Answer to Question #131214 in General Chemistry for Nav lubana

Question

A system with effective mass (m) has a potential energy given by ;-
u(x) = Uo (-2 (x/xo)² + (x/xo)⁴)
(b) find the points where the force on the particle is zero
(e) classify these points as stable or unstable.
(f) calculate the value of U(x)/Uo at these equilibrium points.

Answer 1

$U(x) = U_o (-2 (\frac{x}{x_o})^2 + (\frac{x}{x_o})^4)$

$(b)$ For force on the particle to be zero, $F=-\frac{dU}{dx}=0$

$-\frac{dU}{dx}=-U_o[-4(\frac{x}{x_o})(\frac{1}{x_o})+4(\frac{x}{x_o})^3(\frac{1}{x_o})]=0\implies$ $(\frac{x}{x_o})(1-(\frac{x}{x_o})^2)=0$

( As $U_o,x_o,4$ are constants)

$\implies \frac{x}{x_o}=0\implies x=0$ OR

$\implies (\frac{x}{x_o})^2=1 \implies x=+x_o$ OR $x=-x_o$ .

$(e)$ To determine stable or unstable equilibrium, we check $-\frac{d^2U}{dx^2}$ .

If it is greater than zero,it is unstable equilibrium.

If it is lesser than zero,it is stable equilibrium.

If it is zero,it is neutral equilibrium.

$-\frac{d^2U}{dx^2}=-U_o[-\frac{4}{x_o^2}+12(\frac{x}{x_o})^2(\frac{1}{x_o})^2]$

For $x=0,$ $-\frac{d^2U}{dx^2}=-U_o[-\frac{4}{x_o^2}+12(\frac{x}{x_o})^2(\frac{1}{x_o})^2]=$ $-U_o[-\frac{4}{x_o^2}+12(\frac{0}{x_o})^2(\frac{1}{x_o})^2]=\frac{4U_o}{x_o^2} >0$ .

So,it is unstable equilibrium.

For $x=+x_0,$

$-\frac{d^2U}{dx^2}=-U_o[-\frac{4}{x_o^2}+12(\frac{x}{x_o})^2(\frac{1}{x_o})^2]$ $=-U_o[-\frac{4}{x_o^2}+12(\frac{x_0}{x_o})^2(\frac{1}{x_o})^2]=-\frac{8U_O}{(x_o)^2}<0$

So,it is stable equilibrium.

For $x=-x_0,$

$-\frac{d^2U}{dx^2}=-U_o[-\frac{4}{x_o^2}+12(\frac{x}{x_o})^2(\frac{1}{x_o})^2]$ $=-U_o[-\frac{4}{x_o^2}+12(\frac{-x_0}{x_o})^2(\frac{1}{x_o})^2]=-\frac{8U_O}{(x_o)^2}<0$

So,it is stable equilibrium.

$(f)$ $U(x) = U_o (-2 (\frac{x}{x_o})^2 + (\frac{x}{x_o})^4)$

For $x=0,$

$U(x) = U_o (-2 (\frac{x}{x_o})^2 + (\frac{x}{x_o})^4)$ $=U_o (-2 (\frac{0}{x_o})^2 + (\frac{0}{x_o})^4)=0$

$\implies \frac{U}{U_o}=\frac{0}{U_o}=0$

For $x=+x_0,$

$U(x) = U_o (-2 (\frac{x}{x_o})^2 + (\frac{x}{x_o})^4)=$ $U_o (-2 (\frac{+x_o}{x_o})^2 + (\frac{+x_o}{x_o})^4)=-U_o$

$\implies \frac{U}{U_o}=\frac{-U_o}{U_o}=-1$

For $x=-x_0,$

$U(x) = U_o (-2 (\frac{x}{x_o})^2 + (\frac{x}{x_o})^4)=$ $U_o (-2 (\frac{-x_o}{x_o})^2 + (\frac{-x_o}{x_o})^4)=-U_o$

$\implies \frac{U}{U_o}=\frac{-U_o}{U_o}=-1$