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Q.

If $E_{1}$ is the energy of an electron in first orbit of ${He}^{+}$ , the energy $E_{2}$ of an electron in nth orbit of hydrogen atom will be

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a

$E_{1} / 2 n$

b

$E_{1} / 2 n^{2}$

c

$E_{1} / 4 n^{2}$

d

$E_{1} / 8 n^{2}$

answer is C.

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Detailed Solution

Since $E_{n} \propto - Z^{2} / n^{2}$ ,

we have $E_{1} ({He}^{+}) \propto - \frac{2^{2}}{1^{2}} = - 4$

and $E_{n} (H) = - \frac{1^{2}}{n^{2}} = - \frac{1}{n^{2}}$ .

Hence $E_{n} (H) = \frac{E_{1} / 4}{n^{2}} = \frac{E_{1}}{4 n^{2}}$

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