$(a)\;5.2 eV\qquad(b)\;4.17 eV\qquad(c)\;3.27 eV\qquad(d)\;6.2 eV$

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Answer : (c) 3.27 eV

Explanation :

The light contains two different frequencies . The one with larger frequency will cause photo electrons with largest kinetic energy .

This larger frequency is

$\nu=\large\frac{w}{2 \pi}=\large\frac{8\times10^{15}}{2 \pi}\;s^{-1}$

The maximum kinetic energy of photo electrons is $\;K_{max}=h \nu-W$

$=(4.14\times10^{-15}eV -s)\times(\large\frac{8\times10^{15}}{2 \pi}s^{-1})-2eV$

$=5.27eV-2eV=3.27eV\;.$

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