times 10^{-19} \text {J}$$

$$ E=-13.60569243 \text {eV}$$

$$1 \text {Hartree}=4.359744722207185 \times 10^{-18} \text {J}$$

$$E=-0.5 \text {Hartree}$$

Then for He+ where Z=2 and n=1,

$$E=- \left( \frac{Z^{2}m_ee^4}{32\pi^2\epsilon_0^2\hbar^2} \right) \frac{1}{n^2}$$

$$ E=- \left ( \frac { (2)^{2} (9.10938356 \times10^ {-31} \text{kg}) (1.60217662 \times10^{-19} \text{C})^{4}} {32\pi^2 (8.8541878128 \times10^{-12} \text{Fm})^{2}(1.054571817 \times10^{-34} \text{Js})^{2}} \right) \frac{1}{1^2}$$

$$ E=-8.71948901 \times10^{-18} \text {J}$$

$$ E=-54.42276978 \text {eV}$$

$$E=-2 \text {Hartree}$$ Finally for Li2+ where Z=3 and n=1,

$$E=- \left( \frac{Z^{2}m_ee^4}{32\pi^2\epsilon_0^2\hbar^2} \right) \frac{1}{n^2}$$

$$ E=- \left ( \frac {(3)^{2} (9.10938356 \times10^ {-31} \text{kg}) (1.60217662 \times10^{-19} \text{C})^{4}} {32\pi^2 (8.8541878128 \times10^{-12} \text{Fm})^{2}(1.054571817 \times10^{-34} \text{Js})^{2}} \right) \frac{1}{1^2}$$

$$ E=-1.96188503 \times10^{-17} \text {J}$$

$$ E=-122.4512322 \text {eV}$$

$$E=-4.5 \text{Hartree}$$

The ground state energy for the hydrogen atom is equal to -0.5 Hartree, for He+ it is -2 Hartree and for Li2+ it is -4.5 Hartree. The energies are increasing in a non-linear fashion which can be attributed to the Z2 term in the numerator. This means as Z increases more energy will be required to ionize a one electron atom.

title: Quantum Chemistry/Example 6


text:
Quantum Chemistry/Example 7

Example Problem 7
Using calculus, derive the most probable radius of an electron 1s orbital.

Solution
To determine the equation for the most probable radius of an electron in an 1s orbital we must calculate the value of r at the maximum point on the probability distribution. This is done by setting the derivative of radial probability to zero and solving for radius.

The general formula for the radial probability is.

$$\text{Radial Probabilty} = P(r)= R_{nl}^*(r)R_{nl}(r)\cdot r^2 = (R_{nl}(r))^2 \cdot r^2$$

After inserting the wavefunction of a 1s orbital we can simplify the radial probability equation as follows. $$

$$       P_{1s}(r)= \left(2\left(\frac{Z}{a_o}\right)^{\frac{3}{2}}\cdot e^{\frac{-Z}{a_o}r}\right)^2 \cdot r^2$$

$$       P_{1s}(r)= 4\left(\frac{Z}{a_o}\right)^3\cdot e^{\frac{-2Z}{a_o}r} \cdot r^2$$

$$       P_{1s}(r)= \frac{4Z^3}{a_{o}^3}\cdot e^{\frac{-2Z}{a_o}r} \cdot r^2$$

The derivative of the simplified probability function must then be calculated.

$$       \frac{\delta}{\delta r}\left(P_{1s}(r)\right) = \frac{\delta}{\delta r}\left(\frac{4Z^3}{a_{o}^3}\cdot e^{\frac{-2Z}{a_o}r} \cdot r^2 \right)$$

The constants can be excluded from the rest of the equation by moving them to the front of the derivative.

$$        \frac{\delta}{\delta r}\left(P_{1s}(r)\right)=\frac{4Z^3}{a_{o}^3}\frac{\delta}{\delta r}\left( e^{\frac{-2Z}{a_o}r} \cdot r^2 \right)$$

We see that the function is composed of the product of two smaller functions of the radius, multiplied by constants. Therefore we can apply the derivative product rule to solve for the equation of the derivative of the probability function.

