question_answer

Find the frequency of light which ejects electrons from a metal surface fully stopped by a retarding potential of 3V. The photoelectric effect begin in this metal at frequency of \[6\times {{10}^{14}}\] per second.

A) \[1.32\times {{10}^{15}}Hz\]

B)        \[3.28\times {{10}^{14}}Hz\]

C) \[6.22\times {{10}^{15}}Hz\]

D)        \[2.22\times {{10}^{11}}Hz\]

Correct Answer: A

Solution :

[a] The work function of the metal \[{{W}_{0}}=h{{f}_{0}}=(6.63\times {{10}^{-34}})\times (6\times {{10}^{14}})\] \[=3.98\times {{10}^{19}}J\] If f is the frequency of the light, then \[hc={{W}_{0}}+e{{V}_{0}}\] \[=\frac{3.98\times {{10}^{-19}}+1.6\times {{10}^{-19}}\times 3}{6.63\times {{10}^{-34}}}=1.32\times {{10}^{15}}Hz\]