r¯, average distance, also described as expectation value of the distance of the electron from the nucleus, is different from rmax. For ls orbital of H-atom; rmax=a0,r¯=32a0. For 2s orbital of H-atom rmax=0.77a0 and 5.23a0.Find r¯1 for rmax=5.23a0. Also find r¯2for ls orbital of Li2+ ion . Hence, find the value of r¯1×r¯2a02.

r¯=a02Z3n2−l(l+1)r¯1=a02×13×22−0=6a0r¯2=a02×33×12−0=a02r¯1×r¯2=3a02⇒r¯1×r¯2a02=3

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$\bar{r},$ average distance, also described as expectation value of the distance of the electron from the nucleus, is different from $r_{max}$ . For ls orbital of H-atom; $r_{max} = a_{0}, \bar{r} = \frac{3}{2} a_{0}$ . For 2s orbital of H-atom $r_{max} = 0.77 a_{0}$ and $5.23 a_{0} .$ Find ${\bar{r}}_{1}$ for $r_{max} = 5.23 a_{0} .$ Also find ${\bar{r}}_{2}$ for ls orbital of ${Li}^{2 +} ion$ . Hence, find the value of $\frac{({\bar{r}}_{1} \times {\bar{r}}_{2})}{a_{0}^{2}}$ .

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answer is 3.

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Detailed Solution

$\begin{array}{l} \bar{r} = \frac{a_{0}}{2 Z} \{3 n^{2} - l (l + 1)\} \\ {\bar{r}}_{1} = \frac{a_{0}}{2 \times 1} \{3 \times 2^{2} - 0\} = 6 a_{0} \\ {\bar{r}}_{2} = \frac{a_{0}}{2 \times 3} \{3 \times 1^{2} - 0\} = \frac{a_{0}}{2} \\ {\bar{r}}_{1} \times {\bar{r}}_{2} = 3 a_{0}^{2} \Rightarrow \frac{{\bar{r}}_{1} \times {\bar{r}}_{2}}{a_{0}^{2}} = 3 \end{array}$