$$

$$        \frac{\delta}{\delta r}\left(P_{1s}(r)\right)=\frac{4Z^3}{a_{o}^3}\left(\frac{\delta}{\delta r}\left( e^{\frac{-2Z}{a_o}r}\right) \cdot r^2 + \left(\left( e^{\frac{-2Z}{a_o}r}\right) \cdot \frac{\delta}{\delta r} r^2\right) \right)$$

$$        \frac{\delta}{\delta r}\left(P_{1s}(r)\right)=\frac{4Z^3}{a_{o}^3}\left(\left(\frac{-2Z}{a_o}\right)\left( e^{\frac{-2Z}{a_o}r}\right) \cdot r^2 + \left(\left( e^{\frac{-2Z}{a_o}r}\right) \cdot  2r\right) \right)$$

Simplifying and moving the constants now will be beneficial to simplifying the final equation

$$        \frac{\delta}{\delta r}\left(P_{1s}(r)\right)=\frac{8Z^3}{a_{o}^3}\left(e^{\frac{-2Z}{a_o}r}\right)\left(\left(\frac{-Z}{a_o}\right) \cdot r^2 + r \right)$$

$$        \frac{\delta}{\delta r}\left(P_{1s}(r)\right)=0=\frac{8Z^3}{a_{o}^3}\left(e^{\frac{-2Z}{a_o}r}\right)\left(\left(\frac{-Z}{a_o}\right) \cdot r^2 + r \right)$$

After setting the derivative of radial probability to zero as seen above it can be seen that the only way that the equation can be equal to zero is if the polynomial part of the formula is equal to zero. This is the case because $$        e^x\neq0$$ and the first part of the equation is constant which means it cannot be zero. Therefore... $$        0 =\left(\frac{-Z}{a_o}\right) \cdot r^2 + r $$

The equation for the exact value of the most probable radius in a 1s orbital is then calculated using the quadradic equation.

$$

Therefore let $$        x=r  $$,    $$         a =\frac{-Z}{a_o}  $$,     $$         b=1  $$   and  $$         c=0  $$ $$        r = \frac{-b\pm\left(b^2 - 4ac\right)^{\frac{1}{2}}}{2a} $$

$$        r= \frac{-(1)\pm\left((1)^2 - 4\left(\frac{-Z}{a_o}\right)(0)\right)^{\frac{1}{2}}}{2\left(\frac{-Z}{a_o}\right)} $$

$$        r= \frac{-1\pm\left(1-0\right)^{\frac{1}{2}}}{\left(\frac{-2Z}{a_o}\right)} $$

$$        r= \frac{-1\pm1}{\frac{-2Z}{a_o}} $$

The most probable radius cannot be a negative distance or zero meaning the numerator must be negative so that it is cancelled out by the negative value of the denominator. This means that the plus or minus operator in the numerator must be minus. $$        r= \frac{-1-1}{\frac{-2Z}{a_o}} $$

$$        r= \frac{1}{\frac{Z}{a_o}} $$

$$        r=r_{\text{most probable}}= \frac{a_o}{Z} $$

In conclusion the most probable radius for an electron in an 1s orbital is $$        r_{\text{most probable}}= \frac{a_o}{Z} $$ where that radius is proportional to the bohr radius $$         \left(a_o=\frac{4\pi\epsilon_o\hbar^2}{m_{e}e^2}=5.2917721090\cdot10^{-11}m\right) $$ and inversely proportional to the nuclear charge of the nucleus$$         \left(Z=\text{number of protons in nucleus of atom}\right) $$. This means that for an atom with one proton like hydrogen [H] the most probable radius is $$        1\cdot a_o=5.2917721090\cdot10^{-11}m $$  while for the helium ion [He+] which also only has one electron in the 1s orbital the most probable radius is $$         0.5\cdot a_o= 2.645886055\cdot10^{-11}m $$.

title: Quantum Chemistry/Example 7


text:
Pokémon Crystal/New Game

The game will ask if you are a boy or a girl, which determines the gender of the player character. This is mainly cosmetic. The player is then asked to set the time of the in game clock. You should probably set this to match your real world time.

Afterwards, Professor Oak introduces himself to the player and briefly explains the setting of the game. The player is then able to introduce themselves in turn by giving a name, either by entering a custom name or by choosing one of 4 presets. Custom names can be up to seven characters (Letters) long. Names can include both upper and lower case letters, as well as a few symbols, though not Arabic numerals.

The player is now able to begin the game in earnest, starting in the room of the player character. At this point the player character can now be moved, and the in game menu can now be accessed. This is a good time to save the game progress.

