A) 4
B) 8
C) 16
D) 2
Correct Answer: A
Solution :
| Intensity of light is inversely proportional to square of distance. |
| i.e., \[I\propto \,\,\frac{1}{{{r}^{2}}}\] |
| or \[\frac{{{I}_{2}}}{{{I}_{1}}}=\frac{{{({{r}_{1}})}^{2}}}{{{({{r}_{2}})}^{2}}}\] |
| Given, \[Given,{{r}_{1}}=0.5\,m,\,{{r}_{2}}=1.0\,m\] |
| Therefore, \[\frac{{{I}_{2}}}{{{I}_{1}}}=\frac{{{(0.5)}^{2}}}{{{(1)}^{2}}}=\frac{1}{4}\] |
| Now, since number of photoelectrons emitted per second is directly proportional to intensity, so number of electrons emitted would decrease by factor of 4. |
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