Chapter 17, Problem 27P

(a)

To determine

The electric potential at the electron’s orbit due to the proton.

Answer to Problem 27P

The electric potential at the electron’s orbit due to the proton is $27.17\text{V}$ .

Explanation of Solution

Given info:

Radius of the electron is $r=0.53\times {10}^{-10}\text{m}$ .

Charge on an electron is $e=-1.6\times {10}^{-19}\text{C}$ .

Charge on a proton is $e=1.6\times {10}^{-19}\text{C}$ .

Mass of an electron is $m_e=9.11\times {10}^{-31}\text{kg}$ .

Formula used:

The electric potential at electron is,

  $V_e=\frac{ke}{r}$

The Coulomb’s constant is $k$ and its value is $9\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^2$ .

Calculation:

Substituting the given values

  $\begin{array}{c}{V}_e=\frac{\left(9\times {10}^9\right)\left(1.6\times {10}^{-19}\right)}{0.53\times {10}^{-10}}\text{V}\\ =27.17\text{V}\end{array}$

Conclusion:

Thus, the electric potential at the electron’s orbit due to the proton is $27.17\text{V}$ .

(b)

To determine

The kinetic energy of the electron.

Answer to Problem 27P

The charge on the capacitor is $4.24\times {10}^{-11}\text{C}$ .

Explanation of Solution

Given info:

Radius of the electron is $r=0.53\times {10}^{-10}\text{m}$ .

Charge on an electron is $e=-1.6\times {10}^{-19}\text{C}$ .

Charge on an proton is $e=1.6\times {10}^{-19}\text{C}$ .

Mass of an electron is $m_e=9.11\times {10}^{-31}\text{kg}$ .

Formula used:

The centripetal force on the electron is equal to the electrostatic force of attraction so,

  $\begin{array}{c}\frac{m_e{v}^2}{r}=\frac{k{e}^2}{r^2}\\ m_e{v}^2=\frac{k{e}^2}{r}\\ \frac{1}{2}{m}_e{v}^2=\frac{1}{2}\frac{k{e}^2}{r}\\ KE=\frac{1}{2}\frac{k{e}^2}{r}\end{array}$

Calculation:

Substituting the given values, we get

  $\begin{array}{c}KE=\frac{\left(9\times {10}^9\right){\left(1.6\times {10}^{-19}\right)}^2}{2\times 0.53\times {10}^{-10}}\text{J}\\ =2.17\times {10}^{-18}\text{J}\\ =\left(2.17\times {10}^{-18}\text{J}\right)\times \left(\frac{1\text{eV}}{1.6\times {10}^{-19}\text{J}}\right)\\ =13.56\text{eV}\end{array}$

Conclusion:

Thus, the kinetic energy of the electron is $13.56\text{eV}$ .

(c)

To determine

The total energy of the electron in its orbit.

Answer to Problem 27P

The total energy of the electron is $-13.56\text{eV}$ .

Explanation of Solution

Given info:

Radius of the electron is $r=0.53\times {10}^{-10}\text{m}$ .

Charge on an electron is $e=-1.6\times {10}^{-19}\text{C}$ .

Charge on a proton is $e=1.6\times {10}^{-19}\text{C}$ .

Mass of an electron is $m_e=9.11\times {10}^{-31}\text{kg}$ .

Formula used:

The total energy of the electron is,

  $\begin{array}{c}E=KE+PE\\ =KE+e{V}_e\end{array}$

Calculation:

Substituting the given values,

  $\begin{array}{c}E=2.17\times {10}^{-18}\text{J}+\left(-1.6\times {10}^{-19}\times 27.17\right)\text{J}\\ =-2.177\times {10}^{-18}\text{J}\\ =\left(-2.177\times {10}^{-18}\text{J}\right)\times \left(\frac{1\text{eV}}{1.6\times {10}^{-19}\text{J}}\right)\\ =-13.56\text{eV}\end{array}$

The negative sign indicates it is a bound system.

Conclusion:

Thus, the total energy of the electron is $-13.56\text{eV}$ .

(d)

To determine

The ionization energy.

Answer to Problem 27P

The ionization energy is $13.56\text{eV}$ .

Explanation of Solution

Given info:

Radius of the electron is $r=0.53\times {10}^{-10}\text{m}$ .

Charge on an electron is $e=-1.6\times {10}^{-19}\text{C}$ .

Charge on an proton is $e=1.6\times {10}^{-19}\text{C}$ .

Mass of an electron is $m_e=9.11\times {10}^{-31}\text{kg}$ .

Formula used:

The ionization energy is,

  $IE=\left|E\right|$

Calculation:

Substituting the given values,

  $\begin{array}{c}IE=\left|-13.56\text{eV}\right|\\ =13.56\text{eV}\end{array}$

Here, $\left|E\right|$ is the energy that is required to remove the electron from the atom to infinity.

Conclusion:

Thus, the ionization energy is $13.56\text{eV}$ .