The electric quadruple moment of a nuclear charge distribution which is of the shape of an ellipsoid is given by $p = \frac{2 Ze}{5} (b^{2} - a^{2})$ .
Total nuclear charge = Ze.
Average nuclear radius = $R_{0}^{3} = a^{2} b; b = R_{0} + δ R_{0}$ ; b is semi major axis and a is semi minor axis.
Mark the CORRECT statement(s) under condition $δ R_{0} < < R_{0}$ :

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$p = \frac{6 Ze}{5} R_{0}^{2} (\frac{δ R_{0}}{R_{0}})$

$b^{2} - a^{2} ≅ 3 R_{0}^{2} (\frac{δ R_{0}}{R_{0}})$

$a^{2} ≅ R_{0}^{2} (1 - \frac{δ R_{0}}{R_{0}})$

$p = \frac{2 Ze}{5} R_{0}^{2} (\frac{δ R_{0}}{R_{0}})$

answer is A, B, C.

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Detailed Solution

$\begin{array}{l} p = \frac{2 Ze}{p} (b^{2} - a^{2}) . . . . . . (1) \\ p = \frac{2 Ze}{5} (\frac{{(R_{0} + δ R_{0})}^{3} - R_{0}^{3}}{b}) \\ p = \frac{2 Ze}{5} (\frac{3 R_{0} δ R_{0} (R_{0} + δ R_{0})}{b}) \\ p = \frac{6 Ze}{5} R_{0}^{2} (\frac{δ R_{0}}{R_{0}}) . . . . . . (2) \end{array}$

By comparing (i) and (ii)

$\begin{array}{l} b^{2} - a^{2} ≅ 3 R_{0}^{2} (\frac{δ R_{0}}{R_{0}}) \\ {(R_{0} + δ R_{0})}^{2} - a^{2} = 3 R_{0} δ R_{0} \\ a^{2} = R_{0}^{2} + δ R_{0}^{2} + 2 R_{0} δ R_{0} - 3 R_{0} δ R_{0} \\ = R_{0}^{2} - R_{0} δ R_{0} \{δ R_{0}^{2} \to 0\} \\ = R_{0}^{2} (1 - \frac{δ R_{0}}{R_{0}}) \end{array}$