# De Broglie Wavelength at Relativistic velocity

1. Feb 10, 2015

### Avatrin

1. The problem statement, all variables and given/known data
A particle has charge e and masse m0. It is accelerated by a charge V to a relativistic velocity. Show that its de Broglie wavelength is:
$$\lambda = \frac{h}{\sqrt{2m_0eV}} (1+\frac{eV}{2m_0c^2})^{-\frac{1}{2}}$$

2. Relevant equations
$$\lambda = \frac{h}{p}$$
Conservation of energy can be used. Our potential energy is:
$$PE = eV$$
Kinetic energy is:
$$KE = \frac{1}{2}m_0v^2$$

3. The attempt at a solution
I tried using:
$$eV = \frac{1}{2}m_0v^2$$
That gave me:
$$\sqrt{\frac{2eV}{m_0}} = v$$

This matches what is in the denominator below Planck's constant (when multiplied with m0). However, the expression in the parenthesis is what doesn't make sense to me. My answer is:

$$\lambda = \frac{h}{\sqrt{2m_0eV}} (1-\frac{2eV}{m_0c^2})^{\frac{1}{2}}$$
Why is this wrong?

2. Feb 10, 2015

### lightgrav

is only correct for NON-relativistic speeds ...

3. Feb 10, 2015

### BvU

Here you find the right expression for p to use. It works. I think the difference is a factor gamma somewhere. I can't make out how you go from your expression for v to your last answer for $\lambda$.

4. Feb 11, 2015

### Avatrin

I found a way that gave me the correct solution. However, I do not understand it.

$$eV = \sqrt{p^2c^2 + m_0^2c^4} - m_0c^2$$
This gave me the correct answer for p:
$$p = ((\frac{eV}{c})^2 + 2em_0V)^{\frac{1}{2}} = \sqrt{2m_0eV}\sqrt{1 + \frac{eV}{2m_0c^2}}$$

However, since the particle is at rest at first, why shouldn't the left hand side of the expression above be eV + m0c2?

5. Feb 11, 2015

### lightgrav

?? where do you think the "- mc2" on the right-hand side came from?

6. Feb 11, 2015

### Avatrin

7. Feb 11, 2015

### BvU

There was a Show button on the page where the link in post #2 pointed. Sorry I didn't point that out (or gave that as a link directly)