# De Broglie Wavelength at Relativistic velocity

## Homework Statement

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:
$$\lambda = \frac{h}{\sqrt{2m_0eV}} (1+\frac{eV}{2m_0c^2})^{-\frac{1}{2}}$$

## Homework Equations

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

## 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?

lightgrav
Homework Helper
Kinetic energy is:
KE=12m0v2​
KE = \frac{1}{2}m_0v^2
is only correct for NON-relativistic speeds ...

• Avatrin
BvU
Homework Helper
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##.

• 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?

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

BvU