Chapter 43, Problem 61AP

(a)

To determine

The equilibrium position $x_0$ of the particle.

Answer to Problem 61AP

The equilibrium position $x_0$ of the particle is $0.350\text{nm}$.

Explanation of Solution

Write the expression for the potential energy of the particle field system.

  $U\left(x\right)=\frac{A}{x^3}-\frac{B}{x}$                                                                                                          (I)

Here, $U\left(x\right)$ is the potential energy of the particle field system, $x$ is the position of the particle, and $A$ and $B$ are constants.

At equilibrium distance, force on the particle become zero or potential energy reaches minimum value.

Write the condition for the equilibrium distance.

  ${\frac{dU}{dx}|}_{x=x_0}=0$                                                                                                               (II)

Here, $x_0$ is the equilibrium position of the particle.

Take derivative of equation (I) with respect to distance get equilibrium position.

  $\begin{array}{c}\frac{dU}{dx}=\frac{d\left(\frac{A}{x^3}-\frac{B}{x}\right)}{dx}\\ \frac{dU}{dx}=-\frac{3A}{x^4}+\frac{B}{x^2}\end{array}$

Conclusion:

Apply condition of equilibrium given in equation (II) to get $x_0$.

  $\begin{array}{c}{-\frac{3A}{x^4}+\frac{B}{x^2}|}_{x=x_0}=0\\ -\frac{3A}{x_0^4}+\frac{B}{x_0^2}=0\\ \frac{B}{x_0^2}=\frac{3A}{x_0^4}\\ x_0=\sqrt{\frac{3A}{B}}\end{array}$

Substitute $0.150\text{eV}\cdot {\text{nm}}^3$ for $A$ and $3.68\text{eV}\cdot \text{nm}$ for $B$ in above equation to get $U\left(x\right)$.

  $\begin{array}{c}{x}_0=\sqrt{\frac{3\left(0.150\text{eV}\cdot {\text{nm}}^3\right)}{3.68\text{eV}\cdot \text{nm}}}\\ =0.350\text{nm}\end{array}$

Therefore, the equilibrium position $x_0$ of the particle is $0.350\text{nm}$.

(b)

To determine

The depth $U_0$ of the potential well.

Answer to Problem 61AP

The depth $U_0$ of the potential well is $-7.02\text{eV}$.

Explanation of Solution

The depth of the potential well is the potential energy of the particle at equilibrium position.

Write the expression for the depth of the potential well.

  $U_0={U|}_{x=x_0}$                                                                                                             (III)

Here, $U_0$ is the depth of the potential well.

The equilibrium distance $x_0$ of the particle is $0.350\text{nm}$.

Use equation (I) in (III) to get depth of the potential well.

  $U_0=\frac{A}{x_0^3}-\frac{B}{x_0^{}}$                                                                                                          (IV)

Substitute $\sqrt{\frac{3A}{B}}$ for $x_0$ in equation (IV) to get $U_0$.

  $\begin{array}{c}{U}_0=\frac{A}{{\left(\sqrt{\frac{3A}{B}}\right)}^3}-\frac{B}{\sqrt{\frac{3A}{B}}}\\ =\frac{A{B}^{\frac{3}{2}}}{3^{\frac{3}{2}}{A}^{\frac{3}{2}}}-\frac{B{B}^{\frac{1}{2}}}{3^{\frac{1}{2}}{A}^{\frac{1}{2}}}\\ =\frac{B^{\frac{3}{2}}}{3^{\frac{3}{2}}{A}^{\frac{1}{2}}}-\frac{3\times B^{\frac{3}{2}}}{3\times 3^{\frac{1}{2}}{A}^{\frac{1}{2}}}\end{array}$

Rearrange above equation to get $U_0$.

  $U_0=\frac{-2B^{\frac{3}{2}}}{3^{\frac{3}{2}}{A}^{\frac{1}{2}}}$                                                                                                              (V)

Conclusion:

Substitute $3.68\text{eV}\cdot \text{nm}$ for $B$ and $0.150\text{eV}\cdot {\text{nm}}^3$ for $A$ in (V) to get $U_0$.

