# Homework Help: De Broglie Wave Energy

1. Mar 8, 2013

### Von Neumann

Problem:

a. Calculate the energy in eV of an electron with a wavelength of 1 fm.

b. Make the same calculation for a neutron.

Solution (so far):

a. λ=h/p=(hc)/(pc)=(1240 MeV fm)/(pc)=1fm

so, pc=1240 MeV

E=√[(pc)^2+E_0^2]

=√[(1240 MeV)^2+(.511MeV)^2]

∴E=1.24 GeV

This is the same answer as the back of my book, so I'm assuming this is the correct method of solution. However, I do the same thing for the neutron and my answer does not agree.

b. E=√[(1240 MeV)^2+(940 MeV)^2]

∴E=1560 Mev

My book says the correct answer is 616 MeV.

I don't see how an energy like that is even possible. Solving the following for pc,

E^2=(pc)^2+E_0^2

pc=√[E^2-E_0^2]

When you plug in the "correct" answer of E=616 MeV you get,

pc=√[(616 MeV)^2-(940 MeV)^2]

You certainly cannot take a square root of a negative number and get a meaningful answer. Any suggestions?

2. Mar 8, 2013

### rude man

Fact: when we talk of a particle with x eV of energy we are talking about its KINETIC energy, not its total energy.

So - compute the neutron's total energy E, subtract rest energy E_0, which gets you the K.E. , and you get what?

3. Mar 8, 2013

### Von Neumann

Rude man,

You will always catch my stupid mistakes, and for that I thank you.

It hadn't occurred to me that the kinetic energy is what was being asked for.

4. Mar 8, 2013

### rude man

Don't worry about it, it's just the convention!