A) decrease by a factor of 4
B) increase by a factor of 4
C) decrease by a factor of 2
D) increase by a factor of 2
Correct Answer: B
Solution :
The ratio of intensities, \[\frac{{{l}_{2}}}{{{l}_{1}}}=\frac{{{({{r}_{1}})}^{2}}}{{{({{r}_{2}})}^{2}}}\] where,distance from source. \[\frac{{{l}_{2}}}{{{l}_{1}}}=\frac{{{(1)}^{2}}}{{{\left( \frac{1}{2} \right)}^{2}}}\] \[{{l}_{2}}=4\,{{l}_{1}}\] Now, since number of electrons emitted per second is directly proportional to intensity, so number of electrons emitted by photocathode would increase by a factor of 4.You need to login to perform this action.
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