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

Answer 1

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

$\frac{I_{{He}^{+}}}{I_{{Li}^{2+}}}=\frac{Z_1^2}{Z_2^2}$

$\frac{19.6\times {10}^{-18}}{I_{{Li}^{2+}}}=\frac{4}{9}$

$I_{{Li}^{2+}}=\frac{9}{4}\times 19.6\times {10}^{-18}=44.1\times {10}^{-18}$
$=4.41\times {10}^{-17}J{atom}^{-1}$

$E_{{Li}^{2+}}=-4.41\times {10}^{-17}J{atom}^{-1}$