Question:

The difference between the radii of \(3^{rd}\) and \(4^{th}\) orbits of \(\mathrm{Li^{2+}}\) is \(\Delta R_1\). The difference between the radii of \(3^{rd}\) and \(4^{th}\) orbits of \(\mathrm{He^{+}}\) is \(\Delta R_2\). Ratio \(\Delta R_1 : \Delta R_2\) is :

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Bohr radius \(r_n = 0.529\,n^2/Z\) angstrom, so the difference \(r_4 - r_3\) is proportional to \(1/Z\).
Updated On: Nov 5, 2026
  • 8 : 3
  • 3 : 2
  • 2 : 3
  • 3 : 8
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The Correct Option is C

Solution and Explanation

\(r_n = 0.529 \times \dfrac{n^2}{Z}\) angstrom.
For \(\mathrm{Li^{2+}}\) (\(Z=3\)): \(\Delta R_1 = 0.529\,\dfrac{16-9}{3} = 0.529 \times \dfrac{7}{3}\).
For \(\mathrm{He^{+}}\) (\(Z=2\)): \(\Delta R_2 = 0.529\,\dfrac{16-9}{2} = 0.529 \times \dfrac{7}{2}\).
\(\Delta R_1 : \Delta R_2 = \dfrac{7}{3} : \dfrac{7}{2} = 2 : 3\).
Answer: (3) 2 : 3
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