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Mass of a photon of frequency $v$ is given by:
\[\begin{align}
  & A.\dfrac{hv}{c} \\
 & B.\dfrac{h}{\lambda } \\
 & C.\dfrac{hv}{{{c}^{2}}} \\
 & D.\dfrac{h{{v}^{2}}}{c} \\
\end{align}\]

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Last updated date: 17th Jun 2024
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Answer
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Hint: Consider a light ray of frequency $\nu$, then the energy of the wave is given from the photoelectric effect. Similarly, the energy due to the particles of mass $m$ which are present in the light rays is given from the mass energy equivalence, using the two we can solve the question.
Formula: $E=hv$ and $E=hv$

Complete answer:
We know that the light rays have dual nature and can be represented as waves of some frequency and as particles of some mass.
Then from the photo-electric effect the energy of the waves is given as $E=hv$, where $E$ is the energy of the photon which is produced due to light rays of frequency $v$ and $h$ is the Planck’s constant.
Similarly, the energy of the particle nature is given from mass energy relation, that the energy of any particle like photon in rest is the product of the mass $m$ of the particle with the speed of the light $c$ $E=mc^{2}$
Comparing the two equations, we get, $hv=mc^{2}$
$\therefore m=\dfrac{hv}{c^{2}}$

Hence the correct answer is option \[D.\dfrac{h{{v}^{2}}}{c}\].

Additional information:
We know from photo-electric effect, that when an incident beam of light falls on a metal, it radiates or releases photons. The energy of these photons depends on the frequency of the light, and in turn depends on the wavelength of the light source.
We also know that the intensity of the photons emitted depends on the wavelength. But however it is independent of the intensity of the light waves. Also there is almost no time lag between the light incident on the metal and the photon being emitted.

Note:
Here, we know that due to the dual nature of the light, then the energy of the photon is given in two ways. Thus, we are comparing the energy of a photon given by the photoelectric effect and the mass energy relationship for the same particle, i.e. the photon.