Question

The ionisation energy of ${He}^{+}$ is $19.6\times {10}^{-18}Jperatom$. Calculate the energy of the first stationary state of ${Li}^{2+}$.

Answer 1

Solution:-

Ionization energy of ${He}^{+}=19.6\times {10}^{-18}J/atom\left(\text{Given}\right)$

As we know that,

$I.E.=E_1\times \frac{Z^2}{n^2}$

$\therefore E_1\times \frac{{\left(2\right)}^2}{{\left(1\right)}^2}=19.6\times {10}^{-18}$

$\Rightarrow E_1=\frac{19.6\times {10}^{-18}}{4}=4.9\times {10}^{-18}J$

Therefore,

$E_{{Li}^{+2}}=4.9\times {10}^{-18}\times \frac{{\left(3\right)}^2}{{\left(1\right)}^2}$

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

Hence the energy of first stationary state pf ${Li}^{+}$ is $4.41\times {10}^{-17}J$.