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(B) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6 	ext{ eV}}{n^2}$.For the first transition (seco

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$3: 1$

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(B) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6 	ext{ eV}}{n^2}$.For the first transition (second to first level): $n_i = 2$ to $n_f = 1$.The energy of the emitted photon is $\Delta E_1 = E_2 - E_1 = -\frac{13.6}{2^2} - (-\frac{13.6}{1^2}) = 13.6(1 - \frac{1}{4}) = 13.6 	imes \frac{3}{4} 	ext{ eV}$.For the second transition (highest to second level): $n_i = \infty$ to $n_f = 2$.The energy of the emitted photon is $\Delta E_2 = E_{\infty} - E_2 = 0 - (-\frac{13.6}{2^2}) = \frac{13.6}{4} 	ext{ eV}$.The ratio of the energies is $\frac{\Delta E_1}{\Delta E_2} = \frac{13.6 	imes \frac{3}{4}}{13.6 	imes \frac{1}{4}} = \frac{3}{1}$.Thus,the ratio is $3: 1$.

The ratio of energies of photons produced due to the transition of an electron in a hydrogen atom from its $(a)$ second to first energy level and $(b)$ highest energy level to the second level is:

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