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In an experiment on photoelectric emission from a metallic surface, wavelength of incident light is $2 \times 10^{-7} \mathrm{~m}$ and stopping potential is $2.5 \mathrm{~V}$. The threshold frequency of the metal (in $\mathrm{Hz}$ ) approximately (charge of electron $e=1.6 \times 10^{-19} \mathrm{C}$, Planck's constant $\left.h=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{s}\right)$
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The correct answer is:
$9 \times 10^{14}$
$e V_0=h v-h v_0$
$\therefore$ Threshold frequency,
$\begin{aligned} v_0 & =v-\frac{e V_0}{h} \\ & =\frac{c}{\lambda}-\frac{e V_0}{h} \\ v_0 & =\frac{3 \times 10^8}{2 \times 10^{-7}}-\frac{1.6 \times 10^{-19} \times 2.5}{6.6 \times 10^{-34}} \\ & =9.0 \times 10^{14} \mathrm{~Hz}\end{aligned}$
$\therefore$ Threshold frequency,
$\begin{aligned} v_0 & =v-\frac{e V_0}{h} \\ & =\frac{c}{\lambda}-\frac{e V_0}{h} \\ v_0 & =\frac{3 \times 10^8}{2 \times 10^{-7}}-\frac{1.6 \times 10^{-19} \times 2.5}{6.6 \times 10^{-34}} \\ & =9.0 \times 10^{14} \mathrm{~Hz}\end{aligned}$
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