The ratio of energies of photons produced due | vedclass

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

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$3$

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

The ratio of energies of photons produced due to the transition of an electron in a hydrogen atom from its $(i)$ second to first energy level and (ii) highest energy level to second energy level is: (in $3:1$)

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