Question

Ionisation energy of $H{e}^{+}$ is $8.71\times {10}^{-18}Jato{m}^{-1}$. The energy of the first stationary state (n = 1) of $L{i}^{2+}$ is:

• A. $-4.41\times {10}^{-16}Jato{m}^{-1}$
• B. $-1.96\times {10}^{-17}Jato{m}^{-1}$
• C. $-2.76\times {10}^{-15}Jato{m}^{-1}$
• D. $6.72\times {10}^{-17}Jato{m}^{-1}$

Answer 1

The correct option is B $-1.96\times {10}^{-17}Jato{m}^{-1}$

Given,
$\text{Ionisation energy of}H{e}^{+}=19.6\times {10}^{-18}Jato{m}^{-1}$

We know,
$E_n\propto \frac{Z^2}{n^2}$
Here, for both the cases n = 1.
Hence, $E_n\propto Z^2$
$\therefore \frac{E_{H{e}^{+}}}{E_{L{i}^{2+}}}=\frac{2^2}{3^2}$
$\Rightarrow \frac{-8.71\times {10}^{-18}}{E_{L{i}^{2+}}}=\frac{4}{9}$
$\Rightarrow E_{L{i}^{2+}}=-\frac{9}{4}[8.71\times {10}^{-18}]$
$=-1.96\times {10}^{-17}J/atom$