Question

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?

Short answer

(a) The wavelength of the incident gamma rays is$2.43\text{\hspace{0.17em}pm}$

(b) The wavelength of the gamma rays scattered at 90 degrees to the incident beam is $4.86\text{\hspace{0.17em}pm}$.

(c) The photon energy for the ray scattered in the direction is $0.255\text{\hspace{0.17em}MeV}$.

Step-by-step solution

The wavelength of the incident gamma rays.

(a)

Use the value of $hc=1240\text{\hspace{0.17em}eV}\cdot \text{nm}$then it gives;

$\begin{array}{c}\lambda =\frac{hc}{E}\\ =\frac{1240\text{\hspace{0.17em}nm}\cdot \text{eV}}{\left(0.511\text{\hspace{0.17em}MeV}\right)}\\ =2.43\times {10}^{-3}\text{nm}\\ =2.43\text{\hspace{0.17em}pm}\end{array}$

Hence, the wavelength of the incident gamma rays is$2.43\text{\hspace{0.17em}pm}$ .

The wavelength of the gamma rays scattered at 90∘ to the incident beam.

(b)

The wavelength of the gamma rays is calculated as follows:

$\begin{array}{c}{\lambda }^{\text{'}}=\lambda +\Delta \lambda \\ =\lambda +\frac{h}{mc}(1-\cos \varphi )\\ =2.43+\left(2.34\right)(1-\cos {90}^{\circ })\\ =4.86\text{\hspace{0.17em}pm}\end{array}$

Hence, the wavelength of the gamma rays scattered at 90 degrees to the incident beam is $4.86\text{\hspace{0.17em}pm}$.

The photon energy of the ray scattered in the direction.

(c)

The scattered photons have energy equal to as follows;

$\begin{array}{c}{E}^{\text{'}}=E\left(\frac{\lambda }{{\lambda }^{\text{'}}}\right)\\ =\left(0.511\text{\hspace{0.17em}\hspace{0.17em}MeV}\right)\left(\frac{2.43}{4.86}\right)\\ =0.255\text{\hspace{0.17em}MeV}\end{array}$

Hence, the photon energy for the ray scattered in the direction is$0.255\text{\hspace{0.17em}MeV}$.