An electron in ground state of hydrogen has an angular momentum $L_{1}$ and an electron in the first excited state of Lithium has an angular momentum $L_{2}$ . Then

L_{1} = L_{2}

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L_{1} = 4 L_{2}

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L_{1} = \frac{L_{2}}{2}

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L_{1} = 2 L_{2}

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Solution

The correct option is C $L_{1} = \frac{L_{2}}{2}$
$L = \frac{n h}{2 π}$ Hence $L_{1} \propto n_{1}$
$L_{2} \propto n_{2}$
$\frac{L_{2}}{L_{1}} = \frac{n_{2}}{n_{1}} = \frac{2}{1}$ $L_{1} = \frac{L_{2}}{2}$

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