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Light of frequency $10^{15} \mathrm{~Hz}$ falls on a metal surface of work function $2.5 \mathrm{eV}$. The stopping potential of photoelectrons (in V) is
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Verified Answer
The correct answer is:
$1.6$
Frequency of incident light,
$$
v=10^{15} \mathrm{~Hz}
$$
Work function, $\phi_{0}=25 \mathrm{eV}$
Energy of incident photon,
$$
\begin{aligned}
E &=h \mathrm{v} \\
&=6.62 \times 10^{-34} \times 10^{15} \\
&=6.62 \times 10^{-19} \mathrm{~J} \\
&=\frac{6.62 \times 10^{-19}}{1.6 \times 10^{-19}} \mathrm{eV} \\
&=4.1 \mathrm{eV}
\end{aligned}
$$
According to Einstein's photoelectric equation,
$$
\begin{array}{ll}
\Rightarrow & K_{\max }=h \mathrm{v}-\phi_{0} \\
\Rightarrow & e V_{0}=h \mathrm{v}-\phi_{0} \\
\Rightarrow & e V_{0}=4.1 \mathrm{eV}-25 \mathrm{eV} \\
\Rightarrow & e V_{0}=1.6 \mathrm{eV} \\
\Rightarrow & V_{0}=1.6 \mathrm{~V}
\end{array}
$$
$$
v=10^{15} \mathrm{~Hz}
$$
Work function, $\phi_{0}=25 \mathrm{eV}$
Energy of incident photon,
$$
\begin{aligned}
E &=h \mathrm{v} \\
&=6.62 \times 10^{-34} \times 10^{15} \\
&=6.62 \times 10^{-19} \mathrm{~J} \\
&=\frac{6.62 \times 10^{-19}}{1.6 \times 10^{-19}} \mathrm{eV} \\
&=4.1 \mathrm{eV}
\end{aligned}
$$
According to Einstein's photoelectric equation,
$$
\begin{array}{ll}
\Rightarrow & K_{\max }=h \mathrm{v}-\phi_{0} \\
\Rightarrow & e V_{0}=h \mathrm{v}-\phi_{0} \\
\Rightarrow & e V_{0}=4.1 \mathrm{eV}-25 \mathrm{eV} \\
\Rightarrow & e V_{0}=1.6 \mathrm{eV} \\
\Rightarrow & V_{0}=1.6 \mathrm{~V}
\end{array}
$$
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