Energy of singly ionized helium atom in an excited state is equal to the energy of a hydrogen atom in first excited state. Then angular momentum of the electron of the helium ion is

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$\frac{3 h}{π}$

$\frac{h}{2 π}$

$\frac{4 h}{π}$

$\frac{2 h}{π}$

answer is B.

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Detailed Solution

(2)

Energy of an hydrogen - like ion is given by

$E_{n} = - \frac{13.6 Z^{2}}{n^{2}} e V .$

$\begin{array}{l} ∴ - \frac{13.6 \times 2^{2}}{n^{2}} = - \frac{13.6 \times 1^{2}}{2^{2}} \\ \Rightarrow n = 4 \\ ∴ L = 4 \times \frac{h}{2 π} = \frac{2 h}{π} . \end{array}$

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