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

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$ 1.05 	imes 10^{-34} 	ext{ Js} $

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(B) The energy of the $ n $-th orbit in a hydrogen atom is given by $ E_n = -\frac{13.6}{n^2} 	ext{ eV} $.For the ground state $( n_1 = 1 )$,$ E_1 = -13.6 	ext{ eV} $.When the atom absorbs $ 10.2 	ext{ eV} $,the new energy is $ E_2 = -13.6 + 10.2 = -3.4 	ext{ eV} $.Since $ E_2 = -\frac{13.6}{n_2^2} = -3.4 	ext{ eV} $,we find $ n_2^2 = 4 $,so $ n_2 = 2 $.The orbital angular momentum is given by $ L = \frac{nh}{2\pi} $.The increase in angular momentum is $ \Delta L = L_2 - L_1 = \frac{n_2 h}{2\pi} - \frac{n_1 h}{2\pi} = \frac{(n_2 - n_1)h}{2\pi} $.Substituting $ n_2 = 2 $,$ n_1 = 1 $,and $ h = 6.626 	imes 10^{-34} 	ext{ Js} $:$ \Delta L = \frac{(2 - 1) 	imes 6.626 	imes 10^{-34}}{2 	imes 3.14159} \approx 1.054 	imes 10^{-34} 	ext{ Js} $.Thus,the increase is $ 1.05 	imes 10^{-34} 	ext{ Js} $.

$A$ hydrogen atom in the ground state absorbs $ 10.2 \text{ eV} $ of energy. The orbital angular momentum of the electron is increased by:

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