De Broglie Wavelength at Relativistic velocity

[Avatrin]

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]

Kinetic energy is:

KE=12m0v2​

KE = \frac{1}{2}m_0v^2

is only correct for NON-relativistic speeds ...

 

[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 $\lambda$.

 

[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]

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

 

[Avatrin]

I got a flawed impression by just looking at the equations, i.e. without reading the accompanying text, on this website:
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/releng.html
However, I think I get it now.

 

[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)