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(D) The energy of the electron in the $n^{	ext{th}}$ orbit is given by $E_n = \frac{-13.6}{n^2} 	ext{ eV}$.For the ground state $(n=1)$, $E_1 = \fra

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$8$

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(D) The energy of the electron in the $n^{	ext{th}}$ orbit is given by $E_n = \frac{-13.6}{n^2} 	ext{ eV}$.For the ground state $(n=1)$, $E_1 = \frac{-13.6}{1^2} = -13.6 	ext{ eV}$.For the third orbit $(n=3)$, $E_3 = \frac{-13.6}{3^2} = \frac{-13.6}{9} \approx -1.51 	ext{ eV}$.The energy of the emitted photon is $\Delta E = E_3 - E_1 = -1.51 - (-13.6) = 12.09 	ext{ eV}$.According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectron is $K_{	ext{max}} = \Delta E - \phi$, where $\phi$ is the work function.Given $\phi = 4.1 	ext{ eV}$, we have $K_{	ext{max}} = 12.09 - 4.1 = 7.99 	ext{ eV}$.The stopping potential $V_s$ is related to $K_{	ext{max}}$ by $K_{	ext{max}} = e V_s$, so $V_s \approx 8 	ext{ V}$.

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