# Calculating de Broglie wavelength

Hey, trying to finish an assignment but having some trouble with it. I will show all my work. The topic is on wave/particle dualty, uncertainty principle (second year modern physics course).

So the question is:

Calculate the de Broglie wavelength of a 5.7 MeV α particle emitted from an atomic nucleus whose diameter is approximately $1.6 \times 10^{-14} m$.

So I know the de broglie wavelength is $\lambda = h / p$. But we dont know p in this case, however we do know the energy. So to relate energy and momentum I use the following formula:

$$E^2 = p^2c^2+m^2c^4$$

using $m = 6.644 \times 10^{-27}$ and using $E = 9.13\times 10^{-13}$ gives me a complex solution. (Check WolframAlpha)

Now I've tried leaving the energy in terms of eV and MeV and they give me real solutions however the answer is not correct.

The answer is 6.02 fentometers.

Anyone have any advice?

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I am having trouble understanding why the diameter of the the nucleus is given. Maybe it is to be compared later with $\lambda$. – ja72 Dec 3 '10 at 0:13

## 2 Answers

When the kinetic energy is given (5.7 MeV) and the mass is known, calculating momentum is very straightforward. I would like to point to the part of the assignment that you missed which is the nucleus diameter. It tells you something about the position of the particle and thus prevents you from knowing the momentum (and hence the de Broglie wavelength) all too well.

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The assignment is asking to calculate the debroglie. every formula I've tried dosnt give me the right results. What am I missing? – masfenix Dec 4 '10 at 1:19
You are missing a formula to convert kinetic energy to momentum which is very simple to figure out if you remember that E=mv^2/2 and p=mv. Taking h/p gives me 6.015 fm. – gigacyan Dec 7 '10 at 13:46

Normally I'd knock on you for simply asking a homework question, but in this case you're dealing with a poorly-worded problem. The rest mass of an alpha particle is 3.7 GeV; you can't have an alpha particle with less energy than its rest mass.

The 5.7 MeV is supposed to be interpreted as the kinetic energy of the alpha particle, not its total energy. I guess you can take it from there.

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The rest mass is technically 3.7 GeV/c^2, but yeah... – Noldorin Dec 3 '10 at 2:03
@Noldorin Depends on what units you use. Energy is a completely legitimate unit for mass. – Lagerbaer Dec 3 '10 at 3:04

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@Lagerbaer: Incorrect. The units are already specified. The quantity is dimensionally wrong unless you specify `/c^2` too. – Noldorin Dec 3 '10 at 3:14

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@masfenix G = Giga = 10^9; M = mega = 10^6. The rest mass of the particle is about a thousand times the kinetic energy in this case. You'll have to think about the implications there for what equations you should use. – Mark Eichenlaub Dec 3 '10 at 7:08

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I think you can define a consistent unit system in which energy and mass are measured in the same unit (and trust me, I'm the biggest stickler for units I know so I don't say this lightly) - these would be the natural units with $c=1$. Of course, in a certain sense it's not as useful a system as SI or CGS, but it's so darn convenient that everyone caves eventually ;-) – David Zaslavsky♦ Dec 3 '10 at 8:44
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