A) 4 mA and 1 V
B) 12mA and IV
C) 3mA and IV
D) 12mA and 0.5V
E) 3mA and 0.5V
Correct Answer: E
Solution :
The cut-off voltage or stopping potential measure maximum kinetic energy of the electron. It depends on the frequency of incident light whereas the current depends on the number of photons incident. Hence, cut-off voltage will be 0.5 V. Now by inverse square law, \[12\propto \frac{1}{{{(0.2)}^{2}}}\]or\[I\propto \frac{1}{{{(0.4)}^{2}}}\] \[\therefore \] \[\frac{I}{12}=\frac{{{(0.2)}^{2}}}{{{(0.4)}^{2}}}=\frac{1}{4}\] Or \[I=\frac{12}{4}=3\,mA\]
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