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5.Work, Energy, Power and Collision
normal
If the potential energy of a gas molecule is
$U = \frac{M}{{{r^6}}} - \frac{N}{{{r^{12}}}}$,
$M$ and $N$ being positive constants, then the potential energy at equilibrium must be
A
Zero
B
$M^2/4N$
C
$NM^2/4$
D
$MN^2/4$
Solution
Given in question,
$U=\frac{M}{r^{6}}-\frac{N}{r^{12}}$
$\therefore F=\frac{-d u}{d r}=\frac{-d}{d r}\left(\frac{M}{r^{6}}-\frac{N}{r^{12}}\right)$
$=-\left(\frac{-6 M}{r^{7}}+\frac{12 N}{r^{13}}\right)=\left(\frac{6 M}{r^{7}}-\frac{12 N}{r^{13}}\right)$
For equilibrium position, Force $F=0$
$\therefore \frac{6 M}{r^{7}}=\frac{12 N}{r^{13}}$ or $r^{6}=\frac{2 N}{M}$
Hence, $U=\frac{M}{(2 N / M)}-\frac{N}{(2 N / M)^{2}}=\frac{M^{2}}{4 N}$
So the potential energy at equilibrium must be $U=\frac{M^{2}}{4 N}$
Standard 11
Physics
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