A) \[\frac{25}{3}\times {{10}^{19}}J,0.4amp\]
B) \[\frac{25}{4}\times {{10}^{19}}J,6.2amp\]
C) \[\frac{25}{2}\times {{10}^{19}}J,6.2amp\]
D) None of these
Correct Answer: A
Solution :
[a] \[{{P}_{in}}=25W,\lambda =6000\overset{\text{o}}{\mathop{\text{A}}}\,=6000\times {{10}^{-10}}m\] \[nhv=P\] \[\Rightarrow \] Number of photons emitted/sec, \[n=\frac{P}{\frac{hc}{\lambda }}=\frac{P\lambda }{hc}=\frac{25\times 6600\times {{10}^{-10}}}{6.64\times {{10}^{-34}}\times 3\times {{10}^{8}}}\] \[=8.28\times {{10}^{19}}=\frac{25}{3}\times {{10}^{19}}\] 3% of emitted photons are producing current \[\therefore I=\frac{3}{100}\times ne=\frac{3}{100}\times \frac{25}{3}\times {{10}^{19}}\times 1.6\times {{10}^{-19}}\] \[=0.4A\]
You need to login to perform this action.
You will be redirected in
3 sec
