If $1^{s t}$ ionisation energy of Helium $(H e)$ atom is $25.6 e V / a t o m$ then sum of $I E_{1}$ and $I E_{2}$ per atom of Helium $(H e)$ would be:

51.2 e V

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60 e V

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80.04 e V

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85.7 e V

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Solution

The correct option is D $80.04 e V$
The first $I E$ of $H e$ is much higher than that of its' second $I E$ as the first $e^{-}$ removed is fully filled shell, i.e; $1 s^{2} .$
So, $I E_{1}$ of $H e = 25.6 e V$
Then $H e^{+}$ is isoelectronic to hydrogen atom with $1 s^{1}$ configuration.
$△ H_{E_{2}} = | E_{1} (H e^{+}) |$
$= ∣ ∣ ∣ - 2^{2} \times \frac{13.6 e V}{1^{2}} ∣ ∣ ∣$
$= 54.44 e V$
$∴ I E_{1} + I E_{2} = 25.6 + 54.4 = 80.04 e V$

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