Chapter 17, Problem 126P

(a)

To determine

The electric forces on the electron and on the proton.

Answer to Problem 126P

The electric force on the electron is $82.4\text{ nN radially inward toward the proton}$ and that on proton is $82.4\text{ nN toward the electron}$ .

Explanation of Solution

Write the equation for the magnitude of the electric force using Coulomb’s law.

  $F=\frac{k\left|q_1\right|\left|q_2\right|}{r^2}$        (I)

Here, $F$ is the magnitude of the electric force, $\left|q_1\right|$ is the magnitude of the charge of the electron, $\left|q_2\right|$ is the magnitude of the charge of the and $r$ is the distance between the charges.

Conclusion:

The value of $k$ is $8.988\times {10}^9\text{ N}\cdot {\text{m}}^2$ and the magnitude of charge of both electron and proton is $1.602\times {10}^{-19}\text{ C}$ and have opposite signs.

Substitute $8.988\times {10}^9\text{ N}\cdot {\text{m}}^2$ for $k$ , $1.602\times {10}^{-19}\text{ C}$ for $\left|q_1\right|$ , $\left|q_2\right|$ and $0.0529\text{ nm}$ for $r$ in equation (I) to find $F$ .

  $\begin{array}{c}F=\frac{\left(8.988\times {10}^9\text{ N}\cdot {\text{m}}^2\right)\left(1.602\times {10}^{-19}\text{ C}\right)\left(1.602\times {10}^{-19}\text{ C}\right)}{{\left(0.0529\text{ nm}\frac{1\text{ m}}{{10}^9\text{ nm}}\right)}^2}\\ =82.4\times {10}^{-9}\text{ N}\\ =82.4\times {10}^{-9}\text{ N}\frac{1\text{ nN}}{{10}^{-9}\text{ N}}\\ =82.4\text{ nN}\end{array}$

The force on electron will be toward the proton and the force on proton will be toward the electron.

Therefore, the electric force on the electron is $82.4\text{ nN radially inward toward the proton}$ and that on proton is $82.4\text{ nN toward the electron}$ .

(b)

To determine

The electron’s acceleration and speed.

Answer to Problem 126P

The electron’s acceleration is $9.05\times {10}^{22}{\text{ m/s}}^2\text{ radially inward}$ and speed is $2.19\times {10}^6\text{ m/s}$ .

Explanation of Solution

The electric force is the net force acting on the electron.

Write the equation for the net force.

  ${\overrightarrow{F}}_{\text{net}}=m\overrightarrow{a}$        (II)

Here, ${\overrightarrow{F}}_{\text{net}}$ is the net force, $m$ is the mass of the electron and $\overrightarrow{a}$ is the acceleration.

Rewrite the above equation for $\overrightarrow{a}$ .

  $\overrightarrow{a}=\frac{{\overrightarrow{F}}_{\text{net}}}{m}$        (III)

The net force provides the centripetal force for the motion of the electron.

Write the equation for the centripetal force on the electron.

  ${\overrightarrow{F}}_{\text{net}}=m\frac{v^2}{r}$        (IV)

Here, $v$ is the speed of the electron.

Equate equations (II) and (IV) and rewrite it for $v$ .

  $\begin{array}{c}m\overrightarrow{a}=m\frac{v^2}{r}\\ a=\frac{v^2}{r}\\ v^2=ar\\ v=\sqrt{ar}\end{array}$        (V)

Conclusion:

The mass of electron is $9.109\times {10}^{-31}\text{ kg}$ .

Substitute $82.4\times {10}^{-9}\text{ N radially inward}$ for ${\overrightarrow{F}}_{\text{net}}$ and $9.109\times {10}^{-31}\text{ kg}$ for $m$ in equation (III) to find $\overrightarrow{a}$ .

  $\begin{array}{c}\overrightarrow{a}=\frac{82.4\times {10}^{-9}\text{ N radially inward}}{9.109\times {10}^{-31}\text{ kg}}\\ =9.05\times {10}^{22}{\text{ m/s}}^2\text{ radially inward}\end{array}$

Substitute $9.05\times {10}^{22}{\text{ m/s}}^2$ for $a$ and $0.0529\text{ nm}$ for $r$ in equation to find $v$ .

  $\begin{array}{c}v=\sqrt{\left(9.05\times {10}^{22}{\text{ m/s}}^2\right)\left(0.0529\text{ nm}\frac{1\text{ m}}{{10}^9\text{ nm}}\right)}\\ =2.188\times {10}^6\text{ m/s}\\ =2.19\times {10}^6\text{ m/s}\end{array}$

Therefore, the electron’s acceleration is $9.05\times {10}^{22}{\text{ m/s}}^2\text{ radially inward}$ and speed is $2.19\times {10}^6\text{ m/s}$ .

(c)

To determine

The minimum amount of energy required to ionize the atom if it stars in the ground state.

Answer to Problem 126P

The minimum amount of energy required to ionize the atom if it stars in the ground state is $2.18\times {10}^{-18}\text{ J}$ .

Explanation of Solution

The minimum energy required to ionize the atom will be equal to the total energy of the atom. The total energy of the atom is the sum of the kinetic energy of the electron and the electric potential energy of the atom.

Write the equation for the total energy of atom.

  $E=K+U$        (VI)

Here, $E$ is the total energy of the atom, $K$ is the kinetic energy of the electron and $U$ is the electric potential energy.

Write the equation for $K$ .

  $K=\frac{1}{2}m{v}^2$        (VII)

Write the equation for $U$ .

  $U=\frac{k{q}_1{q}_2}{r}$        (VIII)

Conclusion:

Substitute $9.109\times {10}^{-31}\text{ kg}$ for $m$ and $2.188\times {10}^6\text{ m/s}$ for $v$ in equation (VII) to find $K$ .

  $\begin{array}{c}K=\frac{1}{2}\left(9.109\times {10}^{-31}\text{ kg}\right){\left(2.188\times {10}^6\text{ m/s}\right)}^2\\ =2.180\times {10}^{-18}\text{ J}\end{array}$

Substitute $8.988\times {10}^9\text{ N}\cdot {\text{m}}^2$ for $k$ , $-1.602\times {10}^{-19}\text{ C}$ for $q_1$ , $1.602\times {10}^{-19}\text{ C}$ for $q_2$ and $0.0529\text{ nm}$ for $r$ in equation (VIII) to find $U$ .

  $\begin{array}{c}U=\frac{\left(8.988\times {10}^9\text{ N}\cdot {\text{m}}^2\right)\left(-1.602\times {10}^{-19}\text{ C}\right)\left(1.602\times {10}^{-19}\text{ C}\right)}{\left(0.0529\text{ nm}\frac{1\text{ m}}{{10}^9\text{ nm}}\right)}\\ =-4.360\times {10}^{-18}\text{ J}\end{array}$

Substitute $2.180\times {10}^{-18}\text{ J}$ for $K$ and $-4.360\times {10}^{-18}\text{ J}$ for $U$ in equation (VI) to find $E$ .

  $\begin{array}{c}E=2.180\times {10}^{-18}\text{ J}+\left(-4.360\times {10}^{-18}\text{ J}\right)\\ =2.18\times {10}^{-18}\text{ J}\end{array}$

Therefore, the minimum amount of energy required to ionize the atom if it stars in the ground state is $2.18\times {10}^{-18}\text{ J}$ .