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(A) The energy of an electron in the $n$-th orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \ eV$.For the first orbit $(n_1 = 1)$,$E_1 =

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$1.215 	imes 10^{-7} \ m$

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(A) The energy of an electron in the $n$-th orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \ eV$.For the first orbit $(n_1 = 1)$,$E_1 = -13.6 \ eV$.For the second orbit $(n_2 = 2)$,$E_2 = -\frac{13.6}{2^2} = -3.4 \ eV$.The energy difference is $\Delta E = E_2 - E_1 = -3.4 - (-13.6) = 10.2 \ eV$.Converting this to Joules: $\Delta E = 10.2 	imes 1.6 	imes 10^{-19} \ J = 1.632 	imes 10^{-18} \ J$.Using the relation $\Delta E = \frac{hc}{\lambda}$,where $h = 6.63 	imes 10^{-34} \ J \cdot s$ and $c = 3 	imes 10^8 \ m/s$:$\lambda = \frac{hc}{\Delta E} = \frac{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}{1.632 	imes 10^{-18}} \ m \approx 1.215 	imes 10^{-7} \ m$.

The wavelength of light emitted when an electron transitions from the second orbit to the first orbit in a hydrogen atom is:

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