# A beam of light consists of 4 wavelength $\;4000A^{0},4800A^{0},6000A^{0}\; and\;7000A^{0}\;,$ each of intensity $\;1.5\times10^{-3}Wm^{-2}\;.$The beam falls normally on an area $\;10^{-4} m^2\;$ of a clean metallic surface of work function 1.9eV . Assuming no loss of light energy the no.of photoelectrons liberated per second is

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

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 .
$=\large\frac{I_{1}A_{1}}{\varepsilon_{1}}+\large\frac{I_{2}A_{2}}{\varepsilon_{2}}+\large\frac{I_{3}A_{3}}{\varepsilon_{3}}$
$=IA\;(\large\frac{1}{\varepsilon_{1}}+\large\frac{1}{\varepsilon_{2}}+\large\frac{1}{\varepsilon_{3}})$
$=\large\frac{(1.5\times10^{-3})\;(10^{-4})}{1.6\times10^{-19}}\;(\large\frac{1}{3.1}+\large\frac{1}{2.58}+\large\frac{1}{2.06})$
$=1.12\times10^{12}\;.$