The room offers a good place to experiment with the game controls. Here are some things one can do to acclimate themselves to the game control scheme here:
 * Inspect the radio to listen to a broadcast.
 * Inspect the PC to manage item storage and room decorations.
 * Inspect the television or bookcase for flavor text.
 * Inspect the town map to see the regional map. The name is slightly misleading: This map is a map of the towns in the region, not a detailed map of any specific towns.

In the upper right hand corner of the room is are some stairs. Walk down the stairs to descend to the first floor where the mother of the player character is waiting at the table. Speak to her to learn about Professor Elm, as well as to receive the Pokégear. The Pokégear is essentially a PDA with phone functions, and could be compared to a smartphone from the 2010's or 2020's in terms of in game purpose. Initially the Pokégear functions solely as a mobile phone, though later on additional functions can be earned.

Talk to the other person in the room for some flavor text, and then head out through the red tiles in the southeast corner of the room, leading to New Bark Town.

Questions

 * In the late 1990's and early 2000's, consumer mobile phones were a new and emerging technology. Does the Pokégear represent a cultural zeitgeist about mobile phones of the time? How does it compare with later developments in mobile phones and smart devices?
 * The room of the player features a radio, a television, a desktop computer, and other amenities. This is a fairly accurate depiction of a middle to upper class bedroom of the era. How does it compare to a modern bedroom today? What appliances are still present, and what have been merged into other devices or been eliminated?

title: Pokémon Crystal/New Game


text:
Quantum Chemistry/Example 8

Write a question and its solution that quantitatively demonstrates the Heisenberg uncertainty principle holds for the J=0 state of a quantum rigid rotator

Rigid rotor model
In quantum chemistry, the rigid rotor model is used to describe the rotations in molecules, such as HCl. The assumptions used in the rigid rotor model is that the rotating molecules is rigid, and the changes in bond length that naturally occur in a molecule (such as vibrational) is insignificant compared to the bond length re, and thus, negligible.

The total energy of the system is the summation of the potential energy and the kinetic energy. The potential energy of a rigid rotor is 0 given the assumption that the rigid rotor bond length is constant. As such, the total energy of the system is equal to KE, which is equal to the angular momentum. The energy level (EJ) of a linear rigid rotor model (such as HCl) is given by the equation:

$$E_J = \frac{\hbar}{2I} J(J+1)$$

In which I is the inertia, based on the reduced mass of the diatomic and bond length, and J is the quantum energy level. The rigid rotor model is 3 dimensional, and for ease of calculations, rather than 2 sets of masses that are used (m1 and m2), 1 reduced mass is used instead (μ).

And in a spherical model of the rigid rotor, there are 2 variables that are used to determine position, the angels θ and ϕ, given that r is the constant bond length re.

Thus, the position of the reduced mass is given by the wave function:

$$Y(\theta,\varphi)$$

Heisenberg Uncertainty Principle
The Heisenberg uncertainty principle states that the exact position and momentum of a particle cannot be determines at one given point, and the more precisely either is determined, the less certain the other would be.

However, even thou the exact position and momentum cannot be calculated at a given moment, they can be related, and that relation is:

$$\bigtriangleup \phi \bigtriangleup L_z \geq \frac{\hbar}{2}$$

In which $$\hbar$$ is the reduced Planck constant, $$\bigtriangleup \phi$$  is the uncertainty of position, and $$\bigtriangleup L_z$$ is the uncertainty of momentum.

However, in the rigid rotor model, this equation does not work. The Heisenberg inequality is recalculated:

$$\bigtriangleup \phi \bigtriangleup L_z \geq \frac{1}{2} |<[L_z,\phi]>| = \frac{\hbar}{2} [1-2\pi |\psi( 0 )^2 ]$$

$$\bigtriangleup \phi \bigtriangleup L_z \geq \frac{\hbar}{2} [1-2\pi |\psi( 0 )^2 ]$$

Question: using the rigid rotor model at J=0, what is the moment of angular momentum and position? Does the rigid rotor follow Heisenberg’s principle at J=0?
In the ground state zero-point, J = 0, and thus, the energy level of the rigid rotor is 0.

Given that the energy is 0, the angular momentum is known (Lz = 0).

