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?
A. \[\dfrac{{{v^2}}}{{2\left( {\dfrac{m}{e}} \right)}}\]
B. \[\dfrac{{{v^2}}}{{2\left( {\dfrac{e}{m}} \right)}}\]
C. \[\dfrac{{{v^2}}}{{\left( {\dfrac{e}{m}} \right)}}\]
D. \[\dfrac{{{v^2}}}{{\left( {\dfrac{m}{e}} \right)}}\]

Answer 1

HintThe relation between maximum kinetic energy of photoelectron and stopping potential is given by the equation $\dfrac{1}{2}mv_{\max }^2 = e{V_0}$ where $m$ is the mass of photoelectron, ${v_{\max }}$ is its maximum velocity and ${V_0}$ is the stopping potential.

 Complete step-by-step solution:
Einstein and Millikan are the two physicist who described the photoelectric effect using a formula that is the relation between the maximum kinetic energy $\left( {{K_{\max }}} \right)$ of the photoelectrons and the frequency of the absorbed photons $\left( \nu \right)$ and the threshold frequency $\left( {{\nu _0}} \right)$ of the photoemissive surface.
${K_{\max }} = h\left( {\nu - {\nu _0}} \right)$ where $h$ is plank’s constant.
The maximum kinetic energy $\left( {{K_{\max }}} \right)$ of the photoelectrons (with charge e) can be determined from the stopping potential as
$\Rightarrow {V_0} = \dfrac{W}{q} = \dfrac{{{K_{\max }}}}{e}$. So,
$\Rightarrow {K_{\max }} = e{V_0}$
On further simplifying we have
$\Rightarrow \dfrac{1}{2}mv_{\max }^2 = e{V_0}$ where $m$ is the mass of photoelectron, ${v_{\max }}$ is its maximum velocity and ${V_0}$ is the stopping potential.
As given in the question, the maximum velocity of the photon ${v_{\max }} = v$. So,
$\Rightarrow \dfrac{1}{2}m{v^2} = e{V_0}$
On simplifying we have
\[\Rightarrow {V_0} = \dfrac{{{v^2}}}{{2\left( {\dfrac{e}{m}} \right)}}\]
Hence, option B is correct.

Note: Stopping potential states that the potential required to stop the emission of electrons from a metal surface when a beam of light having energy greater than the work potential of metal is incident on it.
When this negative voltage increases, it prevents all the highest-energy electrons from reaching the collector. When there is no current passing through the tube it means the negative voltage has reached the maximum value to slow down and stop the most energetic photoelectrons of kinetic energy ${K_{\max }}$ . This negative value of the retarding voltage is called the stopping potential or cut off potential ${V_0}$. Since the work done by the retarding potential in stopping the electron of charge e is $e{V_0}$ , the following formula must hold ${K_{\max }} = e{V_0}$