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 $E$. The photoelectron gets deflected by $d$ when it emerges out from the capacitor. The stopping potential for this electron is $\frac{x E l^{2}}{8 d} eV$. Find $x$ (neglecting the effect of gravity).

Answer 1

Let the velocity be $v$.
$\because$ Photoelectron will follow parabolic path,
$\therefore d=\frac{1}{2} a t^{2}$
Now, acceleration, $a=\frac{e E}{m_{e}}$,
where $e$ and $m_{e}$ are charge and mass of electron
$\therefore d=\frac{1}{2} \frac{e E}{m_{e}} t^{2}$
$t=\sqrt{\frac{2 d m_{e}}{e E}}$
$\therefore l=v t$
$\Rightarrow v=\frac{l}{t}=l \sqrt{\frac{e E}{2 d m_{e}}}$
$\because e V_{s}=K E_{\max }$
$\Rightarrow V_{\dot{s}}=\frac{1}{2} \frac{m_{e}}{e}\left(\frac{l^{2} e E}{2 d m_{e}}\right)=\frac{l^{2} E}{4 d}$