For hydrogen-like atoms,the average distance | vedclass

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(D) The average distance of an electron from the nucleus in a hydrogen-like atom is given by the formula: $\langle r \rangle = \frac{a_0 n^2}{Z} \left

Vedclass · Accepted Answer

$4f$

Vedclass · Answer

(D) The average distance of an electron from the nucleus in a hydrogen-like atom is given by the formula: $\langle r \rangle = \frac{a_0 n^2}{Z} \left[ 1 + \frac{1}{2} \left( 1 - \frac{l(l+1)}{n^2} \right) \right]$.For a fixed orbit $n$,the average distance $\langle r \rangle$ depends on the azimuthal quantum number $l$.As $l$ increases,the term $\frac{l(l+1)}{n^2}$ increases,which makes the term in the bracket $\left[ 1 + \frac{1}{2} \left( 1 - \frac{l(l+1)}{n^2} \right) \right]$ decrease.Therefore,for a constant $n$,the average distance $\langle r \rangle$ is minimum for the orbital with the highest value of $l$.For the $4^{th}$ orbit $(n=4)$,the possible values of $l$ are $0, 1, 2, 3$ corresponding to $4s, 4p, 4d, 4f$ orbitals respectively.Since $l$ is maximum for the $4f$ orbital $(l=3)$,the average distance is minimum for $4f$.

For hydrogen-like atoms,the average distance of an electron from the nucleus for the orbitals belonging to the $4^{th}$ orbit is minimum for:

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