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If the work function for a certain metal is $3.2 \times 10^{-19}$ joule and it is illuminated with light of frequency $8 \times 10^{14} \mathrm{~Hz}$. The maximum kinetic energy of the photoelectrons would be $\left(h=6.63 \times 10^{-34} \mathrm{Js}\right)$.
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The correct answer is:
$2.1 \times 10^{-19} \mathrm{~J}$
$\begin{aligned} & \text { K. E.max }=\text { incident energy - work function } \\ & =\mathrm{h} \gamma-3.2 \times 10^{-19} \\ & =6.63 \times 10^{-34} \times 8 \times 10^{14}-3.2 \times 10^{-19} \\ & =5.3 \times 10^{-19}-3.2 \times 10^{-19} \\ & =2.1 \times 10^{-19} \mathrm{~J}\end{aligned}$
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