A) \[12\times {{10}^{18}}\]
B) \[10\times {{10}^{18}}\]
C) \[12\times {{10}^{17}}\]
D) \[12\times {{10}^{16}}\]
Correct Answer: C
Solution :
Energy of each photon is \[E=\frac{hc}{\lambda }\] \[=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{6000\times {{10}^{-10}}}\] \[=3.3\times {{10}^{-19}}J\] Now, power of source \[P=IA=39.6\times 1=39.6\,\text{watt}\] Number of photons emitted per second \[=\frac{p}{E}=\frac{39.6}{3.3\times {{10}^{-9}}}\] \[=12\times {{10}^{19}}\] Only 1% of photon eject elements, so no. of photoelectrons \[=12\times {{10}^{19}}\times \frac{1}{100}\]\[=12\times {{10}^{17}}\]You need to login to perform this action.
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