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 $\frac{xE{l}^2}{8d}eV$. Find $x$ (neglecting the effect of gravity).

Answer 1

Answer: 2

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