Question

Ionisation energy of $H{e}^{+}$ is 19.6 $\times {10}^{-18}Jato{m}^{-1}$. The energy of the first stationary state $(n=1)$ of $L{i}^{2+}$ is:

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

Answer 1

The correct option is B $-4.41\times {10}^{-17}Jato{m}^{-1}$

$I{E}_{H{e}^{+}}=13.6\times Z_{H{e}^{+}}^2(\frac{1}{1^2}-\frac{1}{{\infty }^2})$ where $Z=2$
Hence, $13.6\times Z_{H{e}^{+}}^2=19.6\times {10}^{-18}Jato{m}^{-1}$
$E_{1\left(L{i}^{2+}\right)}=-13.6Z_{L{i}^{2+}}^2\left(\frac{1}{1^2}\right)=-I{E}_{H{e}^{+}}\times \frac{9}{4}$

$E_{1\left(L{i}^{2+}\right)}=-4.41\times {10}^{-17}J/atom$