Question 372: 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

The stopping potential, denoted as $V_s$, is the potential difference required to stop the fastest photoelectrons emitted from a metal surface. It is directly associated with the maximum kinetic energy (KE) these photoelectrons have when emitted. The relationship between kinetic energy and stopping potential is given by the equation:
$ KE = eV_s $
Where:
$KE$ is the maximum kinetic energy of the photoelectron.
$e$ is the charge of the electron (approximately $1.6 	imes 10^{-19}$ Coulombs).
$V_s$ is the stopping potential.
The maximum kinetic energy of the photoelectron can also be written in terms of its mass $m$ and its maximum velocity $v$ as follows:
$ KE = \frac{1}{2}mv^2 $
Combining the two equations gives:
$ eV_s = \frac{1}{2}mv^2 $
To find the stopping potential $V_s$, we rearrange the equation to solve for $V_s$:
$ V_s = \frac{\frac{1}{2}mv^2}{e} = \frac{v^2}{2\left(\frac{e}{m}ight)} $
Therefore, the correct option that represents the stopping potential in terms of $v$, $e$, and $m$ is:
Option A: $\frac{v^2}{2\left(\frac{e}{m}ight)}$

Answer

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

Answer

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

Answer

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

Answer

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

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

Choose the correct answer: