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(A) The energy of a photon emitted during a transition is given by $E = 13.6 Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight) 	ext{ eV}$.For hydr

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$4$ to $3$

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(A) The energy of a photon emitted during a transition is given by $E = 13.6 Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight) 	ext{ eV}$.For hydrogen $(Z=1)$,the transition from $n=4$ to $n=2$ is:$E_H = 13.6 	imes 1^2 \left( \frac{1}{2^2} - \frac{1}{4^2} ight) = 13.6 \left( \frac{1}{4} - \frac{1}{16} ight) = 13.6 \left( \frac{3}{16} ight) = 2.55 	ext{ eV}$.Given that the frequency (and thus energy) of the $Li^{2+}$ transition $(Z=3)$ is such that $E_H = \frac{3}{7} E_{Li}$,we have $E_{Li} = \frac{7}{3} E_H = \frac{7}{3} 	imes 2.55 = 5.95 	ext{ eV}$.For $Li^{2+}$,$E_{Li} = 13.6 	imes 3^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight) = 122.4 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight) = 5.95 	ext{ eV}$.$\left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight) = \frac{5.95}{122.4} \approx 0.0486$.Testing transitions:For $n_2=4$ to $n_1=3$: $\frac{1}{9} - \frac{1}{16} = 0.111 - 0.0625 = 0.0486$.Thus,the transition is $4$ to $3$.

The frequency of the light emitted when an electron transitions from the $n=4$ to $n=2$ level in a hydrogen atom is $\frac{3}{7}$ times the frequency of a transition in a $Li^{2+}$ ion. Which transition in the $Li^{2+}$ ion corresponds to this?

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