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

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$3: 4$

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

Find the ratio of energies of photons produced due to the transition of an electron in a hydrogen atom from its $(i)$ second permitted energy level to the first level,and $(ii)$ the highest permitted energy level to the first permitted level.

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