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(D) The energy of a photon emitted during a transition from $n_i$ to $n_f$ is given by $E = 13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} ight) 	ex

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From $n=4$ to $n=3$

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(D) The energy of a photon emitted during a transition from $n_i$ to $n_f$ is given by $E = 13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} ight) 	ext{ eV}$.For the transition $n=3$ to $n=2$, the energy is $E_{3-2} = 13.6 \left( \frac{1}{4} - \frac{1}{9} ight) = 13.6 \left( \frac{5}{36} ight) \approx 1.89 	ext{ eV}$.Since photoelectrons are just emitted, the work function of the metal is $\Phi = 1.89 	ext{ eV}$.For the photoelectric effect to occur, the energy of the incident photon must be greater than or equal to the work function $(E \ge \Phi)$.Let's calculate the energy for the given options:$A$: $n=2$ to $n=1$: $E = 13.6 (1 - 1/4) = 10.2 	ext{ eV} > 1.89 	ext{ eV}$ (Possible).$B$: $n=3$ to $n=1$: $E = 13.6 (1 - 1/9) = 12.09 	ext{ eV} > 1.89 	ext{ eV}$ (Possible).$C$: $n=5$ to $n=2$: $E = 13.6 (1/4 - 1/25) = 13.6 (21/100) = 2.856 	ext{ eV} > 1.89 	ext{ eV}$ (Possible).$D$: $n=4$ to $n=3$: $E = 13.6 (1/9 - 1/16) = 13.6 (7/144) \approx 0.66 	ext{ eV} Therefore, the photoelectric effect is not possible for the transition from $n=4$ to $n=3$.

Light emitted during the de-excitation of an electron from $n=3$ to $n=2$ is incident on a metal, and photoelectrons are just emitted from that metal. In which of the following de-excitations is the photoelectric effect $NOT$ possible?

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