If there are 2 nodal surfaces in the third ex | vedclass

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Question

(B) The third excited state corresponds to the principal quantum number $n = 4$.The number of radial nodes is given by the formula $n - \ell - 1 = 2$.

Vedclass · Accepted Answer

$\sqrt{2} \, \hbar$

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(B) The third excited state corresponds to the principal quantum number $n = 4$.The number of radial nodes is given by the formula $n - \ell - 1 = 2$.Substituting $n = 4$ into the equation: $4 - \ell - 1 = 2$.Solving for $\ell$: $3 - \ell = 2$,which gives $\ell = 1$.The orbital angular momentum is calculated using the formula $\sqrt{\ell(\ell + 1)} \, \hbar$.Substituting $\ell = 1$: $\sqrt{1(1 + 1)} \, \hbar = \sqrt{2} \, \hbar$.

If there are $2$ nodal surfaces in the third excited state,find the orbital angular momentum.

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