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

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$n^3$

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(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6 	ext{ eV}}{n^2}$.When an electron transitions from $n+1$ to $n$,the energy of the emitted photon is $\Delta E = E_{n+1} - E_n = 13.6 \left( \frac{1}{n^2} - \frac{1}{(n+1)^2} ight) 	ext{ eV}$.This simplifies to $\Delta E = 13.6 \left( \frac{(n+1)^2 - n^2}{n^2(n+1)^2} ight) = 13.6 \left( \frac{2n+1}{n^2(n+1)^2} ight)$.For large $n$,$2n+1 \approx 2n$ and $n^2(n+1)^2 \approx n^4$.Thus,$\Delta E \approx 13.6 \left( \frac{2n}{n^4} ight) = \frac{27.2}{n^3}$.Since the energy of the photon is related to wavelength by $\Delta E = \frac{hc}{\lambda}$,we have $\lambda = \frac{hc}{\Delta E}$.Substituting $\Delta E \propto \frac{1}{n^3}$,we get $\lambda \propto n^3$.

An electron in a hydrogen atom makes a transition from the $(n+1)^{th}$ orbit to the $n^{th}$ orbit. For large $n$,the wavelength of the emitted radiation is proportional to:

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