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The energy of hydrogen atom in nth orbits is $E_{n}$ , then the energy of nth orbit of single ionized helium atom will be :

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$4 E_{n}$

$E_{n} / 4$

$E_{n} / 2$

$2 E_{n}$

answer is A.

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Detailed Solution

Because energy in $n^{t h}$ orbit $E_{n} \propto Z^{2}$ for hydrogen Z = 1, and for singly ionized helium Z = 2

$\frac{E_{2}}{E_{1}} = {(\frac{Z_{2}}{Z_{1}})}^{2} = {(\frac{2}{1})}^{2} = 4$

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