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The threshold frequency for a photosensitive metal is $3.3\times {{10}^{14}}Hz$. If light of frequency $8.2\times {{10}^{14}}Hz$ is incident on this metal. The cut-off voltage for the photoelectric emission is nearly-
(A).$2V$
(B). $3V$
(C). $5V$
(D). $1V$

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Last updated date: 29th Feb 2024
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Hint: A light is made to fall on a metal and electrons are ejected from it as photons having energy collide with electrons and transfer their energy to them due to which they eject out. We can use Einstein’s equation for photoelectric emission and substitute corresponding values to calculate the cut-off voltage. Convert the units as required.
Formula used:
$K=h\nu -W$

Complete answer:
When light falls on a metal, the photons of the light possesses energy and hence when they collide with the electrons of the metal, they transfer their energy due to which the electrons are ejected out.
The minimum energy required to eject an electron is known as the work function of the metal.
Einstein’s equation for photoelectric effect is given as-
$K=h\nu -W$
Here, $K$ is the kinetic energy
$h$ is Planck’s constant
$\nu $ is the frequency of light
$W$ is the work function
From the above equation,
$\begin{align}
  & eV=h\nu -h{{\nu }_{0}} \\
 & \Rightarrow V=\dfrac{h(\nu -{{\nu }_{0}})}{e} \\
\end{align}$
Given, $e=1.6\times {{10}^{-19}}$, ${{\nu }_{0}}=3.3\times {{10}^{14}}Hz$, $\nu =8.2\times {{10}^{14}}Hz$, $h=6.6\times {{10}^{-34}}$
In the above equation, we substitute given values to get,
$\begin{align}
  & V=\dfrac{6.6\times {{10}^{-34}}(8.2\times {{10}^{14}}-3.2\times {{10}^{14}})}{1.6\times {{10}^{-19}}} \\
 & \Rightarrow V=10.56\times {{10}^{-1}}(8.2-3.2) \\
 & \Rightarrow V=5.28V \\
 & \therefore V\approx 5V \\
\end{align}$
Therefore, the cut-off voltage for photoelectron emission is nearly equal to $5V$.

Hence, the correct option is (C).

Note:
The voltage is applied opposite to the motion of electrons to bring them to rest. That is why; this voltage is known as the cut-off voltage. The energy of the photons is directly proportional to the frequency of light and inversely proportional to the wavelength. It is independent of the intensity of light.
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