De Broglie Wavelength at Relativistic velocity

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  • #1
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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:
[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 m0). 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?
 

Answers and Replies

  • #2
lightgrav
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Kinetic energy is:
KE=12m0v2​
KE = \frac{1}{2}m_0v^2
is only correct for NON-relativistic speeds ...
 
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  • #3
BvU
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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##.
 
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  • #4
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I found a way that gave me the correct solution. However, I do not understand it.

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

However, since the particle is at rest at first, why shouldn't the left hand side of the expression above be eV + m0c2?
 
  • #5
lightgrav
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?? where do you think the "- mc2" on the right-hand side came from?
 
  • #7
BvU
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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)
 

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