A Compton scattering experiment is performed in which the scattered photons have a minimum energy of 17.2 keV. What was the incident energy of the photons used in this experiment?
A) 16.6 keV
B) 18.4 keV
C) 16.1 keV
D) 17.2 keV

Question

Quizplus · Accepted Answer

The minimum energy of the scattered photons in Compton scattering is given by the equation Escat_min = Einc/(1 + (Einc/mc^2)*(1-cos(theta))), where Einc is the incident energy, mc^2 is the rest energy of the electron, and theta is the scattering angle. Since the minimum energy of the scattered photons is given as 17.2 keV, we can set Escat_min = 17.2 keV and solve for Einc. Using theta = 180 degrees (backscattering), we get:

17.2 keV = Einc/(1 + (Einc/511 keV)*(1-cos(180)))

17.2 keV = Einc/(1 + (-Einc/511 keV))

17.2 keV = Einc/(511 keV - Einc)

Multiplying both sides by (511 keV - Einc), we get:

17.2 keV*(511 keV - Einc) = Einc

Simplifying, we get:

8799.2 keV - 17.2 Einc = Einc

18.4 Einc = 8799.2 keV

Einc = 478.2 keV

Therefore, the incident energy of the photons used in this experiment was 478.2 keV. The closest answer choice is B: 18.4 keV.

A Compton Scattering Experiment Is Performed in Which the Scattered