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## Homework Statement

A particle has charge e and masse m

_{0}. It is accelerated by a charge V to a relativistic velocity. Show that its de Broglie wavelength is:

[tex] \lambda = \frac{h}{\sqrt{2m_0eV}} (1+\frac{eV}{2m_0c^2})^{-\frac{1}{2}}[/tex]

## Homework Equations

[tex] \lambda = \frac{h}{p} [/tex]

Conservation of energy can be used. Our potential energy is:

[tex] PE = eV [/tex]

Kinetic energy is:

[tex] KE = \frac{1}{2}m_0v^2 [/tex]

## The Attempt at a Solution

I tried using:

[tex] eV = \frac{1}{2}m_0v^2 [/tex]

That gave me:

[tex] \sqrt{\frac{2eV}{m_0}} = v [/tex]

This matches what is in the denominator below Planck's constant (when multiplied with m

_{0}). However, the expression in the parenthesis is what doesn't make sense to me. My answer is:

[tex] \lambda = \frac{h}{\sqrt{2m_0eV}} (1-\frac{2eV}{m_0c^2})^{\frac{1}{2}}[/tex]

Why is this wrong?