The probability density of the position can be obtained using the wavefunction:

Average of position = 0, as such, the variance is calculated:

$$\delta \phi = \sqrt{<\varphi^2>-<\varphi>^2}$$

$$<\varphi> = \int_{0}^{2\pi} (Y^*)(\varphi)(Y) d\varphi$$

Y = $$i^{m+|m|} \sqrt{\frac{2J+1}{2}\frac{(J-|m|)!}{(J+|m|)!}}.P_0^0 cos(\theta)$$

at m=0 J=0

Y = $$\frac{1}{\sqrt{4\pi}}$$

$$<\varphi> = \int_{0}^{2\pi} (\frac{1}{\sqrt{4\pi}})(\varphi)(\frac{1}{\sqrt{4\pi}}) d\varphi$$

$$<\varphi> = (\frac{1}{\sqrt{4\pi}})^2 \int_{0}^{2\pi} (\varphi) d\varphi$$

$$<\varphi> = (\frac{1}) 2\pi^2 =\frac{\pi}{2}$$

$$<\varphi>^2 =\frac{\pi^2}{4}$$

$$<\varphi^2> = \int_{0}^{2\pi} (Y^*)(\varphi)^2(Y) d\varphi$$

$$<\varphi^2> = \int_{0}^{2\pi} (\frac{1}{\sqrt{4\pi}})(\varphi^2)(\frac{1}{\sqrt{4\pi}}) d\varphi$$

$$<\varphi^2> = (\frac{1}{\sqrt{4\pi}})^2 \int_{0}^{2\pi} (\varphi^2) d\varphi$$

$$<\varphi^2> = (\frac{1}) \frac{8\pi^3}{3} =\frac{2\pi^2}{3}$$

$$\delta \phi = \sqrt{<\varphi^2>-<\varphi>^2} = \sqrt{\frac{2\pi^2}{3}-\frac{\pi^2}{4}} =\pm\frac{\sqrt{15}}{6}\pi $$

if $$\delta \phi $$ is plotted on a sphere

Where m = integers between J and – J.

 * m| = number of longitudinal node

J = number of latitudinal nodes

At ground state, J = 0, |m| = 0. Thus, probability of position is equally spread across the sphere.

at J=0,

$$ \frac{\hbar}{2} [1-2\pi |\psi( 0 )^2 ] = 0$$

thus

$$\bigtriangleup \phi \bigtriangleup L_z \geq \frac{\hbar}{2} [1-2\pi |\psi( 0 )^2 ]$$

$$(\pm\frac{\sqrt{15}}{6}\pi) (0) \geq 0$$

is satisfied

Thus, this shows that at ground state, J=0, the rigid rotor model follows the Heisenberg uncertainty.

title: Quantum Chemistry/Example 8


text:
Quantum Chemistry/Example 9

Write a question and its solution that calculates the locations of the nodes of an electron in a 2s orbital.

The Question
The question is to find the location of the radial node in a 2s electron for a hydrogen atom. To find the node one can start by analyzing, generally, how many nodes one should expect to see in a 2s electron system. There are two equations that give the number of nodes present in an orbital, the angular node equation and radial node equations: 1. Radial Nodes: $$ n-\ell-1=n_r $$ 2. Angular Nodes:

$$ \ell=n_a $$

Therefore ℓ must be determined, and based on the table 1 data one can determine ℓ is equal to 0. And the n is equal to the $$ n $$ which is 2, this comes from the number before the orbital type which tells you the principle quantum number. So how many nodes are there?

First analyze the number of angular nodes:

$$ \ell=n_a $$

$$ 0=n_a $$

Therefore, the number of angular nodes is 0.

Radial nodes:

$$ n_r=n-\ell-1 $$

$$ n_r=2-0-1 $$

$$ n_r=1 $$

Therefore there is one radial node present in a 2s orbital, resulting in the question becoming where is the location of that radial node?

The Wavefunction

Now the next main step is to determine what wavefunction describes the wavefunction this scenario of the 2s electron. The wavefunction can be found online which is$$:^1$$

$$ \psi_{2s}={\frac{1}{4(2\pi)^\frac{1}{2}}}\left( \frac{1}{a_0} \right)^{3/2}\left( 2-\frac{r}{a_0} \right)\text{e}^{-r/(2a_0)} $$

In the equation the $$\psi_{2s}$$ is the wavefunction for the 2s electron, 𝒓 is the radius, and $$a_0$$ is the Bohr radius.

