A certain photon after being scattered from a free electron at rest moves at an angle of 120° with respect to the incident direction. If the wavelength of the incident photon is 0.591 nm, what is the energy of the scattered photon? A) 1.05 × 10^-16 J B) 2.10 × 10^-16 J C) 5.71 × 10^-16 J D) 3.34 × 10^-16 J E) 4.21 × 10^-16 J

Question

Quizplus · Accepted Answer

We can use the Compton scattering formula: $\Delta\lambda=\dfrac{h}{mc}(1-\cos heta)$ where $\Delta\lambda$ is the change in wavelength, $h$ is Planck's constant, $m$ is the mass of the electron, $c$ is the speed of light, and $ heta$ is the scattering angle. We can re-arrange the formula to solve for the change in wavelength: $\Delta\lambda=\lambda'-\lambda=\dfrac{h}{mc}(1-\cos heta)$ where $\lambda'$ is the wavelength of the scattered photon and $\lambda$ is the wavelength of the incident photon. Plugging in the given values, we get: $\Delta\lambda=\lambda'-\lambda=\dfrac{h}{mc}(1-\cos heta)=\dfrac{6.626 imes10^{-34} ext{ J}\cdot ext{s}}{9.109 imes10^{-31} ext{ kg}\cdot 3 imes10^8 ext{ m/s}}(1-\cos 120^\circ)=2.42 imes10^{-12} ext{ m}$ We can then use the formula for energy of a photon: $E=hf=\dfrac{hc}{\lambda}$ where $f$ is the frequency of the photon. We can solve for the energy of the scattered photon by using the change in wavelength: $E'=\dfrac{hc}{\lambda'}=\dfrac{hc}{\lambda-\Delta\lambda}=\dfrac{hc}{0.591 imes10^{-9} ext{ m}-2.42 imes10^{-12} ext{ m}}=3.34 imes10^{-16} ext{ J}$ Therefore, the answer is D) 3.34 × 10^-16 J.

A Certain Photon After Being Scattered from a Free Electron