Question

In a Compton collision, if the wavelength of the photon is 1.00 pm and its deflected angle is 60.0°, what is the speed of the scattered electron? The answer can't be greater than c (3x10*8 m/s)

Answer 1

let lamda is the initial wavelength of photon.

wavelength of deflected photon,

lamda' = lamda + h/(m*c)*(1 - cos(theta)

= 1*10^-12 + 6.626*10^-34/(9.1*10^-31*3*10^8)*(1 - cos(60))

= 2.21*10^-12 m

We assume initially the photon is moving towards +x axis

let px and Py are the components of momentum of electron.

Now apply conservation of momentum in x-direction

h/lamda = h/lamda'*cos(60) + Px

Px = h/lamda - h/lamda'*cos(60))

= 6.626*10^-34/(1*10^-12) - 6.626*10^-34/(2.21*10^-12)*cos(60)

= 5.127*10^-22 kg.m/s

apply conservaton of momentum in y-direction

0 = h/lamda' - Py

Py = h/lamda'*sin(60)

= 6.626*10^-34/(2.21*10^-12)*sin(60)

= 2.596*10^-22 kg.m/s

total momentum of elctron, P = sqrt(Px^2 + Py^2)

= sqrt(5.127^2 + 2.596^2)*10^-22

= 5.746*10^-22 kg.m/s

let v is the speed of electron.

we know, the formula for relativistic momentum, p = m*v

= mo*v/sqrt(1 - (v/c)^2))

5.746*10^-22 = 9.1*10^-31*v/sqrt(1 - (v/(3*10^8))^2)

on solving the above equation we get

v = 2.71*10^8 m/s <<<<<<<<----------------Answer