Question

The momentum of a photon of frequency $\nu $ is
$\eqalign{
  & {\text{A) }}h\dfrac{\nu }{c} \cr
  & {\text{B) }}h\nu c \cr
  & {\text{C) }}\dfrac{h}{{c\nu }} \cr
  & {\text{D) }}\dfrac{\nu }{{ch}} \cr} $

Answer 1

Hint: To solve this problem use knowledge of the relationship between velocity, frequency and wavelength of light. Use the concept of wave-particle duality and its mathematical expression to solve for the required answer.

Formula used:
$c = \lambda \nu $
$\lambda = \dfrac{h}{p}$

Complete answer:
Light has wave nature, as proved by an English physicist named Thomas Young via his experiment called Young’s Interference Experiment. And thus satisfies the following equation:
$c = \lambda \nu \cdots \cdots \cdots \left( 1 \right)$
where c represents the velocity of light,
$\lambda $ represents the wavelength of light,
and $\nu $ represents the frequency.

Additionally, we also know that light has particle nature also. This phenomenon was proved by Albert Einstein via discovering the photoelectric effect and we call the quanta of light as photons. Every photon will satisfy the De Broglie relation as given by:
$\lambda = \dfrac{h}{p} \cdots \cdots \cdots \cdots \left( 2 \right)$
where
$\lambda $ represents the wavelength of photon,
$p = mv$ represents the momentum of the photon,
$h$ represents the Planck's constant.

Now, substituting value of $\lambda $ from equation (2) in equation (1), we get
$\eqalign{
  & c = \dfrac{{h\nu }}{p} \cr
  & \Rightarrow p = \dfrac{{h\nu }}{c} \cr} $

So, the correct answer is “Option A”.

Additional Information:
The rest mass of the photons is zero, and so is their rest momentum.
Everything in the universe whether big or small follows the wave-particle duality.
At the subatomic level, the wavelength of subatomic particles becomes pronounced as it is comparable to the size of the particles.
As light shows particle nature, similarly wave nature is observed for electrons, which show phenomena of diffraction and interference.

Note:
An alternative method to solve this question is to use Einstein’s mass energy equivalence i.e. given by:
$E = m{c^2}$
here take $mc = p$as the momentum of the photon, so we have:
$E = pc \cdots \cdots \cdots \left( 3 \right)$
And we also know that, $E = h\nu \cdots \cdots \cdots \left( 4 \right)$
Equating equation (3) and (4) we will get the same value of p.