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}}C\text{ }k{{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}}C\,k{{g}^{-1}} $ Maximum kinetic energy of photoelectron $ \frac{1}{2}mv_{\max }^{2}=e{{V}_{s}} $ where $ {{V}_{s}} $ is the stopping potential. $ \Rightarrow $ $ {{V}_{s}}=\frac{mv_{\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=4\,V $