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 $n^{t h}$ energy level, then

$R (Z_{H})^{2} (\frac{1}{1 ^{2}} - \frac{1}{2 ^{2}}) = R \frac{( Z _{L i} ) ^{2}}{n ^{2}}$

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

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

$n = 12 \approx 3.5$

Hence the least quantum number must be $4$ .

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