Question 378: Dual Nature of Matter and Radiation

Question

The maximum velocity of the photoelectron emitted by the metal surface is ' $v$ '. Charge and mass of the photoelectron is denoted by ' $e$ ' and ' $m$ ' respectively. The stopping potential in volt is

Accepted Answer

Given, maximum velocity of a photoelectron $=v$
charge of photoelectron $=e$
and mass of photoelectron $=m$.
Let, the stopping potential of the photoelectron $=V$,
Then, the maximum kinetic energy
$$\begin{aligned}
& 	ext { Then, the maximum kinetic energy } \
& \qquad K E_{\max }=e V \Rightarrow \frac{1}{2} m v^2=e V\left(\because K E=\frac{1}{2} m v^2ight) \
& \Rightarrow \quad V=\frac{v^2 m}{2 e} 	ext { or } V=\frac{v^2}{2\left(\frac{e}{m}ight)}
\end{aligned}$$

Answer

$\frac{v^2}{2\left(\frac{m}{e}\right)}$

Answer

$\frac{v^2}{2\left(\frac{e}{m}\right)}$

Answer

$\frac{v^2}{\left(\frac{e}{m}\right)}$

Answer

$\frac{v^2}{\left(\frac{m}{e}\right)}$

Dual Nature of Matter and Radiation

The maximum velocity of the photoelectron emitted by the metal surface is ' v '. Charge and mass of the photoelectron is denoted by ' e ' and ' m ' respectively. The stopping potential in volt is

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