If first ionisation energy of hydrogen is $E$ , then the ionisation energy of ${H e}^{+}$ would be:

E

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2 E

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0.5 E

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4 E

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Solution

The correct option is D $4 E$
$I E = \frac{E z^{2}}{n^{2}} = 0$
i.e for hydrogen $z = 1$
$∴ I . E = E$
For helium, $z = 2$ and for first ionization the value of n=1.
$I . E = E z^{2} = E (2)^{2} = 4 E$

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