  $\begin{array}{c}{U}_0=\frac{-2{\left(3.68\text{eV}\cdot \text{nm}\right)}^{\frac{3}{2}}}{3^{\frac{3}{2}}{\left(0.150\text{eV}\cdot {\text{nm}}^3\right)}^{\frac{1}{2}}}\\ =-7.02\text{eV}\end{array}$

Therefore, the depth $U_0$ of the potential well is $-7.02\text{eV}$.

(c)

To determine

The maximum force along the negative $x$ direction that the particle experiences when it moves along the $x$ axis.

Answer to Problem 61AP

The maximum force along the negative $x$ direction that the particle experiences when it moves along the $x$ axis is $-120\text{nN}$.

Explanation of Solution

Write the expression for the force acting on the particle along $x$ direction.

  $F_x=-\frac{dU}{dx}$                                                                                                              (VI)

Here, $F_x$ is the force along $x$ direction.

Write the condition for the maximum force.

  ${\frac{d{F}_x}{dx}|}_{x=x_m}=0$                                                                                                           (VII)

Here, $x_m$ is the position of maximum force.

Conclusion:

Substitute $-\frac{3A}{x^4}+\frac{B}{x^2}$ for $\frac{dU}{dx}$ in equation (VI) to get $F_x$.

  $\begin{array}{c}{F}_x=-\left(-\frac{3A}{x^4}+\frac{B}{x^2}\right)\\ =\frac{3A}{x^4}-\frac{B}{x^2}\end{array}$                                                                                             (VIII)

Differentiate above equation to get derivative of force.

  $\begin{array}{c}\frac{d\left(\frac{3A}{x^4}-\frac{B}{x^2}\right)}{dx}=-\frac{4\times 3A}{x^5}+\frac{2\times B}{x^3}\\ =-\frac{12A}{x^5}+\frac{2B}{x^3}\end{array}$

Apply condition of maximum force given in equation (VI) in above equation to get position at which maximum force obtained.

  $\begin{array}{c}-\frac{12A}{x^5}+\frac{2B}{x^3}=0\\ \frac{2B}{x^3}=\frac{12A}{x^5}\\ x_m^2=\frac{12A}{2B}\end{array}$

Rearrange above equation to get $x_m$.

  $x_m=\sqrt{\frac{6A}{B}}$

Substitute $\sqrt{\frac{6A}{B}}$ for $x$ in equation(VIII) to get maximum force acting along $x$ direction.

  $F_{\text{max}}\frac{3A}{{\left(\sqrt{\frac{6A}{B}}\right)}^4}-\frac{B}{{\left(\sqrt{\frac{6A}{B}}\right)}^2}$

Here, $F_{\text{max}}$ is the maximum force acting on the particle.

Rearrange above equation to get $F_{\text{max}}$.

  $\begin{array}{c}{F}_{\text{max}}=\frac{3A}{{\left(\sqrt{\frac{6A}{B}}\right)}^4}-\frac{B}{{\left(\sqrt{\frac{6A}{B}}\right)}^2}\\ =\frac{3B^2}{6^{2}A}-\frac{B^2}{6A}\\ =-\frac{3B^2}{36A}\\ F_{\text{max}}=-\frac{B^2}{12A}\end{array}$

Substitute $3.68\text{eV}\cdot \text{nm}$ for $B$ and $0.150\text{eV}\cdot {\text{nm}}^3$ for $A$ in above equation to get $F_{\text{max}}$.

  $\begin{array}{c}{F}_{\text{max}}=-\frac{{\left(3.68\text{eV}\cdot \text{nm}\right)}^2}{12\left(0.150\text{eV}\cdot {\text{nm}}^3\right)}\\ =-7.52\text{eV}/\text{nm}\times \left(\frac{1.60\times {10}^{-19}\text{J}}{1\text{eV}}\right)\left(\frac{1\text{nm}}{{10}^{-9}\text{m}}\right)\\ =-1.20\times {10}^{-9}\text{N}\times \frac{1\text{nN}}{{10}^{-9}\text{N}}\\ =-1.20\text{nN}\end{array}$

Since the force is along $x$ direction it can be written as $\overrightarrow{\text{F}}=-1.20\widehat{\text{i}}\text{nN}$, where $\overrightarrow{\text{F}}$ is the maximum force on the particle.

Therefore, the maximum force along the negative $x$ direction that the particle experiences when it moves along the $x$ axis is $-120\text{nN}$.