Question

The electric field of a certain radiation is given by the equation $E = 200\left[ {\sin \left( {4\pi \times {{10}^{10}}} \right)t + \sin \left( {4\pi \times {{10}^{15}}} \right)t} \right]$ falls in a metal surface having work function $2.0eV$. The maximum kinetic energy (in eV) of the photoelectrons is [use Planck’s constant $\left( h \right) = 6.63 \times {10^{ - 34}}J - s$ and electron charge $\left( {E = 1.6 \times {{10}^{ - 19}}C} \right)$
(A) $3.3$
(B) $4.3$
(C) $5.3$
(D) $6.3$
(E) $7.3$

Answer 1

Hint
From Einstein’s photoelectric equation, we get the equation for the kinetic energy of the electrons in terms of the frequency and the work function. Thus we can find the frequency from the equation $\upsilon = \dfrac{\omega }{{2\pi }}$ and $\omega $is the fundamental angular frequency of the equation of electric field.
In this solution we will be using the following formula,
$\Rightarrow K.E{._{\max }} = h\upsilon - {\phi _o}$
$\Rightarrow K.E{._{\max }}$ is the maximum kinetic energy of the photo electrons,
${\phi _o}$ is the work-function of the metal and
and $\upsilon $is the frequency of the photons.
$\Rightarrow \upsilon = \dfrac{\omega }{{2\pi }}$
where $\omega $ is the angular frequency of the electric field.

Complete step by step answer
When the photons of certain energy fall on a surface of a metal then the photoelectrons of a certain kinetic energy is ejected from the surface. The kinetic energy of these ejected photoelectrons depends on the work function of the metal and the energy of the photons.
The kinetic energy is given by the formula,
$\Rightarrow K.E{._{\max }} = h\upsilon - {\phi _o}$
In the question we are given the equation of the electric field that falls on the surface of the metal. It is given by,
$\Rightarrow E = 200\left[ {\sin \left( {4\pi \times {{10}^{10}}} \right)t + \sin \left( {4\pi \times {{10}^{15}}} \right)t} \right]$
The fundamental frequency of this equation is the LCM of the both components of angular frequency.
Therefore by the LCM of $\left( {4\pi \times {{10}^{10}}} \right)$ and $\left( {4\pi \times {{10}^{15}}} \right)$ we get, $\left( {4\pi \times {{10}^{15}}} \right)$
Hence the fundamental angular frequency is, $\omega = 4\pi \times {10^{15}}rad/\sec $
So from this we can find the frequency of the photons as,
$\Rightarrow \upsilon = \dfrac{\omega }{{2\pi }}$
So substituting the values we get the frequency of the photons as,
$\Rightarrow \upsilon = \dfrac{{4\pi \times {{10}^{15}}}}{{2\pi }}$
This gives us the value of the frequency as,
$\Rightarrow \upsilon = 2 \times {10^{15}}Hz$
The value of the Planck’s constant is given in the question as, $\left( h \right) = 6.63 \times {10^{ - 34}}J - s$ and the work function of the metal as ${\phi _o} = 2.0eV = 2 \times 1.6 \times {10^{ - 19}}V$, since the value of $e = 1.6 \times {10^{ - 19}}C$ is given in the question.
So substituting all these values in the photoelectric equation we get
$\Rightarrow K.E{._{\max }} = \left( {6.63 \times {{10}^{ - 34}} \times 2 \times {{10}^{15}}} \right) - \left( {2 \times 1.6 \times {{10}^{ - 19}}} \right)$
Calculating the terms in the bracket we get,
$\Rightarrow K.E{._{\max }} = \left( {1.326 \times {{10}^{ - 18}}} \right) - \left( {0.32 \times {{10}^{ - 18}}} \right)$
So taking ${10^{ - 18}}$ common from both the terms we get
$\Rightarrow K.E{._{\max }} = \left( {1.326 - 0.32} \right) \times {10^{ - 18}}$
This gives us the maximum kinetic energy as,
$\Rightarrow K.E{._{\max }} = 1.006 \times {10^{ - 18}}J$
But we need to find the answer in eV. So we divide this value by $e = 1.6 \times {10^{ - 19}}C$ and hence get,
$\Rightarrow K.E{._{\max }} = \dfrac{{1.006 \times {{10}^{ - 18}}}}{{1.6 \times {{10}^{ - 19}}}}eV$
On calculating we get the maximum kinetic energy of the photoelectrons as,
$\Rightarrow K.E{._{\max }} = 6.3eV$
So the correct answer will be option (D).

Note
The emission of electrons from the surface of a metal when any electromagnetic radiation such as light falls on the surface of the metal is called photoelectric emission. On decreasing the intensity of light, the kinetic energy of the photoelectrons decreases.