Based on the equation we can then solve for the position of the electron. The best way to do this is to find where is the equation going to be equal to zero and what term that contains the position causes this. The first part $${\frac{1}{4(2\pi)^\frac{1}{2}}}$$ is a constant thus won't change with the radius, the position, of the electron. So the only place that will change with the position of the electron are the 𝒓 terms. With $$\text{e}^{-r/(2a_0)}$$ the 𝒓 term can be any number and the term won't be zero unless the 𝒓 is approaching infinity, while the $$\left( 2-\frac{r}{a_0} \right)$$ can potentially be equal to zero since it has 2 subtracted by the position term. Therefore one can set this term equal to zero and solve for 𝒓.

$$

0= 2-\frac{r}{a_0} $$

$$ 2=\frac{r}{a_0} $$

$$ 2a_0=r

$$

Therefore, we get the solution to the position of the radial node which is 𝒓 $$=2a_0$$, so when 𝒓 $$=2a_0$$ the probability of the electron being there is 0 all around the nucleus creating a node. The $$a_0$$ has a length of 52.9 pm which means the node is 105.8 pm in radius away from the nucleus.

In Conclusion

In conclusion the approach to the problem of finding nodes for an electron in an orbital boils down to first finding the number of theoretical nodes, then determining the wavefunction, analyzing the wavefunction's variables and solving for 0. After this is all done you will have the solution to the radial node locations. Further problems can be solved as well, because they are follow up questions that are made easier to solve after finding the nodes, like the position of the electron in it's most probable state. This problems follows the solution of arranging the P=$$\psi_{2s}$$$$\left(\psi_{2s}^*\right)$$$$\left(r^2\right)$$ then finding the derivative of the wavefunction and then simplifying it. The final step is to find the zero points, where the 𝒓 is equal to zero which will give the most probable locations of the electron. The practical applications of finding the nodal locations can help with understanding how orbitals work which can help with making molecular orbital diagrams and SALCs that can be used to determine the way atoms and molecules bond. Other applications include understanding the energy levels of the bonds and orbitals to predict possible interactions between molecules and atoms, for research purposes and chemical engineering.

Reference

1) Branson, J. The Radial Wavefunction Solutions. https://quantummechanics.ucsd.edu/ph130a/130_notes/node233.html (accessed November 16, 2021).

By Dmitry Ivanov

title: Quantum Chemistry/Example 9


text:
Quantum Chemistry/Example 10

Question
Use Planck's radiation law to find the surface temperature of the Sun when its maximum intensity of EM radiation is emitted at 504 nm. Ensure the temperature units are in Kelvins.

Getting the correct function
A blackbody material is defined by its ability to absorb ALL radiation that falls onto it. When the blackbody is at constant temperature, the distribution of its emission frequency can be determined by assuming its only direct relation is to temperature. Furthermore, the frequency of the electromagnetic (EM) radiation can also be measured in units of wavenumber $$\text{cm}^{-1}$$.

$$\begin{align} \rho (\nu, T) &= \frac{2h\nu^{2}}{c^{2}}\left \langle E(\nu, T) \right \rangle &&\rho (\lambda, T)= \frac{2hc^{2}}{\lambda^{5}}\left \langle E(\nu, T) \right \rangle \end{align} $$

Planck's approach to defining $\langle E(v,T) \rangle$ was deriving for a closed form Harmonic Oscillator was based on Boltzmann's distribution. The resulting form of Planck's law can then be applied to the question now that all parameters are known except for the parameter in question. The model below shows the radiation intensity distribution (i.e., area of the curve) of EM with respect to frequency (in Hertz; Hz) and temperature (in Kelvins; K). $$\begin{align} d \rho(v,T)&= \rho (\nu, T) d\nu \text{, where } \rho (\nu, T)=\left ( \frac{2h\nu^{3}}{c^{2}} \right) \left[ \exp \left( \frac{h \nu}{k_{B}T}\right)\right]^{-1} \\ & \text{Note that } \left |\frac{d\nu}{d\lambda} \right | =\frac{c}{\lambda ^{2}} \\ d\rho(\lambda, T)&= \rho (\lambda , T) d\lambda \text{, where }\rho (\lambda, T) = \left ( \frac{2hc^{2}}{\lambda^{5}} \right) \left[\exp \left( \frac{hc}{k_{B}T\lambda}\right)\right]^{-1} \end{align} $$Planck's law for a blackbody material predicts the behaviour of its quantitative properties (such as frequency, wavelength, and temperature) when the environment are in extreme conditions. $$

Solving for the temperature
The maximum value of the independent variable of any function can be found by equating the function's derivative to zero. Only $$T$$ is unknown and is with respect to the independent variable, $$\lambda$$. Thus, Planck's law will be derived with respect to $$\lambda$$.

