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Relativistic or Classical? Calculation of De Broglie Wavelength

  • Thread starter EIRE2003
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  • #1
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Relativistic or Classical?? Calculation of De Broglie Wavelength

Hi there,

Okay, the question is: Calculate the De Broglie Wavelength of a 10 MeV proton and a 1 MeV electron.

How does one know whether to use relativistic or classical means?

Relativistic => Kinetic / Rest mass >/ = 10%

For proton: 10^7 eV / 938 x 10^6 eV = 0.0106

For electron 10^6 eV / 0.511 x 10^6 eV = 1.957

Are these relativistic energies? And if so, which equation would be required?

Thanks
 

Answers and Replies

  • #2
Doc Al
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When that ratio of KE to rest energy is small, then a classical treatment is fine. That's not the case for the electron example. But if you're not sure, do it both ways and compare.

What's the relativistic relationship between energy, momentum, and rest energy for a particle?
 
  • #3
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My answer I keep on obtaining for the De Broglie wavelength seems to be ridiculous. 3.2x10^-46

I am using p = sqrt [(Kinetic energy)^2 + (2(rest mass)(kinetic energy))]

and lambda = h / p

rest mass of electron 0.511 MeV (0.511 x 10^6 eV)
 
  • #4
Doc Al
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I am using p = sqrt [(Kinetic energy)^2 + (2(rest mass)(kinetic energy))]
That doesn't look right to me.
 
  • #5
malawi_glenn
Science Advisor
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That doesn't look right to me.

It looks just fine to me altough.

[tex] p = \sqrt{E_k^2+2E_km_0} [/tex]

This is in natural units, i.e [itex] c = \hbar = 1 [/tex]

So you must make sure you have the correct units everywhere, i.e you might want to convert all masses and energies into SI units (Joule, kg etc) before you start to calculate.
 
  • #6
Doc Al
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D'oh! That's why it looked funny--I rarely use natural units.

In SI-friendly form it will look like this:

[tex] p = \frac{\sqrt{E_k^2+2E_km_0c^2}}{c} [/tex]

(I would find p in units of MeV/c and then convert to SI.)
 

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