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**Hint:**Photon is the smallest quanta of energy. According to the particle nature of light, light behaves as particles and this is confirmed by photoelectric effect, but according to wave nature of light, light behaves as waves and this is confirmed by phenomenon like reflection, refraction etc. We can make use of the Planck quantum formula to solve this problem.

**Complete Step by step solution:**

Given the energy of photon is $75eV$

We know energy is given by the formula\[E=h\nu \], where h is the Planck’s constant whose value is \[6.6\times {{10}^{-34}}Js\] and v is the frequency.

When the energy is given in electron-Volts, we have to first convert it into Joules.

$E=75eV \\

\Rightarrow E=75\times 1.6\times {{10}^{-19}}J \\

\therefore E=12\times {{10}^{-18}}J \\$

Now we make use of the above-mentioned formula,

$E=12\times {{10}^{-18}}J \\

\Rightarrow hv=12\times {{10}^{-18}} \\

\Rightarrow 6.6\times {{10}^{-34}}\times v=12\times {{10}^{-18}} \\

\Rightarrow v=\dfrac{12\times {{10}^{-18}}}{6.6\times {{10}^{-34}}} \\

\therefore v=1.82\times {{10}^{16}}Hz \\$

**So, the frequency comes out to be \[1.82\times {{10}^{16}}Hz\]**

**Additional Information:**From photoelectric effect, Einstein equation is given by \[h\nu =h{{\nu }_{0}}+KE\]

Here \[h\nu \]in the energy of the incident radiation and \[h{{\nu }_{0}}\]is the work function of the metal given.

**Note:**Here the energy was given in eV, so we first converted it into Joules and then used the formula. But suppose if we had to find the wavelength and the frequency both then we could have use the formula; \[E=\frac{hc}{\lambda }\], where we would have taken E in eV and the value of hc we would have taken 1240 eV-nm. After finding the wavelength in nm we would have converted it into metres, and then we would have used the formula for finding out the frequency.

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