Question

In a photoelectric set up the most energetic photoelectron from the material is introduced horizontally parallel to the parallel plate capacitor of length $L$ and constant electric field $E$, it gets deflected by $d$ when it emerges from the capacitor. The stopping potential for this electron will be (neglecting the effect of gravity)
A) $\dfrac{{E{l^2}}}{{4d}}eV$
B) $\dfrac{{8d}}{{E{l^2}}}eV$
C) $\dfrac{{4Ed}}{{{r^2}}}eV$
D) None of these

Answer 1

Hint: The most energetic photoelectron from the material is introduced horizontally parallel to the parallel plate capacitor of length $L$ and constant electric field $E$, the force experienced by a charged photoelectron in an electric field is given by \[f = QE\] where $E$ is electric field intensity, which exists in the capacitor and it accelerates the electron in a certain direction and hence causes deflection.

Formula Used:
When the photoelectron travelled between two points. The distance travelled \[\left( s \right)\] by a photoelectron in time \[\left( {\mathbf{t}} \right)\], ​the body having initial velocity \[\left( u \right)\] and acceleration \[\left( a \right)\].
$S = ut + \dfrac{1}{2}a{t^2}$

Complete step by step answer:
If $Q$ is the positive charge. They force works towards the direction of $E$.
$a = \dfrac{F}{M} = \dfrac{{QE}}{M}$
$\implies$ $a = \dfrac{{QE}}{M}$
The acceleration of the charged photoelectron in time \[\left( {\mathbf{t}} \right)\] is:
$\implies$ $a = \left( {\dfrac{{QE}}{M}} \right) \times t$
The initial velocity of the photoelectron is $0$. Hence,
$\implies$ $S = \dfrac{1}{2}\left( {\dfrac{{QE}}{M}} \right){t^2}$
Now, the displacement of photoelectron in the y direction \[\left( {\mathbf{d}} \right)\]:
$\implies$ $d = \dfrac{1}{2}\left( {\dfrac{{QE}}{M}} \right){t^2}$
Time $\left( t \right)$ taken to get deflected by $\left( d \right)$ when emerges out from the capacitor:
$\implies$ $t = \dfrac{L}{v}$
$\implies$ $d = \dfrac{1}{2}\left( {\dfrac{{QE}}{M}} \right)\dfrac{{{L^2}}}{{{v^2}}}$
 $ = \dfrac{{QE{L^2}}}{{2M{v^2}d}}$
For calculating the stopping potential:
 ${V_s} = \dfrac{{{E_k}}}{Q}$
$\implies$ ${E_k} = \dfrac{1}{2}m{v^2}$
$\implies$ $\dfrac{{\dfrac{1}{2}m{v^2}}}{Q} = \dfrac{{QE{L^2}}}{{2M{v^2}d}}$
$\implies$ ${V_s} = \dfrac{{E{L^2}}}{{4d}}$
Hence, the stopping potential for this electron will be $\dfrac{{E{L^2}}}{{4d}}eV$.
$\therefore$ The correct option is A

Note: When a photon particle does fall on a metal plate, it can transfer its whole energy to one of them inside electrons of the metal plate and his own existence ceases. Charged electron gets the rest of the energy in the form of kinetic energy. But not all the electrons get charged/energized from the surface of the metal plate. Some of the electrons get charged inside the metal plate. The atoms collide when they reach the page, due to which they lose some of their energy. Therefore, the energy level of electrons is different when released from the metal plate.
If the energy taken by the electrons is less than the work function of the metal plate then the electron will not get charged/energized
 $W = h{v_0}$
Substituting the value of $W$ in the equation:-
 ${E_k} = h(v - {v_0})$
If the maximum velocity of charged electrons is ${v_{\max }}$ then:-
${E_k} = \dfrac{1}{2}m{v^2} = h(v - {v_0})$
This equation is called Einstein's photoelectric equation.