Xrays having an energy of 260 keV undergo Compton scattering

X-rays having an energy of 260 keV undergo Compton scattering from a target. The scattered rays are detected at 42.0degree relative to the incident rays. Find the Compton shift at this angle. Find the energy of the scattered x-ray. Find the energy of the recoiling electron. A 0.065 nm photon collides with a stationary electron. After the collision, the electron moves forward and the photon recoils backward. Find the momentum and energy of the electron. Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. keV/c Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. eV

Solution

a)

Compton shift is,

Dl = (h/m_ec)(1 -cosq) = (0.00243nm)(1 - cos42°)

=0.0006241 nm

=0.00062*10^-9 m

b)

hc = (6.63×10^-34Js)(3.0×10⁸m/s) =2.0×10^-25Jm

Convert it into eV

hc =(2.0×10^-25Jm)/(1.6×10^-19J/eV) =1.24×10^-6eV m

The wavelength of the incoming x-ray is
l₀ = hc/E₀ =(1.24×10^-6eV m) / (2.60×10⁵eV) =4.76×10^-12m = 0.00476nm
The wavelength of out going wave is
l = l₀ + Dl =0.00476nm + 0.000624 nm = 0.005384 nm.
Photons of this wavelength have energy:
E = hc/l = (1.24×10^-6eVm) / (5.38×10^-12m) = 0.230×10⁶eV= 230 keV.

c)

The energy after the scattering is theenergy of the outgoing x-ray, plus the kinetic energy of therecoiling electron, plus the rest energy of the electron. These must be equal,

therefore E₀ + m_ec²= E + KE + m_ec²

^or

KE = E₀ - E = 260keV - 230 keV = 30keV.