Question

The threshold frequency of the metal of the cathode in a photoelectric cell is $ 1\times {{10}^{15}}Hz. $ When a certain beam of light is incident on the cathode, it is found that a slopping potential 4.144 V is required to reduce the current to zero. The frequency of the incident radiation is

• A. $ 2.5\times {{10}^{15}}Hz $
• B. $ 2\times {{10}^{15}}Hz $
• C. $ 4.144\times {{10}^{15}}Hz $
• D. $ 3\times {{10}^{16}}Hz $

Answer 1

Answer: (B)

$ {{V}_{0}}=\frac{h}{e}(v-{{v}_{0}}) $ or $ (v-{{v}_{0}})=\frac{{{V}_{0}}\times e}{h} $ $ =\frac{4.144\times 1.6\times {{10}^{-19}}}{6.63\times {{10}^{-34}}} $ $ v-1\times {{10}^{5}}={{10}^{15}} $ Frequency $ ={{10}^{15}}+1\times {{10}^{15}} $ $ =2\times {{10}^{15}}Hz $