First, for simplicity, let $ x=\frac{hc}{k_{B}\lambda T} $.

$${d\rho \over d\lambda} = (2hc^{2})\left[ \frac{x}{\lambda^{6}}\frac{e^{x}}{\left( e^{x} -1\right)^{2}} - \frac{5}{\lambda^{6}\left(e^{x}-1\right)} \right] =0$$

The first term consists only of non-zero constants so that leaves:

$$\frac{x}{\lambda^{6}}\frac{e^{x}}{\left( e^{x} -1\right)^{2}} - \frac{5}{\lambda^{6}\left(e^{x}-1\right)}=0$$

$$\begin{align} \frac{x}{\lambda^{6}}\frac{e^{x}}{\left( e^{x} -1\right)^{2}} &= \frac{5}{\lambda^{6}\left(e^{x}-1\right)} \\ \frac{xe^{x}}{e^{x}-1}&=5 \\ \frac{xe^{x}}{e^{x}-1}-5&=0 \\ (x-5)e^{x}+5&=0 \end{align}$$

Subbing $ x=\frac{hc}{k_{B}\lambda T} $ back in gives:

$$\begin{align} \left(\frac{hc}{k_{B}\lambda T}-5\right)e^{x}&=0 && \rightarrow &&& \frac{hc}{k_{B}\lambda T}-5 =0 \end{align}$$

At max intensity, $\lambda _{max} = 504\text{nm} $ . Subbing in all known values and constants (in SI units) then rearranging to solve for T will determine the surface temperature of the Sun under these specific conditions. $$\begin{align} T= \frac{hc}{5k_{B}\lambda_{max}} =\frac{(6.62607\times10^{-34})(2.99792\times10^{8})}{5(1.38065\times10^{-23})(504\times10^{-9})} =5709.432 \text{ }K \end{align} $$

Therefore, the surface temperature of the Sun at a maximum wavelength of 504 nm is just above 5709 Kelvins.

title: Quantum Chemistry/Example 10


text:
Quantum Chemistry/Example 11

Example 11
Use the 1D particle in a box model to estimate the wavelength of light required to excite an electron from a pi to pi* MO in ethene.

Solution:

The energy levels of a particle in a 1D box with a specific quantum number $$ n $$, are as follows.

$$ E_n = \frac{h^2n^2}{8mL^2} $$

In this equation $$ h $$ represents Planck's constant, $$ m $$ is the mass of the particle, and $$ L $$ is the length of the box.

The pi electron in the double bond between the carbon atoms in ethene can be approximated to the particle in a 1D box model. This means that the mass of the particle in this question will be the mass of an electron, and the length of the box corresponds to the bond length between the carbon atoms in the molecule ethene.

Additionally, the energy equation above needs to be transformed into a equation for $$ \Delta E $$ since the electron is moving from one energy level to another.

$$ \Delta E = E_f - E_i = \frac{h^2n_i^2}{8mL^2} - \frac{h^2n_f^2}{8mL^2} = \frac{h^2}{8mL^2}(n_f^2-n_i^2) $$

The change in energy between the pi and pi* MO in ethene can now be calculated knowing that the bond length between doubly bonded carbon atoms is 133pm and the mass of an electron is 9.1093856x10-31 kg. Moving from the ground state n=1 to an excited state of n=2 :

$$ \Delta E = \frac{h^2}{8mL^2}(n_f^2-n_i^2) $$

$$ \Delta E = \frac{(6.626\text{x}10^{-34}\frac{\text{m}^2\text{kg}}{\text{s}})^2}{8(9.109\times 10^{-31}\text{kg})(133\times 10^{-12}\text{m})^2}(2^2-1^2) $$

$$ \Delta E = 1.0217 \times 10^{-17}\text{J} $$

Now that the energy required to excite the electron to the pi* orbital is known, the wavelength of light can be calculated through the following equation, where c is the speed of light in a vacuum and $$ \lambda $$ is the wavelength of light.

$$ \Delta E = \frac{hc}{\lambda} $$

The equation can then be re-arranged to solve for the wavelength of light.

$$ \lambda = \frac{hc}{\Delta E} $$

By plugging in the known constants and the value for $$ \Delta E $$ that has been calculated above, the wavelength can be found.

$$ \lambda = \frac{hc}{\Delta E} $$

$$ \lambda = \frac{(6.626 \times 10^{-34}\frac{\text{m}^2\text{kg}}{\text{s}})(2.998 \times 10^8\frac{\text{m}}{\text{s}})}{1.0217 \times 10^{-17}\text{J}} $$

$$ \lambda = 1.9 \times 10^{-7} \text{m} $$

$$ \lambda = 19 \text{nm} $$

Therefore the wavelength of light required to excite an electron from a pi to pi* molecular orbital in ethene is 19nm.

title: Quantum Chemistry/Example 11


text:
Quantum Chemistry/Example 12

Write a question and its solution that shows the specific selection rule for a quantum rigid rotor.

