De Broglie wavelength of electron and alpha particle with 2 MeV kinetic energy

[Cay]

I'm reposting this thread with some editing suggested by fresh_42:

1. Homework Statement
Calculate the mass, velocity and the de Broglie wavelength of an electron and an alpha particle, given the kinetic energy K = 2MeV

Homework Equations

The Attempt at a Solution

please find attached![/B]

I went through my calculations again and I think I have one problem in determining the formula of relativistic kinetic energy. What I used is: E = m*c^2, where m - the effective mass of the electron. I'm afraid I should've used
E = m*c^2 - m0*c^2, where m = m0/sqrt(1-β^2). Is that correct?

 

[RPinPA]

You're correct about the error. I prefer to use $m$ to refer to the invariant (rest) mass, but to avoid confusion I'll stick with your notation and terminology. So $K = mc^2 - m_0c^2$ or $(m - m_0)c^2$.

Or you could just say the total energy is 2 MeV plus the rest energy of the electron, 0.511 MeV. So then $mc^2 = 2.51$ MeV. Go through all the same steps but using that value of total energy rather than the 2 MeV.

 

[Cay]

You're correct about the error. I prefer to use $m$ to refer to the invariant (rest) mass, but to avoid confusion I'll stick with your notation and terminology. So $K = mc^2 - m_0c^2$ or $(m - m_0)c^2$.

Or you could just say the total energy is 2 MeV plus the rest energy of the electron, 0.511 MeV. So then $mc^2 = 2.51$ MeV. Go through all the same steps but using that value of total energy rather than the 2 MeV.

Thank you for replying!
However, if I apply the relativistic energy formula also for the alpha particle, I get that effective mass is approximately the same as the rest mass. So I would get that β=0, which is not possible. What am I doing wrong?

 

[RPinPA]

Thank you for replying!
However, if I apply the relativistic energy formula also for the alpha particle, I get that effective mass is approximately the same as the rest mass. So I would get that β=0, which is not possible. What am I doing wrong?

An alpha particle is two protons and two neutrons, each of which has a rest energy of approximately 1 GeV. So 2 MeV is a very small fraction of the rest energy and this particle is not relativistic. Use the low-velocity formula $K = (1/2)mv^2$.

To be precise, the rest energy of an alpha is 3.727 GeV, so 2 MeV kinetic energy increases that to 3.729 GeV total energy. If you want to use the relativistic formula, the increase is a factor of 3.729/3.727 = 1.00054 = $1/\sqrt{1-\beta^2}$. $\beta$ is indeed close to zero, meaning as I said this is not a relativistic particle and it is moving at a small fraction of $c$.

In contrast, the electron has a total energy of 2.5 MeV which is nearly 5 times its rest energy, so that is definitely a relativistic particle.

It's often a good idea to first do a sanity check to see "is this a relativistic particle or not?"