Question

Ionisation energy of $He^+$ is $19.6 \times 10^{-18}\, J$ atom$^{-1}$. The energy of the first stationary state $(n= 1)$ of $Li^{2+}$ is

• A. $4.41 \times 10^{-16}\,J\,atom^{-1}$
• B. $-4.41 \times 10^{-17}\,J\,atom^{-1}$
• C. $-2.2 \times 10^{-15}\,J\,atom^{-1}$
• D. $8.82 \times 10^{-17}\,J\,atom^{-1}$

Answer 1

Answer: (B)

$I.E= \frac{Z^{2}}{n^{2}}\times13.6 eV...\left(i\right)$
or $\frac{I_{1}}{I_{2}} = \frac{Z^{2}_{1}}{n^{2}_{1}}\times\frac{n^{2}_{2}}{Z^{2}_{2}} ...\left(ii\right)$
Given $I_{1} = -19.6 \times 10^{-18}, Z_{1} = 2,$
$n_{1} = 1, Z_{2 }= 3$ and $n_{2} = 1$
Substituting these values in equation $\left(ii\right)$.
$-\frac{19.6\times10^{-18}}{I_{2}} = \frac{4}{1}\times\frac{1}{9}$
or $ I_{2}=19.6\times 10^{-18}\times \frac{9}{4}$
$= -4.41 \times 10^{-17}\,J/atom$