A hydrogen atom makes a transition from n = 2 | vedclass

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(B) The energy of the photon emitted by the hydrogen atom during the transition from $n = 2$ to $n = 1$ is given by:$E = 13.6 	ext{ eV} 	imes Z^2 \l

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$4$

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(B) The energy of the photon emitted by the hydrogen atom during the transition from $n = 2$ to $n = 1$ is given by:$E = 13.6 	ext{ eV} 	imes Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$For hydrogen $(Z = 1)$,$E = 13.6 	imes 1^2 \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = 13.6 	imes \frac{3}{4} = 10.2 	ext{ eV}$.This photon is absorbed by a doubly ionized lithium atom ($Li^{2+}$,$Z = 3$) in an excited state $n$. The energy required to remove an electron from the $n$-th state of a hydrogen-like ion is:$E_n = 13.6 	ext{ eV} 	imes \frac{Z^2}{n^2} = 13.6 	imes \frac{3^2}{n^2} = 13.6 	imes \frac{9}{n^2}$.For the photon to completely remove the electron,its energy must be greater than or equal to the ionization energy of the state $n$:$10.2 \ge 13.6 	imes \frac{9}{n^2}$$\frac{10.2}{13.6} \ge \frac{9}{n^2}$$0.75 \ge \frac{9}{n^2}$$n^2 \ge \frac{9}{0.75} = 12$$n \ge \sqrt{12} \approx 3.46$.Since $n$ must be an integer,the least quantum number is $n = 4$.

$A$ hydrogen atom makes a transition from $n = 2$ to $n = 1$ and emits a photon. This photon strikes a doubly ionized lithium atom $(Z = 3)$ in an excited state and completely removes the orbiting electron. The least quantum number for the excited state of the ion for this process is:

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