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JEE Main 2019 · Chemistry · Atomic Structure · medium

The photon emitted when the electron of a He^+ ion falls from the fourth to the second energy level has the same wavelength as the photon emitted in…

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The photon emitted when the electron of a He+\mathrm{He^+} ion falls from the fourth to the second energy level has the same wavelength as the photon emitted in which transition of the hydrogen atom?
  1. A. n=3→n=1n=3 \to n=1
  2. B. n=2→n=1n=2 \to n=1
  3. C. n=3→n=2n=3 \to n=2
  4. D. n=4→n=2n=4 \to n=2
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Answer: (B) n=2→n=1n=2 \to n=1

Wave number νˉ=RZ2(1n12−1n22)\bar\nu = RZ^2\left(\dfrac1{n_1^2}-\dfrac1{n_2^2}\right). For He+\mathrm{He^+} (Z=2Z=2), 4→24\to2: νˉ=4R(14−116)=4R×316=3R4\bar\nu = 4R\left(\dfrac14-\dfrac1{16}\right) = 4R\times\dfrac3{16} = \dfrac{3R}{4}. For hydrogen, 2→12\to1: R(1−14)=3R4R\left(1-\dfrac14\right) = \dfrac{3R}{4}, identical. The other options give 8R9\tfrac{8R}{9}, 5R36\tfrac{5R}{36} and 3R16\tfrac{3R}{16}. The wavelength is about 121.5 nm121.5\ \mathrm{nm} (Lyman α\alpha).
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