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Light strikes a metal surface causing photoelectric emission. The wavelength of incident light is $248 \mathrm{~nm}$. If the stopping potential for the ejected electrons is $2.8 \mathrm{eV}$, then the work function of the metal is (Take hc $=1240 \mathrm{eV} . \mathrm{nm})$
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Verified Answer
The correct answer is:
$2.2 \mathrm{eV}$
We have
K.E. = E - W
$$
\mathrm{K} . \mathrm{E}=\frac{\mathrm{hc}}{\lambda}-\mathrm{W}
$$
$$
\begin{aligned}
& \mathrm{W}=\frac{\mathrm{hc}}{\lambda}-\mathrm{K} . \mathrm{E} \\
& =\frac{1240}{278}=2.8
\end{aligned}
$$
The work function of the metal, $\mathrm{W}=2.2 \mathrm{eV}$
K.E. = E - W
$$
\mathrm{K} . \mathrm{E}=\frac{\mathrm{hc}}{\lambda}-\mathrm{W}
$$
$$
\begin{aligned}
& \mathrm{W}=\frac{\mathrm{hc}}{\lambda}-\mathrm{K} . \mathrm{E} \\
& =\frac{1240}{278}=2.8
\end{aligned}
$$
The work function of the metal, $\mathrm{W}=2.2 \mathrm{eV}$
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