VITEEE 2018 on Chemistry Questions and Answers | ExamFriend.in | General questions

36.

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

$4.4\times10^{-16}Jatom^{-1}$

$-4.4\times10^{-17}Jatom^{-1}$

$-2.2\times10^{-15}Jatom^{-1}$

$8.82\times10^{-17}Jatom^{-1}$

Option B

$I.E = \frac{Z^{2}}{n^{2}}\times13.6eV$ ....(i)

$\frac{I_{1}}{I_{2}} = \frac{Z_{1}^{2}}{n_{1}^{2}}\times\frac{n_2^2}{Z_2^2}$ ...(ii)

Given I₁= -19.6 x 10^-18 Z₁ = 2, n₁= 1, Z₂= 3 and n₂ = 1

Substituting these values in equation (ii).

$-\frac{19.6\times10^{-18}}{I_{2}} = \frac{4}{1}\times\frac{1}{9}$

Or I₂ = - $19.6\times10^{-18}\times\frac{9}{4}$ = $-4.4\times10^{-17}Jatom^{-1}$

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