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(C) The orbital angular momentum of an electron in a hydrogen atom is given by $L = \frac{nh}{2 \pi}$.Initially, the electron is in the ground state,

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$10.2$

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(C) The orbital angular momentum of an electron in a hydrogen atom is given by $L = \frac{nh}{2 \pi}$.Initially, the electron is in the ground state, so $n_1 = 1$. The initial angular momentum is $L_1 = \frac{1 \cdot h}{2 \pi} = \frac{h}{2 \pi}$.After absorbing energy $\Delta E$, the angular momentum increases by $\frac{h}{2 \pi}$.Thus, the new angular momentum is $L_2 = L_1 + \frac{h}{2 \pi} = \frac{h}{2 \pi} + \frac{h}{2 \pi} = \frac{2h}{2 \pi} = \frac{2h}{2 \pi}$.Comparing this with $L = \frac{nh}{2 \pi}$, we find the new principal quantum number is $n_2 = 2$.The energy of an electron in the $n$-th state is given by $E_n = -\frac{13.6 \ eV}{n^2}$.For $n_1 = 1$, $E_1 = -13.6 \ eV$.For $n_2 = 2$, $E_2 = -\frac{13.6 \ eV}{2^2} = -\frac{13.6 \ eV}{4} = -3.4 \ eV$.The energy absorbed is $\Delta E = E_2 - E_1 = -3.4 \ eV - (-13.6 \ eV) = 10.2 \ eV$.

$A$ hydrogen atom in the ground state absorbs $\Delta E$ amount of energy. If the orbital angular momentum of the electron is increased by $\frac{h}{2 \pi}$ ($h=$ Planck constant), then the magnitude of $\Delta E$ is (in $eV$)

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