Question

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

Answer 1

By the equation of compton shift we can write

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

here given that

$\lambda = 0.4500 nm = 0.4500*10^{-9} m$

$h = 6.6 * 10^{-34}$

$m_e = 9.1*10^{-31} kg$

$c = 3 * 10^8$

$\theta = 60^o$

now plug in all values to above equation

$\lambda' - 0.4500*10^{-9} = \frac{6.6*10^{-34}}{9.1*10^{-31}*3*10^8}(1 - cos60)$

$\lambda' - 0.4500 * 10^{-9} = 2.42*10^{-12}(1 - 0.5)$

$\lambda' = 0.4512* 10^{-9}$

so the wavelength will be 0.4512 nm

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 abd be brainly...............................................