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Question

(C) The energy of a photon emitted during a transition from energy level $n_2$ to $n_1$ in a hydrogen atom is given by the Rydberg formula: $E = 13.6

Vedclass · Accepted Answer

$25: 7$

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(C) The energy of a photon emitted during a transition from energy level $n_2$ to $n_1$ in a hydrogen atom is given by the Rydberg formula: $E = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight) 	ext{ eV}$.For the first transition ($n_2 = 2$ to $n_1 = 1$): $E_1 = 13.6 \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = 13.6 \left( 1 - \frac{1}{4} ight) = 13.6 	imes \frac{3}{4} 	ext{ eV}$.For the second transition ($n_2 = 5$ to $n_1 = 2$): $E_2 = 13.6 \left( \frac{1}{2^2} - \frac{1}{5^2} ight) = 13.6 \left( \frac{1}{4} - \frac{1}{25} ight) = 13.6 \left( \frac{25 - 4}{100} ight) = 13.6 	imes \frac{21}{100} 	ext{ eV}$.The ratio of the energies is $\frac{E_1}{E_2} = \frac{13.6 	imes (3/4)}{13.6 	imes (21/100)} = \frac{3}{4} 	imes \frac{100}{21} = \frac{1}{1} 	imes \frac{25}{7} = \frac{25}{7}$.

The ratio of energies of photons produced due to transition of an electron in a hydrogen atom from the second energy level to the first energy level and from the fifth energy level to the second energy level is:

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