Calculating de Broglie wavelength

Question

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?

Answer 1

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.

Answer 2

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.