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The energy of an electron in an excited $L i^{+ 2}$ ion is $- 13.6 e V$ . Hence angular momentum of the electron is given by (in $J s$ )

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$4.2 \times 10^{- 34}$

$2.1 \times 10^{- 34}$

$1.05 \times 10^{- 34}$

$3.15 \times 10^{- 34}$

answer is B.

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Detailed Solution

$E n = - 13.6 \times \frac{z^{2}}{n^{2}} \Rightarrow n = 3$

Angular momentum $= \frac{n h}{2 π} = \frac{3 \times 6.62 \times 10^{- 34}}{2 \times 3.14} = 3.15 \times 10^{- 34}$

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