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(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{R_H}{n^2}$,where $R_H = 2.18 	imes 10^{-18} \ J$.The

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$3.08 	imes 10^{15} \ s^{-1}$

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(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{R_H}{n^2}$,where $R_H = 2.18 	imes 10^{-18} \ J$.The energy difference for the transition from $n_2 = 4$ to $n_1 = 1$ is $\Delta E = E_4 - E_1 = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.$\Delta E = 2.18 	imes 10^{-18} \left( \frac{1}{1^2} - \frac{1}{4^2} ight) = 2.18 	imes 10^{-18} \left( 1 - \frac{1}{16} ight) = 2.18 	imes 10^{-18} 	imes \frac{15}{16} \ J$.Since $\Delta E = h
u$,the frequency $
u = \frac{\Delta E}{h}$.$
u = \frac{2.18 	imes 10^{-18} 	imes 15}{16 	imes 6.625 	imes 10^{-34}} = \frac{32.7 	imes 10^{-18}}{106 	imes 10^{-34}} \approx 3.08 	imes 10^{15} \ s^{-1}$.

The frequency of radiation emitted when the electron falls from $n = 4$ to $n = 1$ in a hydrogen atom will be (Given ionization energy of $H = 2.18 \times 10^{-18} \ J \ atom^{-1}$ and $h = 6.625 \times 10^{-34} \ Js$)

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