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JEE Main 2021 · Chemistry · Atomic Structure · medium

A hydrogen-like ion Li^2+ emits a photon of the same wavelength as the first line (longest wavelength) of the Lyman series of atomic hydrogen. The…

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A hydrogen-like ion Li2+\mathrm{Li^{2+}} emits a photon of the same wavelength as the first line (longest wavelength) of the Lyman series of atomic hydrogen. The electronic transition in Li2+\mathrm{Li^{2+}} responsible for this is
  1. A. n=3→n=2n=3\to n=2
  2. B. n=6→n=3n=6\to n=3
  3. C. n=4→n=2n=4\to n=2
  4. D. n=9→n=3n=9\to n=3
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Answer: (B) n=6→n=3n=6\to n=3

The first Lyman line of hydrogen is 2→12\to1: 1λ=R(1−14)=3R4\dfrac{1}{\lambda}=R\left(1-\dfrac14\right)=\dfrac{3R}{4}. For Li2+\mathrm{Li^{2+}} (Z=3Z=3): 1λ=9R(1n12−1n22)\dfrac1\lambda=9R\left(\dfrac1{n_1^2}-\dfrac1{n_2^2}\right). We need 1n12−1n22=112\dfrac1{n_1^2}-\dfrac1{n_2^2}=\dfrac1{12}. Try n1=3n_1=3: 19−1n22=112\dfrac19-\dfrac1{n_2^2}=\dfrac1{12} gives 1n22=136\dfrac1{n_2^2}=\dfrac1{36}, so n2=6n_2=6. Hence 6→36\to3. Check 3→23\to2: 14−19=536≠112\dfrac14-\dfrac19=\dfrac5{36}\ne\dfrac1{12}; 4→24\to2: 316\dfrac3{16} no; 9→39\to3: 19−181=881\dfrac19-\dfrac1{81}=\dfrac8{81} no.
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