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

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$12.1$

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(B) The energy of an electron in the $n$-th orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} 	ext{ eV}$.For the ground state $(n_1 = 1)$, the energy is $E_1 = -13.6 	ext{ eV}$.For the excited state $(n_2 = 3)$, the energy is $E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} \approx -1.51 	ext{ eV}$.The energy required to excite the atom from $n=1$ to $n=3$ is $\Delta E = E_3 - E_1$.$\Delta E = -1.51 - (-13.6) = 12.09 	ext{ eV}$.Rounding this value, we get $\Delta E \approx 12.1 	ext{ eV}$.

An electron collides with a hydrogen atom in its ground state and excites it to the $n=3$ state. The energy given to the hydrogen atom in this inelastic collision (neglecting the recoil of the hydrogen atom) is (in $\text{ eV}$)

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