Discussion Forum



1)

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


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

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

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

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

Answer:

Option B

Explanation:

$I.E = \frac{Z^{2}}{n^{2}}\times13.6eV$  ....(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 = -$19.6\times10^{-18}\times\frac{9}{4}$ = $-4.4\times10^{-17}Jatom^{-1}$