If the distance between two gas molecules bec | vedclass

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(D) London dispersion forces are inversely proportional to the sixth power of the distance $(r)$ between the particles: $F \propto \frac{1}{r^{6}}$.Le

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It becomes $64$ times

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(D) London dispersion forces are inversely proportional to the sixth power of the distance $(r)$ between the particles: $F \propto \frac{1}{r^{6}}$.Let the initial force be $x_{1} = \frac{k}{r^{6}}$.When the distance becomes half,the new distance $r' = \frac{r}{2}$.The new force $x_{2} = \frac{k}{(r/2)^{6}} = \frac{k \cdot 2^{6}}{r^{6}}$.Therefore,$x_{2} = 2^{6} \cdot x_{1} = 64 \cdot x_{1}$.The London forces become $64$ times the original value.

If the distance between two gas molecules becomes half,what change occurs in the London dispersion forces?

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