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Compton Effect

  • Thread starter Khasimir
  • Start date
1
0
Hallo

I don't understand why I can't use

Δλ = λ'- λ = h/(m·c) · ( 1 - cos φ )
and ΔE = h · c / Δλ

for getting the energy ΔE which is transferred to the electron

In school we did it like this: 1/f'- 1/f = h/(m·c) · ( 1 - cos φ )
<-> E' = hf' = hf/(1 + hf(1 - cos φ )/(mc^2))

-> ΔE = E (energy of photon before scattering) - E' (energy of photon after)

I really need help

Christoph
 

Answers and Replies

jtbell
Mentor
15,395
3,179
[tex]\Delta E = E^\prime - E = \frac{hc}{\lambda^\prime} - \frac{hc}{\lambda} = hc \left( \frac{1}{\lambda^\prime} - \frac{1}{\lambda} \right)[/tex]

This does not equal [itex]hc / \Delta \lambda[/itex] because

[tex]\frac{1}{\lambda^\prime} - \frac{1}{\lambda} \ne \frac{1}{\lambda^\prime - \lambda}[/tex]
 

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