A) 2W
B) 5W
C) 7W
D) 10W
Correct Answer: C
Solution :
[c] The energy of incident photons is given by \[hv=e{{V}_{s}}+{{\phi }_{0}}=2+5=7eV\] (\[{{V}_{s}}\] is stopping potential and \[{{\phi }_{0}}\] is work function) Saturation current\[={{10}^{-5}}A\] \[=\frac{\eta P}{hv}e=\frac{{{10}^{-5}}P}{7\times e}e\] (\[\eta \] is photon emission efficiency) \[\therefore P=7W\]
You need to login to perform this action.
You will be redirected in
3 sec
