Question

# Photon of frequency $\nu$ has a momentum associated with it. If c is the velocity of light, then the momentum is:\begin{align} & A.\dfrac{\nu }{c} \\ & B.h\nu c \\ & C.\dfrac{h\nu }{{{c}^{2}}} \\ & D.\dfrac{h\nu }{c} \\ \end{align}

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Hint: The momentum of a photon is very much related to its energy. As we know, the energy of a photon is proportional to its frequency, then the momentum of a photon is in relation with its wavelength. Even though photons are not having mass, they are possessing momentum.

First of all let us discuss the momentum of a photon in detail. The momentum of a photon, p, is calculated in kilogram meters per second, which is equal to Planck's constant, h, divided by the de Broglie wavelength of the light, lambda, calculated in meters. The equation can be written as,
$P=\dfrac{h}{\lambda }$
Particles possess momentum as well as energy. Even though photons have no mass, there has been proof that EM radiation possesses momentum. It is now a well-established concept that photons do have momentum. Particles are carrying momentum as well as energy. Clearly, photons carry momentum in the direction of their motion itself which is away from the sun and some of this momentum is being changed to dust particles in collisions.
We know that,
$P=\dfrac{h}{\lambda }$
In which $\lambda$ is the wavelength.
And also we know that,
$\lambda =\dfrac{c}{\nu }$
Where $c$ is the velocity of light, $\nu$ is frequency of photon.
Substituting this in the above equation will give,
$P=\dfrac{h\nu }{c}$
Therefore the correct answer is option D.

Note: The photons of the light which is reflected from a metal or a dielectric mirror are similar to the incident ones, apart from the varied propagation direction. The loss of light in the metal includes that some fraction of the photons are lost, while the energy of each reflected photon is completely preserved.