Consider an N2 molecule with a bond length of 1.09 Å.

(a) Calculate the energy at the quantum number 3 using the specific selection rule for a rigid rotor.

Solution: This bond length is given (1.09 Å), and the reduced mass (μ) and the inertia (I) must be calculated to determine the energy at the angular momentum quantum number $$J=3$$. The bond length and reduced mass must also be changed to SI units.

$$\mu= \frac {m_1 m_2}{m_1 + m_2}$$

$$\mu= \frac {14.0067u \cdot 14.0067u}{14.0067u + 14.0067u} \cdot(\frac {1.6605\times10^{-27} kg}{1u})$$

$$\mu= 1.1629\times10^{-26}kg$$

Now, calculating the moment of inertia ($$I$$):

$$I = \mu \cdot (r_e)^2$$

$$= (1.1629\times10^{-26}kg) \cdot (1.09\AA\cdot1.00\times10^{-10}m)^2$$

$$= 1.3816\times10^{-46}kg\cdot m^2$$

Since the molecule is linear (N2), the energy level ($$E_J$$) can be calculated with the linear rigid rotor equation:

$$E_J = J(J+1) \cdot (\frac{\hbar^2}{2I} )$$

$$E_J= 3(3+1) \cdot (\frac{(1.05457\times 10^{-34}J\cdot s)^2}{2(1.3816\times10^{-46}kg\cdot m^2)} )$$

$$E_J= 12 \cdot ( 4.0247 \times10^{-23}J)$$

$$E_J= 4.8296\times10^{-22}J$$

(b) Calculate the quantum number if the transition energy is 2.4176 × 10-22J. In reference to part (a), does this value adhere to the specific selection rule? Why or why not?

The linear rigid rotor equation must be rearranged into linear form to solve for $$J$$.

$$E_J = J(J+1) \cdot (\frac{\hbar^2}{2I} )$$

$$J^2+J = \frac{ 2 \cdot E\cdot I }{\hbar^2}$$

$$J^2+J - \frac{ 2 \cdot E\cdot I }{\hbar^2}= 0$$

$$J^2+J - \frac{ (2) \cdot (2.4176\times10^{-22}J)\cdot (1.3816\times10^{-46}kg\cdot m^2) }{(1.05457\times10^{-34}J\cdot s)^2}= 0$$

$$J^2+J - 6.0069 = 0$$

To solve this relationship, the quadratic formula must be utilized:

$$J=  \frac{ -b \pm \sqrt{b^2-4\cdot a\cdot c} }{2\cdot a}$$

$$J=  \frac{ -1 \pm \sqrt{1^2-4\cdot1\cdot-6.0069} }{2\cdot1}$$

$$J=  \frac{ -1 \pm \sqrt{25.0275} }{2}$$

$$J=  \frac{ -1 \pm {4.9973} }{2}$$

$$J=2.0014,-3.0014$$

Rigid rotor quantum numbers cannot be negative, ∴ $$J= 2$$.

This transition adheres to the specific selection rule for a quantum rigid rotor because the change in rotational quantum number is $$\Delta J = \pm1$$.

(c) Calculate the quantum number if the transition energy is 1.2073 × 10-21J. In reference to part (a), does this value adhere to the specific selection rule? Why or why not?

The linear rigid rotor equation must be rearranged into linear form to solve for $$J$$.

$$J^2+J - \frac{ 2 \cdot E\cdot I }{\hbar^2}= 0$$

$$J^2+J - \frac{ (2) \cdot (1.2073\times10^{-21}J)\cdot (1.3816\times10^{-46}kg\cdot m^2) }{(1.05457\times10^{-34}J\cdot s)^2}= 0$$

$$J^2+J - 29.9970 = 0$$

To solve this relationship, the quadratic formula must be utilized:

$$J=  \frac{ -b \pm \sqrt{b^2-4\cdot a\cdot c} }{2\cdot a}$$

$$J=  \frac{ -1 \pm \sqrt{1^2-4\cdot1\cdot-29.9970} }{2\cdot1}$$

$$J=  \frac{ -1 \pm \sqrt{120.98813} }{2}$$

$$J=  \frac{ -1 \pm {10.9994} }{2}$$

$$J=4.9997,-5.9997$$

Rigid rotor quantum numbers cannot be negative, ∴ $$J= 5$$.This transition does not adhere to the specific selection rule for a quantum rigid rotor, because the change in rotational quantum number is not within the $$\Delta J = \pm1$$. Compared to part (a), it has a change of $$\Delta J = \pm2$$.

