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JEE Main 2025 · Chemistry · Atomic Structure · medium

A photon is emitted when an electron in He^+ jumps between two Bohr levels, and its wavelength equals that of the first line (n=3→ n=2) of the Balmer…

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A photon is emitted when an electron in He+\mathrm{He^{+}} jumps between two Bohr levels, and its wavelength equals that of the first line (n=3→n=2n=3\to n=2) of the Balmer series of hydrogen. The transition in He+\mathrm{He^{+}} is
  1. A. n=4→n=2n=4\to n=2
  2. B. n=5→n=3n=5\to n=3
  3. C. n=6→n=4n=6\to n=4
  4. D. n=3→n=2n=3\to n=2
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Answer: (C) n=6→n=4n=6\to n=4

For a hydrogen-like ion νˉ=RZ2(1n12−1n22)\bar\nu = RZ^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right). Hydrogen 3→23\to2: R(14−19)=5R36R\left(\frac14-\frac19\right)=\frac{5R}{36}. For He+\mathrm{He^{+}} (Z=2Z=2) we need 4(1n12−1n22)=5364\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)=\frac{5}{36}, i.e. 1n12−1n22=5144\frac{1}{n_1^2}-\frac{1}{n_2^2}=\frac{5}{144}. Trying n1=4n_1=4: 116−5144=4144=136\frac{1}{16}-\frac{5}{144}=\frac{4}{144}=\frac1{36}, so n2=6n_2=6. Other options give 34\frac34, 16225⋅4\frac{16}{225}\cdot4 and 536⋅4\frac{5}{36}\cdot4, none equal to 536\frac5{36}.
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