Question

Calculate the energy of first stationary state of $L{i}^{2+}$ if ionisation energy of $H{e}^{+}$ is $19.6\times 10^{-18}J$ atom ${}^{-1}$

• A. $176.4\times 10^{-18}Jato{m}^{-1}$
• B. $4.9\times 10^{-18}Jato{m}^{-1}$
• C. $8.7\times 10^{-18}Jato{m}^{-1}$
• D. $44.1\times 10^{-18}Jato{m}^{-1}$

Answer 1

Answer: (D)

$E_1$ i.e. $I.E.$ of $H{e}^{+}=E_1$ of $H\times 2^2$

$E_1$ i.e. I.E. of $L{i}^{2+}=E_1$ of $H\times 3^2$

$\frac{E_1\text{ of }L{i}^{2+}}{E_1\text{ of }H{e}^{+}}=\frac{9}{4}$

$\Rightarrow E_1\text{ of }L{i}^{2+}$

$=\frac{9}{4}\times 19.6\times 10^{-18}$

$=44.1\times 10^{-18}J{\text{ atom }}^{-1}$