Question

The ionisation energy of the ionised sodium atom $N{a}^{+10}$ is:

• A. $13.6eV$
• B. $13.6\times 10eV$
• C. $\left(13.6/11\right)eV$
• D. $13.6\times \left({11}^2\right)eV$

Answer 1

The correct option is D $13.6\times \left({11}^2\right)eV$

$\begin{array}{c}\text{We know that energy for exitation of}{e}^{-}\\ \text{is given by -}\end{array}$

$E=13.6\left(z^2\right)\times [\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

$\text{For ionisation of atom}{Na}^{+10}(z=11)$

$\begin{array}{c}\text{we have,}\\ n_1=1,n_2\to \infty \\ \Rightarrow E=13.6\times (11{)}^2\times [\frac{1}{1}-\frac{1}{\infty }]eV\\ \Rightarrow E=13.6\times (11{)}^{2}eV\\ \text{Hence, option (d) is correct.}\end{array}$