A) 4 mA and IV
B) 12 mA. and IV
C) 3 mA and IV
D) 3mA and 0.5V
Correct Answer: D
Solution :
Inverse Square Law, \[I\propto \frac{1}{{{d}^{2}}}\] \[12\,\propto \frac{1}{{{(0.2)}^{2}}}\,\,\,or\,\,I\,\propto \,\frac{1}{{{(0.2)}^{2}}}\] \[\therefore \,\,\frac{I}{12}\,=\frac{{{(0.2)}^{2}}}{{{(0.4)}^{2}}}\,=\frac{1}{4}\,\Rightarrow I=\frac{12}{4}\,=3\,mA\] Stopping potential will be same because frequency of source and work function of material is same.You need to login to perform this action.
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