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Q.

The ionization potential for the electron in the ground state of the hydrogen atom is 13.6 eV atom^-1. What would be the ionization potential for the electron in the first excited state of Li²⁺ ?

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a

3.4 eV

b

10.2 eV

c

30.6 eV

d

6.8 eV

answer is C.

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Detailed Solution

$ΔE = 13 .6 Z^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) eV {atom}^{- 1}$
For the ionization of Li²⁺(z = 3) from first excited state, n₁ =2 and n₂= $\infty$
$Hence, IP = ΔE = 13 .6 \times 3^{2} \times (\frac{1}{2^{2}} - \frac{1}{\infty^{2}})$
= 30.6 eV

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The ionization potential for the electron in the ground state of the hydrogen atom is 13.6 eV atom-1. What would be the ionization potential for the electron in the first excited state of Li2+ ?