Question

(a) A $\gamma$-ray photon has a momentum of 8.00(10^-21)kg.m/s .What is its wavelength?

(b) Calculate its energy in MeV.

Short answer

a. The wavelength of a photon is $8.27\times 10^{-15}\mathrm{m}$.

b. The energy in MeV of photons is $150\mathrm{M}\mathrm{e}\mathrm{V}$.

Step-by-step solution

Determine the formula for the wavelength

To evaluate the wavelength, of the value $\lambda$, of a photon with a given momentum. Consider the equation of the wavelength.

$\lambda \mathbf{=}\frac{\mathbf{h}}{\rho }$

Find the wavelength of a photon

(a)

Considering the given information,

$\begin{array}{l}p=8\times {10}^{-21}\mathrm{k}\mathrm{g}\cdot {\mathrm{m}}^{-1}\\ h=6.63\times {10}^{-34}\mathrm{J}\cdot {\mathrm{s}}^{-1}\end{array}$

Apply the formula and determine the wavelength of a photon.

$\begin{array}{c}\lambda \approx \frac{6.63\times {10}^{-34}\mathrm{J}}{8\times {10}^{-21}\mathrm{k}{\mathrm{g}}_{\mathrm{g}\times {\mathrm{m}}_{\mathrm{s}}}{\mathrm{s}}^{-1}}\\ \lambda \approx 8.27\times {10}^{-15}\mathrm{m}\end{array}$

Therefore, the required wavelength of a photon is $8.27\times 10^{-15}\mathrm{m}$.

Find the energy in MeV of photons

(b)

Considering the given information,

$\begin{array}{l}c=3\times {10}^8\mathrm{m}\cdot {\mathrm{s}}^{-1}\\ p=8\times {10}^{-21}\mathrm{k}\mathrm{g}\cdot \mathrm{m}\cdot {\mathrm{s}}^{-1}\end{array}$

Apply the formula,

The relationship between energy and wavelength in photons is:

$E\approx pc$

The value of the relation is as follows:

$1\mathrm{e}\mathrm{V}=1.6\times {10}^{-19}\mathrm{J}$

Currently, the photon's energy is,

$\begin{array}{l}E\approx 8\times {10}^{-21}\times 3\times {10}^8\mathrm{J}\\ E\approx 24\times {10}^{-21}\times \frac{1}{1.6\times {10}^{-19}}\mathrm{e}\mathrm{V}\\ \mathrm{E}\approx 15\mathrm{M}\mathrm{e}\mathrm{V}\end{array}$

Therefore, the required energy in MeV of photons is$15\mathrm{M}\mathrm{e}\mathrm{V}$.