Q5 / 201 · NTA Abhyas (NEET) 2020 · Physics · Atomic Physics

The ionisation energy of 10 time ionised sodium atom is

A
$\frac{13.6}{11} e V$
B
$\frac{13.6}{112} e V$
Correct answer
C
$13.6 \times {(11)}^{2} e V$
D
$13.6 e V$

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

.

∴ E_{n} = \frac{- 13.6 \times {(11)}^{2}}{{(1)}^{2}} e V

=

- 13.6 \times {(11)}^{2}

eV

Hence, the correct answer is option: 'C'.

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