Question

An x-ray photon with a wavelength 0.1100 nm collides with a free electron initially at rest. The scattered photon has a wavelength of 0.1122 nm. (a) Calculate the angle 0 between the initial and final directions of the photon. (b) What is the kinetic energy (in ev) and speed (in m/s) of the electron after the collision?

Answer 1

`\theta = 84.63 degree`

v = 8.82 *10^6

Step-by-step explanation:

from compton scattering formula

`\lambda' -\lambda = (h)/(m_e C) {1-COS\theta}`

m_e = mass of electron

h - plank's constant

`0.1122 - 0.1100) * 10^(-9) = (6.626 * 10^[-34)/(9.1*10^(-31) *3*10^(8)) (1-cos\theta)`

`1 - cos\theta = 0.9064`

`cos\theta = 0.09356`

`\theta = cos^(-1) 0.09356`

`\theta = 84.63 degree`

b)

from conservation energy principle

`k =\frac{hc}[{\lambda} -(hc)/(\lambda')]`

`K ={6.626*10^(-34)*3*10^(8)}*[(1)/(0.1100*10^(-9)) - (1)/(0.1122*10^(-9))}]`

`k = 354.33 *10^(-19)` j

`k = 221.4571 ev`

we know that

`k = (1)/(2) m_e v^(2)`

`v^2 = ( 2*354.33 *10-19)/(9.1*10^(-31))`

v = 8.82 *10^6