Question

x-rays of wavelength 0.4500nm are scattered from free electrons in a target.what is the wavelength of photons scattered at 60 degree relative to the incident rays?

Answer 1

by the formula of wavelength shift we can say

$\lambda' - \lambda = \frac{h}{m_e c}(1 - cos\theta)$

initial wavelength is given as

$\lambda = 0.450 nm$

$\theta = 60 degree$

now by above formula

$\lambda' - 0.450nm = \frac{6.6*10^^{-34}}{9.1*10^{-31}*3*10^8}(1 - cos60)$

$\lambda' = 0.450nm + 0.0012nm = 0.4512nm$

Answer 2

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 scatered 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]

hope its help you

thanx and be brainly...............................................