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A photon released by the transition of an electron from the second excited state to the ground state of Hydrogen atom is incident on the surface of a metal of work function $3.1 \mathrm{eV}$. The de Broglie wavelength of the most energetic electron emitted from that metal surface is nearly
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Work function, $\mathrm{W}=3.1 \mathrm{eV}$
$$
\begin{aligned}
& \text { hu }=13.6\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right] \\
& =13.6\left[\frac{1}{1^2}-\frac{1}{3^2}\right]=13.6\left[1-\frac{1}{9}\right]=13.6 \times \frac{8}{9} \\
& =12.1 \mathrm{eV} .
\end{aligned}
$$
From Einstein's photoelectric equation
$$
\begin{aligned}
& \mathrm{hv}=\mathrm{K}_{\max }+\mathrm{W} \\
& \frac{\mathrm{hc}}{\lambda}=\mathrm{hv}-\mathrm{w} \\
& =12.1-3.1 \\
& \Rightarrow \frac{\mathrm{hc}}{\lambda}=9 \mathrm{eV} \\
& \lambda=\frac{66 \times 10^{-34} \times 3 \times 10^8}{9 \times 1.6 \times 10^{-19}}=1.3 \times 10^{-7} \mathrm{~m}
\end{aligned}
$$
$$
\begin{aligned}
& \text { hu }=13.6\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right] \\
& =13.6\left[\frac{1}{1^2}-\frac{1}{3^2}\right]=13.6\left[1-\frac{1}{9}\right]=13.6 \times \frac{8}{9} \\
& =12.1 \mathrm{eV} .
\end{aligned}
$$
From Einstein's photoelectric equation
$$
\begin{aligned}
& \mathrm{hv}=\mathrm{K}_{\max }+\mathrm{W} \\
& \frac{\mathrm{hc}}{\lambda}=\mathrm{hv}-\mathrm{w} \\
& =12.1-3.1 \\
& \Rightarrow \frac{\mathrm{hc}}{\lambda}=9 \mathrm{eV} \\
& \lambda=\frac{66 \times 10^{-34} \times 3 \times 10^8}{9 \times 1.6 \times 10^{-19}}=1.3 \times 10^{-7} \mathrm{~m}
\end{aligned}
$$
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