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Monochromatic radiation emitted when electron on hydrogen atom jumps from first excited to the ground state irradiates a photosensitive material. The stopping potential is measured to be $3.57 \mathrm{~V}$. The threshold frequency of the material is
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The correct answer is:
$1.6 \times 10^{15} \mathrm{~Hz}$
Energy released from emission of electron
$\begin{aligned}
E & =(-3.4)-(-13.6) \\
& =10.2 \mathrm{eV}
\end{aligned}$
From photo electric equation.
Work function
$\begin{aligned}
\phi & =E-e V=h v \\
v & =\frac{E-e V}{h} \\
& =\frac{(10.2-3.57) e}{6.67 \times 10^{-34}} \\
v & =\frac{6.63 \times 1.6 \times 10^{-19}}{6.67 \times 10^{-34}} \\
& =1.6 \times 10^{15} \mathrm{~Hz}
\end{aligned}$
$\begin{aligned}
E & =(-3.4)-(-13.6) \\
& =10.2 \mathrm{eV}
\end{aligned}$
From photo electric equation.
Work function
$\begin{aligned}
\phi & =E-e V=h v \\
v & =\frac{E-e V}{h} \\
& =\frac{(10.2-3.57) e}{6.67 \times 10^{-34}} \\
v & =\frac{6.63 \times 1.6 \times 10^{-19}}{6.67 \times 10^{-34}} \\
& =1.6 \times 10^{15} \mathrm{~Hz}
\end{aligned}$
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