Question 102 of 200
The work function of a metal is 3 eV. The color of the visible light that is required to cause emission of photoelectrons is
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Answer: (B) Blue
$\begin{aligned}
& (\mathrm{KE})_{\max }=\frac{\mathrm{hc}}{\lambda}-\phi \\
& \frac{\mathrm{hc}}{\lambda} \gt \phi[\text { for emission }] \\
& \lambda \lt \frac{\mathrm{hc}}{\phi} \Rightarrow \lambda \lt \frac{1242}{3} \mathrm{~nm}
\end{aligned}$
So blue light option (B)
& (\mathrm{KE})_{\max }=\frac{\mathrm{hc}}{\lambda}-\phi \\
& \frac{\mathrm{hc}}{\lambda} \gt \phi[\text { for emission }] \\
& \lambda \lt \frac{\mathrm{hc}}{\phi} \Rightarrow \lambda \lt \frac{1242}{3} \mathrm{~nm}
\end{aligned}$
So blue light option (B)