Dual Nature Of Light Question 393
Question: Find the number of photon emitted per second by a 25 watt source of monochromatic light of wavelength $ \text{6600 }\overset{o}{\mathop{A}},\text{.} $ What is the photoelectric current assuming 3% efficiency for photoelectric effect?
Options:
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
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ P_{in}=25W,\lambda =6000\overset{o}{\mathop{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 $
BETA
