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The orbital angular momentum of an electron is . Which of the following may be the permissible valueAtomic Structure Chemistry Question

Question

The orbital angular momentum of an electron is $\sqrt{3} \frac{h}{\pi}$. Which of the following may be the permissible value of angular momentum of this electron revolving in unknown Bohr orbit?

Answer: D

💡 Solution & Explanation

$L_{orbital} = \sqrt{l(l+1)}\frac{h}{2\pi} = \sqrt{3}\frac{h}{\pi} = \sqrt{12}\frac{h}{2\pi} \implies l(l+1) = 12 \implies l=3$. The principal quantum number $n$ must be strictly greater than $l$, so $n \ge 4$. The Bohr angular momentum is $L = \frac{nh}{2\pi}$. For minimum valid $n=4$, $L = \frac{4h}{2\pi} = \frac{2h}{\pi}$, which matches option (d). Therefore, correct answer is D.

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