Question

Maximum velocity of the photoelectrons emitted by a metal surface is $1.2\times {10}^{6}m{s}^{-1}$ . Assuming the specific charge of the electron to be $1.8\times {10}^{11}Ck{g}^{-1},$ the value of the stopping potential in volt will be

• A. 2
• B. 3
• C. 4
• D. 6

Answer 1

Answer: (C)

Specific charge of electron, $\frac{e}{m}=1.8\times {10}^{11}Ck{g}^{-1}$ Maximum kinetic energy of photoelectron $\frac{1}{2}m{v}_{\max }^2=e{V}_s$ where $V_s$ is the stopping potential. $\Rightarrow$ $V_s=\frac{m{v}_{\max }^2}{2e}=\frac{v_{\max }^2}{2(e/m)}$ $=\frac{{(1.2\times {10}^6)}^2}{2\times 1.8\times {10}^{11}}$ $=0.4\times 10=4V$