Chapter 38: Q36P (page 1183)

Gamma rays of photon energy 0.511 MeV are directed vonto an aluminium target and are scattered in various directions by loosely bound electrons there.
(a) What is the wavelength of the incident gamma rays?
(b) What is the wavelength of gamma rays scattered at $90^{\circ}$ to the incident beam?
(c) What is the photon energy of the rays scattered in this direction?

Expert verified

(a) The wavelength of the incident gamma rays is $2.43 pm$

(b) The wavelength of the gamma rays scattered at 90 degrees to the incident beam is $4.86 pm$ .

Step by step solution

(a)

Use the value of $h c = 1240 eV \cdot nm$ then it gives;

$\begin{matrix} λ = \frac{h c}{E} \\ = \frac{1240 nm \cdot eV}{(0.511 MeV)} \\ = 2.43 \times 10^{- 3} nm \\ = 2.43 pm \end{matrix}$

Hence, the wavelength of the incident gamma rays is $2.43 pm$ .

(b)

The wavelength of the gamma rays is calculated as follows:

$\begin{matrix} λ^{'} = λ + Δ λ \\ = λ + \frac{h}{m c} (1 - \cos ϕ) \\ = 2.43 + (2.34) (1 - \cos 90^{\circ}) \\ = 4.86 pm \end{matrix}$

Hence, the wavelength of the gamma rays scattered at 90 degrees to the incident beam is $4.86 pm$ .

(c)

The scattered photons have energy equal to as follows;

$\begin{matrix} E^{'} = E (\frac{λ}{λ^{'}}) \\ = (0.511 MeV) (\frac{2.43}{4.86}) \\ = 0.255 MeV \end{matrix}$

Hence, the photon energy for the ray scattered in the direction is $0.255 MeV$ .

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