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\[\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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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,

\[P=\dfrac{h\nu }{c}\]