Question

Photons of wavelength 65.0 pm are Compton-scattered from a free electron which picks up a kinetic energy of 0.84 keV from the collision. What is the wavelength of the scattered photon?

Answer 1

Answer:

λ  = 65.6 pm

Explanation:

Given that

λo = 65 pm

The initial energy of the electron

$E_o=\dfrac{hC}{\lambda_0 }$

Now by putting the values

$E_o=\dfrac{hC}{\lambda_0 }$

$E_o=\dfrac{6.67\times 10^{-34}\times 3\times 10^8}{65\times 10^{-12}}$

$E_o=3.05\times 10^{-15}\ J$

$E_o=\dfrac{3.05\times 10^{-15}}{1.6\times 10^{-19}\times 10^3}\ KeV$

Eo=19.06 KeV

Given that kinetic energy KE= 0.84 KeV

Therefore the final energy

E= Eo - KE

E = 19.06 - 0.84 KeV

E= 18.22 KeV

The wavelength  λ can be find as

$E=\dfrac{hC}{\lambda}$

$\lambda=\dfrac{hC}{E}$

$\lambda=\dfrac{6.67\times 10^{-34}\times 3\times 10^8}{19.06 \times 10^3\times 1.6\times 10^{-19}}$

λ = 6.56 x 10⁻¹¹ m

λ  = 65.6 pm