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A particle (atom) has potential energy $U = \frac{a}{r^{6}} - \frac{b}{r^{12}}$ relative to the inter-atomic distance r, the effective stiffness of the molecular spring is given by $6 x {(\frac{a}{2b})}^{\frac{4}{3}}$ , then the value of x is

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answer is F.

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Detailed Solution

K = {[\frac{d^{2} u}{{dr}^{2}}]}_{r = r_{0}}

Where $r_{0}$ is equilibrium distance between atoms

$F = \frac{- du}{dr} = 0$

$\Rightarrow r_{0}^{6} = \frac{2b}{a} \Rightarrow \frac{d^{2} u}{{dr}^{2}} = \frac{- 42 a}{r^{8}} + \frac{12 \times 13 b}{r^{14}}$

$\Rightarrow K = \frac{42 a}{r_{0}^{8}} + \frac{12 \times 13 b}{r_{0}^{14}}$

$K = 36 a {(\frac{a}{2b})}^{4/3}$

Hence x = 6

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