A) \[1.84\times {{10}^{-6}}A\]
B) \[1.84\times {{10}^{-7}}A\]
C) \[1.84\times {{10}^{-5}}A\]
D) \[1.84\times {{10}^{-4}}A\]
Correct Answer: A
Solution :
From Plancks law, the energy of incident photon is \[E=nhv=\frac{nhc}{\lambda }\] where n is number of photons. \[\Rightarrow \] \[n=\frac{E}{hc/\lambda }\] Given, \[E=1\text{ }mW={{10}^{-3}}W,\] \[h=6.62\times {{10}^{-34}}J-s,\] \[c=3\times {{10}^{8}}J-s,\] \[\lambda =4560{\AA}=4560\times {{10}^{-10}}m\] \[n=\frac{{{10}^{-3}}\times 4.56\times {{10}^{-7}}}{6.62\times {{10}^{-34}}\times 3\times {{10}^{8}}}\] \[=2.29\times {{10}^{15}}\] Given, quantum efficiency = 0.5%. Hence, number of electrons liberated from surface is \[n=n\frac{0.5}{100}=1.15\times {{10}^{13}}\] \[\therefore \] \[l=ne=1.15\times {{10}^{13}}\times 1.6\times {{10}^{-19}}\] \[i=1.84\times {{10}^{-6}}A\]
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