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(A) The energy of the ground state of a hydrogen atom is $E_1 = -13.6 \ eV$. When it absorbs $10.2 \ eV$ of energy,its new energy becomes $E = -13.6 \

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$1.05 	imes 10^{-34} \ J \cdot s$

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(A) The energy of the ground state of a hydrogen atom is $E_1 = -13.6 \ eV$. When it absorbs $10.2 \ eV$ of energy,its new energy becomes $E = -13.6 \ eV + 10.2 \ eV = -3.4 \ eV$. This corresponds to the first excited state,where the principal quantum number $n = 2$. The orbital angular momentum $L$ of an electron in a hydrogen atom is given by $L = \frac{nh}{2\pi}$. In the ground state $(n_1 = 1)$,the angular momentum is $L_1 = \frac{1 \cdot h}{2\pi}$. In the first excited state $(n_2 = 2)$,the angular momentum is $L_2 = \frac{2 \cdot h}{2\pi} = \frac{h}{\pi}$. The increase in orbital angular momentum is $\Delta L = L_2 - L_1 = \frac{2h}{2\pi} - \frac{h}{2\pi} = \frac{h}{2\pi}$. Substituting the value of $h = 6.6 	imes 10^{-34} \ J \cdot s$: $\Delta L = \frac{6.6 	imes 10^{-34}}{2 	imes 3.14} \approx \frac{6.6 	imes 10^{-34}}{6.28} \approx 1.05 	imes 10^{-34} \ J \cdot s$.

$A$ hydrogen atom in its ground state absorbs $10.2 \ eV$ of energy. The orbital angular momentum is increased by (Given Planck constant $h = 6.6 \times 10^{-34} \ J \cdot s$)

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