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Q.

Ionisation energy of He⁺ is 19.6 x 10^-18 J atom^-1. The energy of the first station ary state (n = 1) of Li²⁺ is:

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a

8.82 x 10^-17 J atom^-1

b

4.41 x 10^-16 J atom^-1

c

-4.41 x 10^-17 J atom^-1

d

-2.2 x 10^-15 J atom^-1

answer is C.

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Detailed Solution

$\begin{array}{l} E_{{He}^{+}} = E_{1} (H) \times Z_{1}^{2} \\ E_{{Li}^{2 +}} = E_{1} (H) \times Z_{2}^{2} \\ ∴ E_{{Li}^{2 +}} = E_{{He}^{+}} \times \frac{Z_{2}^{2}}{Z_{1}^{2}} \\ = \frac{19 .6 \times 10^{- 18} \times 3^{2}}{2^{2}} \\ = 4 .41 \times 10^{- 17} J {atom}^{- 1} \end{array}$
Energy in orbit = $- E_{{Li}^{2 +}}$

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