Question

An X-ray photon scatters from a free electron at rest at an angle of 120 ∘ relative to the incident direction. a) If the scattered photon has a wavelength of 0.330 nm, what is the wavelength of the incident photon? b) Determine the energy of the incident photon. c) Determine the energy of the scattered photon. d) Find the kinetic energy of the recoil electron.

Answer 1

Explanation:

It is given that,

An X-ray photon scatters from a free electron at rest at an angle of 120° relative to the incident direction.

(a) The wavelength of scattered photon, $\lambda'=0.33\ nm=0.33\times 10^{-9}\ m$

Let the wavelength of incident photon is $\lambda$

Using the relation,

$\lambda'-\lambda=\dfrac{h}{mc}(1-cos\theta)$

$\lambda'-\dfrac{h}{mc}(1-cos\theta)=\lambda$

$0.33\times 10^{-9}-\dfrac{6.62\times 10^{-34}}{9.1\times 10^{-31}\times 3\times 10^8}(1-cos(120))=\lambda$

$\lambda=3.26\times 10^{-10}\ m$

$\lambda=0.326\ nm$

(b) Energy of incident photon, $E=\dfrac{hc}{\lambda}$

$E=\dfrac{6.62\times 10^{-34}\times 3\times 10^8}{3.26\times 10^{-10}}$

$E=6.09\times 10^{-16}\ J$

(c) Energy of scattered photon, $E'=\dfrac{hc}{\lambda'}$

$E=\dfrac{6.62\times 10^{-34}\times 3\times 10^8}{0.33\times 10^{-9}}$

$E=6.01\times 10^{-16}\ J$

(d) The kinetic energy of the recoil electron, $E''=E-E'$

$E''=6.01\times 10^{-16}\ J-6.09\times 10^{-16}\ J$

$E=8\times 10^{-18}\ J$

Hence, this is the required solution.