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(C) The energy of the emitted photon during a transition from $n_i = 3$ to $n_f = 2$ is given by the Rydberg formula: $E = 13.6 \left[ \frac{1}{n_f^2}

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$1.89$

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(C) The energy of the emitted photon during a transition from $n_i = 3$ to $n_f = 2$ is given by the Rydberg formula: $E = 13.6 \left[ \frac{1}{n_f^2} - \frac{1}{n_i^2} ight] 	ext{ eV}$ Substituting the values $n_f = 2$ and $n_i = 3$: $E = 13.6 \left[ \frac{1}{2^2} - \frac{1}{3^2} ight] 	ext{ eV}$ $E = 13.6 \left[ \frac{1}{4} - \frac{1}{9} ight] 	ext{ eV}$ $E = 13.6 \left[ \frac{9 - 4}{36} ight] 	ext{ eV}$ $E = 13.6 	imes \frac{5}{36} 	ext{ eV} \approx 1.89 	ext{ eV}$ For the photoelectric effect to occur, the energy of the incident photon must be greater than or equal to the work function $(\Phi)$ of the metal. Therefore, the maximum work function is $\Phi_{	ext{max}} = 1.89 	ext{ eV}$.

The light emitted in the transition $n = 3$ to $n = 2$ (where $n$ is the principal quantum number of the state) in hydrogen is called $H_{\alpha}$-light. Find the maximum work function that a metal can have so that $H_{\alpha}$-light can emit photoelectrons from it. (in $\text{ eV}$)

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