Question

A classical particle confined to the positive x-axis experiences a force whose potential energy is-

$\mathbf{U}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{-}\frac{\mathbf{2}}{\mathbf{x}}\mathbf{+}\mathbf{1}$

a) By finding its minimum value and determining its behaviors at $\mathit{x}\mathbf{=}\mathbf{0}$and role="math" localid="1660119698069" $\mathit{x}\mathbf{=}\mathbf{\infty }$, sketch this potential energy.

b) Suppose the particle has energy of $\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{J}$. Find any turning points. Would the particle be bound?

c) Suppose the particle has the energy of $\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{J}$. Find any turning points. Would the particle be bound?

Short answer

(a) Sketch of potential energy is shown below.

(b) When a particle has $0.5\mathrm{J}$energy, the turning point lies at $x=2\pm \sqrt{2}$.

(c) When a particle has $2.0\mathrm{J}$energy, the turning point lies at $x=-1+\sqrt{2}$.

Step-by-step solution

Given Data

The potential energy is: $U\left(x\right)=\frac{1}{x^2}-\frac{2}{x}+1·································\left(1\right)$

Concept of Turning Points

Turning points are points where the slope of the curve becomes zero in a graph. Therefore to find the turning points we equate the first derivative of the dependent quantity with respect to the independent quantity, equal to zero.

Potential Energy

The minimum value of potential energy can be obtained by differentiating the potential energy function with respect to x and then equating it to zero.

$\frac{d\left(U\left(x\right)\right)}{dx}=0$

Using equation (1), the above equation becomes-

$$\begin{gathered} \frac{d\left(U\left(x\right)\right)}{dx}=0 \\ \frac{d\left(\frac{1}{x^2}-\frac{2}{x}+1\right)}{dx}=0 \\ \frac{-2}{x^3}+\frac{2}{x^2}=0 \\ -2x^2=-2x^3 \end{gathered}$$

On solving further

$x=1$

Thus potential energy has minima at $x=1.$

At $x=0$, the potential energy becomes-

$$\begin{gathered} U\left(x\right)=\frac{1}{x^2}-\frac{2}{x}+1 \\ =\frac{1}{0}-\frac{2}{0}+1 \\ =\infty \end{gathered}$$

At $x=\infty$, the potential energy becomes-

localid="1660121084990" $$\begin{gathered} U\left(x\right)=\frac{1}{\infty }-\frac{2}{\infty }+1 \\ =0-0+1 \\ =1 \end{gathered}$$

Based on the values of potential energy calculated above, the potential energy as a function of x is given as-

Finding the Turning point at O.5 J.

(b) if the particle has energy $0.5\mathrm{J}$, from equation (1), we get-

role="math" localid="1660118872128" $$\begin{gathered} 0.5\text{J}=\frac{1}{x^2}-\frac{2}{x}+1 \\ 1\text{J}=\frac{2}{x}-\frac{4}{x}+2 \\ {x}^2=2-4x+2x^2 \\ {x}^2-4x+2=0 \end{gathered}$$

The solution for the quadratic solution is given as-

$$\begin{gathered} x=\frac{4\pm \sqrt{4^2-4\times 2}}{2} \\ =2\pm \sqrt{2} \end{gathered}$$

Therefore, the turning points are at role="math" localid="1660119030923" $x=2\pm \sqrt{2}$. The particle is bounded.

Finding Turning point at 2 J.

(c) if the particle has energy $2.0\mathrm{J}$, from equation (1), we get

role="math" localid="1660119292924" $$\begin{gathered} 2\text{J}=\frac{1}{x^2}-\frac{2}{x}+1 \\ {x}^2+2x-1=0 \\ x=\frac{-2\pm \sqrt{2^2+4\times 1}}{2} \\ x=-1\pm \sqrt{2} \end{gathered}$$

As, x lies on the positive side of x-axis only, negative value is ignored.

$\therefore x=-1+\sqrt{2}$

Therefore, the turning point is at$x=-1+\sqrt{2}$, but the solution is unbounded.