$(a)\;28 \mu A\qquad(b)\;34 \mu A\qquad(c)\;30 \mu A\qquad(d)\;18 \mu A$

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Answer : (b) $\;34 \mu A$

Explanation :

Energy of photon

$\varepsilon_{2}=\large\frac{12400}{1650}=7.51 eV$

No. of photons incident per second

$n_{2}=\large\frac{P_{2}}{\varepsilon_{2}}=\large\frac{5\times10^{-3}}{12\times10^{-19}}=4.17\times10^{15}\;$ per second

No.of electrons emitted per second = $\;\large\frac{5.1}{100}\times4.7\times10^{15}$

$=2.13\times10^{14}\;$ per second

Saturation current in $\;2^{nd}\;$ case

$1=(2.13\times10^{14})\;(1.6\times10^{-19})\;amp$

$=3.4\times10^{-5}\;A$

$=34 \mu A\;.$

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