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Energy of photon $E=hf$ and its momentum is $P=\dfrac{h}{\lambda }$, where $\lambda $ is the wavelength of photon with its assumption speed of light wave is____.
(A). $\dfrac{P}{E}$
(B). $EP$
(C). $\dfrac{E}{P}$
(D). ${{\left( \dfrac{E}{P} \right)}^{2}}$

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Last updated date: 23rd Jun 2024
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Answer
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Hint: Photon energy and its momentum are related as $P=Ec$ where $P$ is momentum of photon, $E$ is the energy associated with photon and $c$ is the estimated speed of light.

Formulae used:
$E=hf$
$P=\dfrac{h}{\lambda }$
$P=\dfrac{E}{c}$
$E$= energy of photon, $h$= planck's constant, $f$= frequency of oscillation of light particles, \[P\]= momentum of photon, $\lambda $= wavelength of photon, $c$= speed of light wave.

Complete step by step answer:
Photon is an elementary particle which is massless and always moves at the speed of light in vacuum, approximately equal to $299792458\text{ }\dfrac{m}{s}$. We can also say photons are the particles which transmit light waves.
According to the relativistic total energy of a particle, energy and momentum of a photon is related as:
${{E}^{2}}={{(Pc)}^{2}}+{{(mc)}^{2}}$
Putting the value of $m=0$ as a photon is a massless particle.
We get, $E=Pc$ or $P=\dfrac{E}{c}$
We get value of $c=\dfrac{E}{P}$
Hence, the correct option is C.

Additional information:
A photon is the smallest discrete quantum of electromagnetic radiation. It is the basic unit of all light waves. Photons are always in motion, and in vacuum, they always travel at a constant speed of $2.998\times {{10}^{8}}\dfrac{m}{s}$. This speed is referred to as speed of light and is denoted by the letter $c$. A photon usually looks like a Sine wave on a graph paper, but in reality we cannot see a photon. A photon is a packet of energy.

Note: Energy and momentum both are related to an object’s mass and velocity or speed. Momentum is a vector quantity which describes the amount of mass in motion, while Energy is a scalar quantity.