The difference between the radii of 3rd and 4th orbitals of Li2+ is ∆R1. The difference between the radii of 3rd and 4th orbits of He+ is ∆R2. Ratio ∆R1 : ∆R2 is

r=0.529n2ZÅFor Li2+rLi2+n=4−rLi2+n=3=0.5293[16−9]=ΔR1For He+rHe+n=4−rHe+n=3=0.5292[16−9]=ΔR2∴ΔR1ΔR2=23

The difference between the radii of 3rd and 4th orbitals of Li2+ is ∆R1. The difference between the radii of 3rd and 4th orbits of He+ is ∆R2. Ratio ∆R1 : ∆R2 is

r=0.529n2ZÅFor Li2+rLi2+n=4−rLi2+n=3=0.5293[16−9]=ΔR1For He+rHe+n=4−rHe+n=3=0.5292[16−9]=ΔR2∴ΔR1ΔR2=23

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The difference between the radii of 3^rd and 4^th orbitals of Li²⁺ is $∆$ R₁. The difference between the radii of 3^rd and 4^th orbits of He⁺ is $∆$ R₂. Ratio $∆$ R₁ : $∆$ R₂ is

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3 : 2

2 : 3

3 : 8

8 : 3

answer is A.

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Detailed Solution

$r = 0 .529 \frac{n^{2}}{Z} Å$
For Li²⁺
${(r_{{Li}^{2 +}})}_{n = 4} - {(r_{{Li}^{2 +}})}_{n = 3} = \frac{0 .529}{3} [16 - 9] = {ΔR}_{1}$
For He⁺
${(r_{{He}^{+}})}_{n = 4} - {(r_{{He}^{+}})}_{n = 3} = \frac{0 .529}{2} [16 - 9] = {ΔR}_{2}$
$∴ \frac{{ΔR}_{1}}{{ΔR}_{2}} = \frac{2}{3}$