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For an electron, with n = 3 has only one radial node. The orbital angular momentum of the electron will be :

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\sqrt{6} \frac{h}{2 π}

\sqrt{2} \frac{h}{2 π}

3 (\frac{h}{2 π})

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detailed solution

Correct option is C

For an electron, with n = 3 has only one radial node. The orbital angular momentum of the electron will be $\sqrt{2} \frac{h}{2 π}$ .

Number of radial nodes $= n - 1 - 1 = 1$

Since, $n = 3 . ∴ 1 = 1$

Orbital angular momentum $= \sqrt{1 (1 + 1)} \frac{h}{2 π} = \sqrt{2} \frac{h}{2 π}$

Hence, the option C is correct.

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