Question

# Potential energy of a system of particles is U=α3r3−β2r2, where r is distance between the particles. Here α and β are positive constants. Which of the following are correct for the system?

A
Equilibrium separation between the particles is αβ
B
For r=αβ, the equilibrium is stable
C
For r=αβ, the equilibrium is unstable
D
Work required to slowly move the particles to infinite separation from initial equilibrium position is βα

Solution

## The correct options are A Equilibrium separation between the particles is αβ B For r=αβ, the equilibrium is stableGiven: U=α3r3−β2r2 For equilibrium condition, F=0 ⇒−dUdr=0 ⇒−[−3α3r4−−2β2r3]=0 ⇒αr4−βr3=0 ⇒req=αβ Now, d2Udr2=+4αr5−3βr4 At r=αβ, d2Udr2=4αβ5α5−3ββ4α4=β5α4 Since d2Udr2>0 ⟹stable equilibrium (as P.E is minimum) Work required to slowly move the particle from r=αβ to ∞ =U∞−Ur=αβ =0−−β36α2 =β36α2 ∴ Option A and B are correct.

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