CBSE 2026 · Region 3 · Set 3 · Q19 · 2 marks
Light of frequency $\displaystyle 5 \cdot 0 \times 10^{14} \mathrm{~Hz}$ is incident on a metal surface. If the maximum speed of photoelectrons emitted is $\displaystyle 6.63 \times 10^{5} \mathrm{~ms}^{-1}$, calculate the threshold frequency for the surface.
Marking-scheme solution
\[\begin{aligned}
K . E & =h v-h v_{0} \\
v_{0} & =\frac{h v-K . E}{h} \\
& =\frac{\left.\left(6.63 \times 10^{-34} \times 5 \times 10^{14}\right)-\left(\dfrac{1}{2} \times 9.1 \times 10^{-31}\right) \times(6.63)^{2} \times 10^{10}\right)}{6.63 \times 10^{-34}} \\
& =\frac{\left(5 \times 10^{-20}\right)-\left(3 \times 10^{-20}\right)}{10^{-34}} \\
& =2 \times 10^{14} \mathrm{~Hz}
\end{aligned}
\]
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