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Ultraviolet light of wavelength $99 \mathrm{~nm}$ falls on a metal plate of work function $1.0 \mathrm{eV}$. If the mass of the election is $9.1 \times 10^{-31} \mathrm{~kg}$, the wavelength of the fastest photoelectron emitted is
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Verified Answer
The correct answer is:
$0.36 \mathrm{~mm}$
Given
$$
\begin{aligned}
& K_{\max }=h v-W_0 \\
& =\frac{\left(6.6 \times 10^{-34}\right) \times\left(3 \times 10^8\right)}{99 \times 10^{-9}}-1.0 \times 1.6 \times 10^{-19} \\
& =2 \times 10^{-18}-1.6 \times 10^{-19} \\
& =1.84 \times 10^{-18} \mathrm{~J}
\end{aligned}
$$
Now,
$$
\begin{aligned}
\lambda & =\frac{h}{\sqrt{2 m K_{\max }}} \\
& =\frac{6.6 \times 10^{-34}}{\sqrt{2 \times\left(9.1 \times 10^{-31}\right) \times 1.84 \times 10^{-18}}} \\
& =0.36 \times 10^{-9} \mathrm{~m}=0.36 \mathrm{~nm}
\end{aligned}
$$
$$
\begin{aligned}
& K_{\max }=h v-W_0 \\
& =\frac{\left(6.6 \times 10^{-34}\right) \times\left(3 \times 10^8\right)}{99 \times 10^{-9}}-1.0 \times 1.6 \times 10^{-19} \\
& =2 \times 10^{-18}-1.6 \times 10^{-19} \\
& =1.84 \times 10^{-18} \mathrm{~J}
\end{aligned}
$$
Now,
$$
\begin{aligned}
\lambda & =\frac{h}{\sqrt{2 m K_{\max }}} \\
& =\frac{6.6 \times 10^{-34}}{\sqrt{2 \times\left(9.1 \times 10^{-31}\right) \times 1.84 \times 10^{-18}}} \\
& =0.36 \times 10^{-9} \mathrm{~m}=0.36 \mathrm{~nm}
\end{aligned}
$$
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