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Atomic Structure

Section: Chemistry

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Question : 8 of 32

Marks: +1, -0

The difference between the radii of

3^{\text{rd}}

2^{\text{nd}}

orbit of H-atom is

x \pm

. The difference between the radii of

4^{\text{th}}

3^{\text{rd}}

\mathrm{Li}^{2+}

y \pm y: x

[AP EAPCET 22 May 2025 S2]

$15:7$
$7:15$
$3:1$
$1:3$

Solution:

The radii of

n

th orbit of hydrogen atom,

r_n = \frac{a_0 n^2}{z}

For hydrogen atom the radii of 3 rd and 2 nd obit respectively can be calculated as

r_3 = \frac{a_0 3^2}{1} = 9 a_0 \quad (z=1)

r_2 = \frac{a_0 2^2}{1} = 4 a_0

x = r_3 - r_2 = 9 a_0 - 4 a_0 = 5 a_0

For

\mathrm{Li}^{2+}

ion, the radii of 3rd and 4th orbit respectively can be calculated as

r_4 = a_0 \frac{4^2}{3} = \frac{16 a_0}{3} \quad (z=3)

r_3 = a_0 \frac{3^2}{3} = 3 a_0

y = r_4 - r_3 = \frac{7 a_0}{3}

y: x = \frac{7 a_0}{3} \times \frac{1}{5 a_0} = 7:15

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