A) \[\frac{{{b}^{2}}}{6a}\]
B) \[\frac{{{b}^{2}}}{2a}\]
C) \[\frac{{{b}^{2}}}{12a}\]
D) \[\frac{{{b}^{2}}}{4a}\]
Correct Answer: D
Solution :
| [d] \[U=\frac{a}{{{x}^{12}}}-\frac{b}{{{x}^{6}}}\] |
| \[{{U}_{x=\infty }}=0\] |
| At equilibrium |
| \[F=0=-\frac{dU}{dx}=-12a{{x}^{-13}}+6b{{x}^{-7}}=0\] |
| \[\Rightarrow \] \[\frac{1}{{{x}^{6}}}=\frac{b}{2a}\] So \[x={{\left( \frac{2a}{b} \right)}^{\frac{1}{6}}}\] |
| \[{{U}_{eq}}=-\frac{{{b}^{2}}}{4a}\] \[D=\left[ 0-\left( -\frac{{{b}^{2}}}{4a} \right) \right]=\frac{{{b}^{2}}}{4a}\] |
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