Question

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

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

Answer 1

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

Since $\frac{I{E}_{L{i}^{2+}}}{I{E}_{H{e}^{+}}}={\left(\frac{Z_{L{i}^{2+}}}{Z_{H{e}^{+}}}\right)}^2=\frac{9}{4}$

$\therefore I{E}_{L{i}^{2+}}=\frac{9}{4}\times I{E}_{H{e}^{+}}=9/4\times 19.6\times {10}^{-18}$

$=4.41\times {10}^{-17}J/atom$

Energy in stationary state $=-IE=-4.41\times {10}^{-17}J/atom$