Atoms & Nuclei

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145 of 239

Question 145 of 239

An electron in a hydrogen atom makes a transition from n=2n=2 to n=1n=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

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Answer: (B) 44

Let us assume that the electron in the lithium-ion is knocked off from the nth{n}^{th} energy level, then
R(ZH)2(112−122)=R(ZLi)2n2R{\left({Z}_{H}\right)}^{2}\left(\dfrac{1}{{1}^{2}}-\dfrac{1}{{2}^{2}}\right)=R\dfrac{{\left({Z}_{Li}\right)}^{2}}{{n}^{2}}
R(1)2×34=R(3)2[1n2]R{\left(1\right)}^{2}\times \dfrac{3}{4}=R{\left(3\right)}^{2}\left[\dfrac{1}{{n}^{2}}\right]
34=9n2\dfrac{3}{4}=\dfrac{9}{{n}^{2}}

n=12≈3.5n=\sqrt{12}\approx 3.5
Hence the least quantum number must be 44.

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