X ray of wavelength 0.4500 nm are scattered from free electrons from a free target. what is the wavelength of the photons scattered at 60 degree relative to the incident rays
Answers
The formula is :
Y' = Y + hc / mc² (1 - Cos Ф)
Where Y' = wavelength of the scattered X ray.
Y = Initial wavelength = 0.4500nm
h = planks constant
c = speed of light
m = Mass of the recoiling particle and in this case it is an electron.
Useful constants :
hc = 1240eV
mc² = 511 keV
Doing the substitution we have :
Y' = 0.45nm + (1240eV / 511000eV) × (1 - Cos 60)
= 0.45nm + 2.426 × 10⁻³ × 0.5
= 0.45nm + 1.213 × 10⁻³
= 0.451213nm
From equation Δλ = h/mc [1-cos?]
= [(6.63×10-34 J.s) (1-cos 60°)]/[(9.11×10-31kg) (3.00×108 m/s)]
= 4.5 × 10-10 m
≈ 450 pm
From the above observation we conclude that, the Compton shift of the scattered rays would be 2.2 pm.
(b) The fractional energy loss frac is,
frac = [E-E'] / E
= [hf – hf '] / hf
= [c/λ – c/λ'] / [c/λ]
= [λ' - λ] / λ'
= Δλ / [λ+Δλ]
Substitution gives,
frac = [450 pm]/[45 pm+45. pm]