Chapter 42, Problem 89CP

(a)

To determine

The allowed distance between positron and electron.

Answer to Problem 89CP

The allowed distance between positron and electron is $\underset{\_}{0.106n^2}$.

Explanation of Solution

Positron and electron are moving in a circle of radius $\frac{r}{2}$ with opposite velocities. Since the mass and velocity of both the particles are same, their angular momentum is also same.

Write the expression for the total angular momentum of the system.

    $\begin{array}{c}{L}_n=2\times \frac{mvr}{2}\\ =mvr\end{array}$                                                                                                        (I)

Here, $L_n$ is the total angular momentum of the system, $m$ is the mass of particles, $v$ is the speed, and $r$ is the distance between electron and positron.

Write the expression for the quantized angular momentum.

    $L_n=n\hslash$                                                                                                                    (II)

Here, $n$ is integer, and $\hslash$ is the reduced Planck’s constant.

Equate expressions (I) and (II) and solve for $v$.

    $\begin{array}{c}mvr=n\hslash \\ v=\frac{n\hslash }{mr}\end{array}$                                                                                                               (III)

Write the expression for the sum of force acting on the particles.

    $\sum F=ma$                                                                                                                (IV)

Write the expression for the electrostatic force acting between electron and positron.

    $\sum F=\frac{k_e{e}^2}{r^2}$                                                                                                              (V)

Here, $k_e$ is the Coulomb’s constant, $e$ is the charge of electron.

Write the expression for the centripetal acceleration of the particle while moving along a circular path of radius of $\frac{r}{2}$.

    $a=\frac{v^2}{\frac{r}{2}}$                                                                                                                    (VI)

Use expressions (VI), (V) and (III) in expression (IV) and solve to find $r$.

    $\begin{array}{c}\frac{k_e{e}^2}{r^2}=\frac{2m{v}^2}{r}\\ \frac{k_e{e}^2}{r}=2m{v}^2\\ =\frac{2m{n}^2{\hslash }^2}{m^2{r}^2}\\ r=\frac{2n^2{\hslash }^2}{m{k}_e{e}^2}\end{array}$                                                                                                     (VII)

Here, $n$, $\hslash$, $m$, $k_e$, and $e$ are constants. Take the product of all these terms to be equal to $a_0$.

    $a_0=\frac{{\hslash }^2}{m{k}_e{e}^2}$                                                                                                          (VIII)

    $\begin{array}{c}r=\frac{2n^2{\hslash }^2}{m{k}_e{e}^2}\\ =2a_0{n}^2\end{array}$                                                                                                               (IX)

Conclusion:

Substitute $1.06\times {10}^{-34}\text{J}\cdot \text{s}$ for $\hslash$, $9.1\times {10}^{-31}\text{kg}$ for $m$, $8.99\times {10}^9{\text{Nm}}^2/{\text{C}}^2$ for $k_e$, and $1.6\times {10}^{-19}\text{C}$ for $e$ in equation (VIII) to find $a_0$.

    $\begin{array}{c}{a}_0=\frac{{\left(1.05\times {10}^{-34}\text{J}\cdot \text{s}\right)}^2}{\left(9.1\times {10}^{-31}\text{kg}\right)\left(8.99\times {10}^9{\text{Nm}}^{\text{2}}{\text{/C}}^{\text{2}}\right){\left(1.602\times {10}^{-19}\text{C}\right)}^2}\\ =0.053\text{nm}\end{array}$

Substitute $0.053\text{nm}$ for $a_0$ in equation (VIII) to find $r$.

    $\begin{array}{c}r=2\left(0.053\text{nm}\right)n^2\\ =0.106n^2\end{array}$

Therefore, the allowed distance between positron and electron is $\underset{\_}{0.106n^2}$.

(b)

To determine

The allowed energies of the system.

Answer to Problem 89CP

The allowed energies of the system is $\underset{\_}{E_n=-\frac{6.80}{n^2}}$.

Explanation of Solution

Write the expression for the kinetic energy of the system.

    $K_t=K_e+K_p$                                                                                                         (X)

Here, $K_t$ is the total kinetic energy, $K_e$ is the kinetic energy of electron, and $K_p$ is the kinetic energy of the positron.

Write the expression for electron.

    $K_e=\frac{1}{2}m{v}^2$                                                                                                             (XI)

Since the mass and speed of both the particles are same, their kinetic energies are also equal.

    $K_p=\frac{1}{2}m{v}^2$                                                                                                           (XII)

Use expression (XI) and (XII) in (X) to find $K_t$.

    $\begin{array}{c}{K}_t=\frac{1}{2}m{v}^2+\frac{1}{2}m{v}^2\\ =m{v}^2\end{array}$                                                                                              (XIII)

Write the expression for the total energy of the system.

    $E=K_t+U$                                                                                                         (XIV)

Here, $E$ is the total energy, and $U$ is the electrostatic potential energy.

Write the expression for the electrostatic potential energy of the system.

    $U=-\frac{k_e{e}^2}{r}$                                                                                                        (XV)

Use expressions (XIII) and (XV) in expression (XIV) to find $E$.

    $E=m{v}^2-\frac{k_e{e}^2}{r}$                                                                                                (XVI)

Use expression (VII).

    $\begin{array}{c}\frac{k_e{e}^2}{r}=2m{v}^2\\ m{v}^2=\frac{k_e{e}^2}{2r}\end{array}$                                                                                                    (XVII)

Use expression (XVII) in (XVI) to find $E$.

    $\begin{array}{c}E=\frac{k_e{e}^2}{2r}-\frac{k_e{e}^2}{r}\\ =-\frac{k_e{e}^2}{2r}\end{array}$                                                                                                (XVIII)

Use expression (VIII) in (XVIII) to find $E$.

    $\begin{array}{c}E=-\frac{k_e{e}^2}{2\left(2a_0{n}^2\right)}\\ =-\frac{k_e{e}^2}{4a_0{n}^2}\end{array}$                                                                                                 (XIX)

Conclusion:

Substitute, $8.99\times {10}^9{\text{Nm}}^2/{\text{C}}^2$ for $k_e$, $1.6\times {10}^{-19}\text{C}$ for $e$, and $0.052\text{nm}$ for $a_0$ in equation (XIX) to find $E$.

    $\begin{array}{c}E=-\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right){\left(1.06\times {10}^{-19}\text{C}\right)}^2}{4\left(0.052\text{nm}\right)n^2}\\ =-\frac{6.80}{n^2}\end{array}$

Therefore, The allowed energies of the system is $\underset{\_}{E=-\frac{6.80}{n^2}}$.