Question

If the energy of the photon is increased by a factor of 4, then its momentum
A. Does not change
B. Decrease by a factor of 4
C. Increase by a factor of 4
D. Decrease by a factor of 2

Answer 1

Hint:The energy of the photon is inversely proportional to the wavelength of the photon and the momentum of the photon is also inversely proportional to the wavelength.

Formula used:
\[E = \dfrac{{hc}}{\lambda }\],
Here, E is the energy of the photon of wavelength \[\lambda \], h is the Plank’s constant and c is the speed of light.
\[p = \dfrac{h}{\lambda }\]
Here p is the momentum of the photon.

Complete step by step solution:
If a photon is having wavelength \[\lambda \]then the energy associated with the photon is given using the expression,
\[E = \dfrac{{hc}}{\lambda }\]
So, the energy of the photon is inversely proportional to the wavelength because the Plank’s constant and the speed of the light is constant.

On writing the wavelength of the photon in terms of energy contained in a photon, we get
\[\lambda = \dfrac{{hc}}{E} \ldots \left( i \right)\]
The momentum of the wave is given as,
\[p = \dfrac{h}{\lambda }\]

On replacing the wavelength with the expression containing energy, we get the relation between the energy and the momentum of the photon as,
\[p = \dfrac{h}{{\left( {\dfrac{{hc}}{E}} \right)}}\]
On simplifying the expression, we get
\[p = \dfrac{E}{c}\]
\[\Rightarrow p \propto E\]
So, the momentum of the photon is directly proportional to the energy of the photon.

When we increase the energy of the photon then the momentum of the photon also increases in the same proportion as energy is increased. It is given that the energy of the is increased by a factor of 4, i.e. \[{E_2} = 4{E_1}\], then we need to find the change in the momentum of the photon.From the relation between the energy of the photon and the momentum of the photon, we get
\[\dfrac{{{p_1}}}{{{p_2}}} = \dfrac{{{E_1}}}{{{E_2}}}\]
\[\Rightarrow {p_2} = \left( {\dfrac{{{E_2}}}{{{E_1}}}} \right){p_1}\]
\[\Rightarrow {p_2} = \left( {\dfrac{{4{E_1}}}{{{E_1}}}} \right){p_1}\]
\[\therefore {p_2} = 4{p_1}\]
Hence, the momentum of the photon is also increased by a factor of 4.

Therefore, the correct option is C.

Note: The energy of the photon is proportional to the frequency of the photon and the momentum of the photon is inversely proportional to the wavelength. From the relation \[c = \nu \lambda \] we see that the frequency is inversely proportional to the wavelength. Hence, the momentum of the photon is directly proportional to the energy of the photon.