An electron in a hydrogen atom makes a transition from n = 2 to n = 1 and emits a photon. This photon strikes a doubly ionized lithium atom which was already in an excited state and completely removes

Question

An electron in a hydrogen atom makes a transition from n = 2 to n = 1 and emits a photon. This photon strikes a doubly ionized lithium atom which was already in an excited state and completely removes the orbiting electron. The least quantum number for the excited state of the lithium-ion for the process is

TestHub · Accepted Answer

Let us assume that the electron in the lithium-ion is knocked off from the $n^{th}$ energy level, then

$R {(Z_{H})}^{2} (\frac{1}{1^{2}} - \frac{1}{2^{2}}) = R \frac{{(Z_{Li})}^{2}}{n^{2}}$

$R {(1)}^{2} \times \frac{3}{4} = R {(3)}^{2} [\frac{1}{n^{2}}]$

$\frac{3}{4} = \frac{9}{n^{2}}$

$n = \sqrt{12} \approx 3.5$

Hence the least quantum number must be $4$ .

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