Q. The ionisation energy of 10 times ionised sodium atom is

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A

$\frac{13.6}{11}eV$

B

$\frac{13.6}{112}eV$

C

$13.6 \, \times \, \left(11\right)^{2 \, }eV$

D

$13.6eV$

Solution:

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 $n=1$ .
$\therefore E_{n}=\frac{- 13.6 \, \times \left(\right. 11 \left.\right)^{2}}{\left(\right. 1 \left.\right)^{2}} \, eV$
= $-13.6\times \left(\right. 11 \left.\right)^{2}$ $eV$
Ionization energy = $-E_{n}=13.6\times \left(\right.11\left(\left.\right)^{2}$

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