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(B) The energy of the photon released during the transition from the second excited state $(n_2 = 3)$ to the ground state $(n_1 = 1)$ is given by:$E =

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$4 \ 	ext{Å}$

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(B) The energy of the photon released during the transition from the second excited state $(n_2 = 3)$ to the ground state $(n_1 = 1)$ is given by:$E = 13.6 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ight] \ eV$$E = 13.6 \left[ \frac{1}{1^2} - \frac{1}{3^2} ight] = 13.6 \left[ 1 - \frac{1}{9} ight] = 13.6 	imes \frac{8}{9} \approx 12.09 \ eV \approx 12.1 \ eV$.Using Einstein's photoelectric equation,the maximum kinetic energy $(K_{\max})$ of the emitted electron is:$K_{\max} = E - W = 12.1 \ eV - 3.1 \ eV = 9.0 \ eV$.The de Broglie wavelength $(\lambda)$ is given by $\lambda = \frac{h}{\sqrt{2mK_{\max}}}$.Using the formula $\lambda \approx \frac{12.27}{\sqrt{K_{\max}}} \ 	ext{Å}$ (where $K_{\max}$ is in $eV$):$\lambda = \frac{12.27}{\sqrt{9}} = \frac{12.27}{3} = 4.09 \ 	ext{Å}$.Thus,the wavelength is nearly $4 \ 	ext{Å}$.

$A$ photon released by the transition of an electron from the second excited state to the ground state of a Hydrogen atom is incident on the surface of a metal with a work function of $3.1 \ eV$. The de Broglie wavelength of the most energetic electron emitted from that metal surface is nearly:

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