The maximum velocity of the photoelectrons emitted by a metal surface is 9 × 1 0 5 m / s . The value of ratio of charge (e) to mass (m) of the photoelectron is 1.8 × 1 0 11 C / kg . The value of stopp

Question

The maximum velocity of the photoelectrons emitted by a metal surface is 9 × 1 0 5 m / s . The value of ratio of charge (e) to mass (m) of the photoelectron is 1.8 × 1 0 11 C / kg . The value of stopping potential in volt is

TestHub · Accepted Answer

The relation between maximum kinetic energy and stopping potential is given by:

$e V_{0} = \frac{1}{2} m v_{m a x}^{2}$

Rearranging for stopping potential $V_{0}$ , we get:

$V_{0} = \frac{v _{m a x}^{2}}{2 ( \frac{e}{m} )}$

Substituting the given values $v_{m a x} = 9 \times 1 0^{5} m/s$ and $\frac{e}{m} = 1.8 \times 1 0^{11} C/kg$ :

$V_{0} = \frac{( 9 \times 1 0 ^{5} ) ^{2}}{2 \times 1.8 \times 1 0 ^{11}}$

$V_{0} = \frac{81 \times 1 0 ^{10}}{3.6 \times 1 0 ^{11}}$

$V_{0} = \frac{8.1 \times 1 0 ^{11}}{3.6 \times 1 0 ^{11}}$

$V_{0} = \frac{8.1}{3.6} = 2.25 V$

Answer: $2.25$

Physics - Modern Physics Question with Solution | TestHub

Options:

Doubts & Discussion