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JEE Main 2020 · Physics · Atoms, Nuclei and Dual Nature of Radiation · easy

A Li^2+ ion emits a photon of the same wavelength as the first line of the Lyman series of hydrogen (n=2→ n=1). The transition in Li^2+ responsible is

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A Li2+\mathrm{Li^{2+}} ion emits a photon of the same wavelength as the first line of the Lyman series of hydrogen (n=2→n=1n=2\to n=1). The transition in Li2+\mathrm{Li^{2+}} responsible is
  1. A. n=3→n=1n=3\to n=1
  2. B. n=4→n=2n=4\to n=2
  3. C. n=6→n=3n=6\to n=3
  4. D. n=5→n=2n=5\to n=2
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Answer: (C) n=6→n=3n=6\to n=3

For hydrogen-like species 1λ=RZ2(1n12−1n22)\dfrac1\lambda = RZ^2\left(\dfrac1{n_1^2} - \dfrac1{n_2^2}\right). For hydrogen Lyman-α\alpha: R(1−14)=34RR(1 - \tfrac14) = \tfrac34 R. For Li2+\mathrm{Li^{2+}} (Z=3Z=3): 9(1n12−1n22)=349\left(\dfrac1{n_1^2} - \dfrac1{n_2^2}\right) = \dfrac34, so 1n12−1n22=112\dfrac1{n_1^2} - \dfrac1{n_2^2} = \dfrac1{12}. Trying integers: n1=3n_1 = 3 gives 19−112=136\dfrac19 - \dfrac1{12} = \dfrac1{36}, so n2=6n_2 = 6. Other n1n_1 give no integer solution, hence n=6→n=3n=6\to n=3.
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