Question

Find the momentum (in kg m/s) of a photon having an energy 1 MeV.
A) ${10^{ - 22}}{\text{kgm/s}}$
B) $0.33 \times {10^6}{\text{kgm/s}}$
C) $5 \times {10^{ - 22}}{\text{kgm/s}}$
D) $7 \times {10^{ - 24}}{\text{kgm/s}}$

Answer 1

Photon is described as the quantum of the light wave. de Broglie’s relation for the wavelength of the photon can then be used to find its momentum as its energy is known. The energy of the photon is given in MeV. This should be converted to Joules to obtain the momentum in the desired unit.

Formulas used:
de Broglie’s relation for the wavelength of the photon is given by, $\lambda = \dfrac{h}{p}$ where $h$ is the Planck's constant and $p$ is the momentum of the photon.
The energy of a photon is given by, $E = \dfrac{{hc}}{\lambda }$ where $h$ is the Planck's constant, $c$ is the velocity of light and $\lambda$ is the wavelength of the photon.

Complete step by step answer.
Step 1: List the information known from the question.
The energy of the photon is $E = 1{\text{MeV}}$ . In Joules it will be $E = {10^6} \times 1.6 \times {10^{ - 19}}{\text{J}}$

Step 2: Express the relation for the energy of the photon in terms of its momentum using de Broglie relation.
de Broglie’s relation for the wavelength of the photon is given by, $\lambda = \dfrac{h}{p}$ -------- (1)
where $h$ is the Planck's constant and $p$ is the momentum of the photon.
The energy of a photon is expressed as $E = \dfrac{{hc}}{\lambda }$ --------- (2)
where $h$ is the Planck's constant, $c$ is the velocity of light and $\lambda$ is the wavelength of the photon.
Substituting equation (1) in (2) we get, $E = \dfrac{{hpc}}{h} = pc$ ---------- (3)

Step 3: Using equation (3) find the momentum of the photon.
Equation (3) gives $E = pc$ or on rearranging we get, $p = \dfrac{E}{c}$
Substituting the values for $E = 1 \times 1.6 \times {10^{ - 19}}{\text{J}}$ and $c = 3 \times {10^8}{\text{m/s}}$ in the above relation we get, $p = \dfrac{{{{10}^6} \times 1.6 \times {{10}^{ - 19}}}}{{3 \times {{10}^8}}} \cong 5 \times {10^{ - 22}}{\text{kgm/s}}$
$\therefore$ the momentum of the photon is $p \cong 5 \times {10^{ - 22}}{\text{kgm/s}}$ .

Thus the correct option is C.

Note: Alternate method-
The photon is considered as a particle moving with the velocity of light. This allows us to express its energy as $E = m{c^2}$
Since the momentum of a particle moving with the velocity of light is given by, $p = mc$ , on expanding the relation for the photon energy we get, $E = \left( {mc} \right) \times c = pc$ .
Then the momentum of the photon can be expressed as $p = \dfrac{E}{c}$
Now on substituting the values for $E = 1 \times 1.6 \times {10^{ - 19}}{\text{J}}$ and $c = 3 \times {10^8}{\text{m/s}}$ in the above relation we get, $p = \dfrac{{{{10}^6} \times 1.6 \times {{10}^{ - 19}}}}{{3 \times {{10}^8}}} \cong 5 \times {10^{ - 22}}{\text{kgm/s}}$
$\therefore$ the momentum of the photon is $p \cong 5 \times {10^{ - 22}}{\text{kgm/s}}$ .