Question

Ionization 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

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

$IE$ of ${He}^{+}$=$19.6\times {10}^{-18}J$ per atom

$IE=E_1\left(forH\right)\times Z^2$

$E_1\times 4=-19.6\times {10}^{-18}J$

$E_1\left(for{Li}^{+2}\right)=E_1\left(forH\right)\times 9$

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

=$-441\times {10}^{-18}$

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