Physics - Modern Physics Question with Solution | TestHub

The energy of$n$th orbit of a hydrogen-like atom is,

$E_n = -13.6 \frac{Z^2}{n^2}$

Here,$Z=11$for Na atom. 10 electrons are removed already. For the last electron to be removed$n=1$.

$\therefore E_n=\frac{-13.6 \times {\left(11\right)}^2}{{\left(1\right)}^2} \mathrm{e}\mathrm{V}$

=$-13.6 \times {\left(11\right)}^2$$\mathrm{eV}$