The ionization energy of H e + ion at minimum position is

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Question: Answered & Verified by Expert

The ionization energy of

H e^{+}

ion at minimum position is

PhysicsAtomic PhysicsNEET

Options:

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

Solution:

2928 Upvotes Verified Answer

The correct answer is:

54.4 e V

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

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