title: Quantum Chemistry/Example 12


text:
Quantum Chemistry/Example 13

Write a question and its solution that shows the specific selection rule for a quantum harmonic oscillator

Calculate the energy for the vibrational transitions $$v\rightarrow v+1$$ and $$v+6\rightarrow v+7$$. If they have the same energy gaps comment on why?

$$\bigtriangleup E v\rightarrow v+1 $$$$=hv_0 \Bigl( (v+1)+\tfrac{1}{2}\Bigr) -hv_0 \Bigl(v+ \tfrac{1}{2}\Bigr) $$

$$=hv_0 \biggl((v+1)+ \tfrac{1}{2} - \Bigl(v+ \tfrac{1}{2}\Bigr) \biggr)$$

$$=hv_0 \biggl(v+1+\tfrac{1}{2}-v-\tfrac{1}{2} \biggr)$$

$$=hv_0$$

$$\bigtriangleup E v+1\rightarrow v + 2 $$$$=hv_0 \Bigl( (v+7)+\tfrac{1}{2}\Bigr) -hv_0 \Bigl((v+6)+ \tfrac{1}{2}\Bigr) $$

$$=hv_0 \biggl((v+7)+ \tfrac{1}{2} - \Bigl((v+6 )+ \tfrac{1}{2}\Bigr) \biggr)$$

$$=hv_0 \biggl(v+7+\tfrac{1}{2}-v-6 -\tfrac{1}{2} \biggr)$$

$$=hv_0$$

The energy of the vibrational transitions from from $$v\rightarrow v+1$$ and $$v+6\rightarrow v+7$$ have the same energy gap $$(hv_0)$$. This is because of the specific selection rule for the quantum harmonic oscillator. The rule states that the molecule is only allowed to move up or down one vibrational energy level for the transition to occur. If the molecule diverges from the rule $$\bigtriangleup V \pm 1$$ then it is considered an overtone, and these are unlikely to occur.

title: Quantum Chemistry/Example 13


text:
Quantum Chemistry/Example 14

Show using calculus the most probable position of a quantum harmonic oscillator in the ground state (n=0)

Question:

What is the most probable position of a quantum harmonic oscillator at the ground state? Calculate this using the probability density equation to find the most probable position at n=0.

$$

Solution:

The Hermite polynomial at n=0 is:

$$H_0(x) = 1$$

The normalization factor at n=0 is:

$$N_0= \frac {1}{\sqrt{2^1 1!}}\left( \frac {\alpha}{\pi}\right) ^{1/4} = \left( \frac {\alpha}{\pi}\right) ^{1/4} $$

α is a constant and is equal to:

$$\alpha =\sqrt{\frac {k\mu}\hbar} $$

The probability distribution at n=0:

$$P_{n=0}(x) = \left( \frac {\alpha}{\pi}\right) ^{1/2} e^{-\alpha x^2 } $$

The most probable position is when the maximum probability distribution is:

$$\frac{\partial P}{\partial x}=0 $$

Applying this partial derivative to the probability distribution gives:

$$\frac{\partial }{\partial x} \left( \frac {\alpha}{\pi}\right) ^{1/2} e^{-\alpha x^2 } =0  $$

The constants can be taken out of the derivative:

$$\left( \frac {\alpha}{\pi}\right) ^{1/2} \frac{\partial }{\partial x} e^{-\alpha x^2 } =0   $$

The derivative gives:

$$\left( \frac {\alpha}{\pi}\right) ^{1/2} [-2\alpha x e^{-\alpha x^2 } ]=0  $$

Since it is equal to zero the constants can be divided out leaving:

$$[-2\alpha x e^{-\alpha x^2 } ]=0  $$

Since all of the parts are multiplied they can be divided out leaving:

$$x=0  $$

The point where the probability distribution is at a maximum for the ground state of n=0 for the quantum harmonic oscillator is 0.

title: Quantum Chemistry/Example 14