If the potential energy of a gas molecule is | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

Given in question,

$U=\frac{M}{r^{6}}-\frac{N}{r^{12}}$

$	herefore F=\frac{-d u}{d r}=\frac{-d}{d r}\left(\frac{M}{r^{6}}-\frac{N}{r^{12}}igh

Vedclass · Accepted Answer

$M^2/4N$

Vedclass · Answer

Given in question,

$U=\frac{M}{r^{6}}-\frac{N}{r^{12}}$

$	herefore F=\frac{-d u}{d r}=\frac{-d}{d r}\left(\frac{M}{r^{6}}-\frac{N}{r^{12}}ight)$

$=-\left(\frac{-6 M}{r^{7}}+\frac{12 N}{r^{13}}ight)=\left(\frac{6 M}{r^{7}}-\frac{12 N}{r^{13}}ight)$

For equilibrium position, Force $F=0$

$	herefore \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 $U=\frac{M^{2}}{4 N}$

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

Similar Questions

A shell of mass $m$ moving with velocity $v$ suddenly breakes into two pieces. The part having mass $\frac{m}{5}$ remains stationary. The velocity of the other part will be

When a ball is freely fallen from a given height it bounces to $80\%$ of its original height. What fraction of its mechanical energy is lost in each bounce ?

When a ball is freely fallen from a given height it bounces to $80\%$ of its original height. What fraction of its mechanical energy is lost in each bounce ?

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

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

Similar Questions

A shell of mass $m$ moving with velocity $v$ suddenly breakes into two pieces. The part having mass $\frac{m}{5}$ remains stationary. The velocity of the other part will be

When a ball is freely fallen from a given height it bounces to $80\%$ of its original height. What fraction of its mechanical energy is lost in each bounce ?

When a ball is freely fallen from a given height it bounces to $80\%$ of its original height. What fraction of its mechanical energy is lost in each bounce ?

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

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