$(a)\;1.12\times10^{14}\qquad(b)\;1.12\times10^{136}\qquad(c)\;1.12\times10^{13}\qquad(d)\;1.12\times10^{12}$

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Answer : (d) $\;1.12\times10^{12}$

Explanation :

$\varepsilon_{1}=\large\frac{12400}{4000}=3.1eV$

$\varepsilon_{2}=\large\frac{12400}{4800}=2.58eV$

$\varepsilon_{3}=\large\frac{12400}{6000}=2.06eV$

$\varepsilon_{4}=\large\frac{12400}{7000}=1.77eV$

Therefore , light of wavelength $\;4000A^{0}\;,4800A^{0}\;and\;6000A^{0}\;$ can only emit photoelectrons .

Therefore , Number of photoelectrons emitted per second = No . of photons incident per second .

$=\large\frac{I_{1}A_{1}}{\varepsilon_{1}}+\large\frac{I_{2}A_{2}}{\varepsilon_{2}}+\large\frac{I_{3}A_{3}}{\varepsilon_{3}}$

$=IA\;(\large\frac{1}{\varepsilon_{1}}+\large\frac{1}{\varepsilon_{2}}+\large\frac{1}{\varepsilon_{3}})$

$=\large\frac{(1.5\times10^{-3})\;(10^{-4})}{1.6\times10^{-19}}\;(\large\frac{1}{3.1}+\large\frac{1}{2.58}+\large\frac{1}{2.06})$

$=1.12\times10^{12}\;.$

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