Q33 / 201 · NTA Abhyas (NEET) 2020 · Physics · Atomic Physics

The ionization energy of $H e^{+}$ ion at minimum position is

A
$13.6 e V$
B
$27.2 e V$
Correct answer
C
$54.4 e V$
D
$68.0 e V$

Solution

Energy of helium ions.

E_{n} = -

\frac{13.6 Z^{2}}{n^{2}} e V

In minimum position,

n

=1

For

H e^{+}, Z = 2

E = \frac{- 13.6 {\times (2)}^{2}}{1} e V

E = 54.4 e V

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

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