A) \[0.48\text{ }\mu A\]
B) resistance value not given
C) zero
D) \[0.96\text{ }\mu A\]
Correct Answer: D
Solution :
Let n be the number of photons per sec absorbed by cell, then \[n\frac{hc}{\lambda }=3\times {{10}^{-3}}\] \[n=6.04\times {{10}^{15}}\] The electron which have been ejected per sec are 0.1% of\[n=6.04\times {{10}^{13}}\]. So, rate of flow of charge\[I=\frac{dq}{dt}=0.1%\]of\[ne=0.96\text{ }\mu A\]
You need to login to perform this action.
You will be redirected in
3 sec
