Chapter 16, Problem 86E

(a)

To determine

The ratio of distance to the radius of a hydrogen atom.

Answer to Problem 86E

The ratio of distance to the radius of a hydrogen atom is $\underset{\_}{0.72}$.

Explanation of Solution

Given that the charge of an electron is $1.6\times {10}^{-19}\text{C}$ and the radius of hydrogen atom is $5.29\times {10}^{-11}\text{C}$.

Write the expression for the dipole moment.

  $p=dq$        (I)

Here, $p$ is the dipole moment, $d$ is the distance, and $q$ is the charge.

Solve Equation (I) for $d$.

  $d=\frac{p}{q}$        (II)

Conclusion:

Substitute $6.13\times {10}^{-30}\text{C}\cdot \text{m}$ for $p$ and $1.6\times {10}^{-19}\text{C}$ for $q$ in Equation (II) to find the ratio of distance to the radius of a hydrogen atom.

  $\begin{array}{c}d=\frac{6.13\times {10}^{-30}\text{C}\cdot \text{m}}{1.6\times {10}^{-19}\text{C}}\\ =3.8\times {10}^{-11}\text{m}\end{array}$

The ratio of the distance to the radius of a hydrogen atom is,

  $\begin{array}{c}\frac{d}{R}=\frac{3.8\times {10}^{-11}\text{m}}{5.29\times {10}^{-11}\text{C}}\\ =0.72\end{array}$

Therefore, the ratio of distance to the radius of a hydrogen atom is $\underset{\_}{0.72}$.

(b)

To determine

The energy needed to flip the dipole.

Answer to Problem 86E

The energy needed to flip the dipole is $\underset{\_}{7.66\times {10}^{-5}\text{eV}}$.

Explanation of Solution

Given that the electric field is ${10}^6\text{V}\cdot {\text{m}}^{-1}$.

The energy required to flip the dipole can be measured by using $\left|-pE\cos \theta +pE\cos 180°\right|$.

  $\left|-pE\cos \theta +pE\cos 180°\right|=2pE$        (III)

Here, $E$ is the electric field.

Conclusion:

Substitute $6.13\times {10}^{-30}\text{C}\cdot \text{m}$ for $p$ and ${10}^6\text{V}\cdot {\text{m}}^{-1}$ for $E$ in Equation (III) to find the energy needed to flip the dipole.

  $\begin{array}{c}\left|-pE\cos \theta +pE\cos 180°\right|=2pE\\ =2\left(6.13\times {10}^{-30}\text{C}\cdot \text{m}\right)\left({10}^6\text{V}\cdot {\text{m}}^{-1}\right)\\ =1.22\times {10}^{-23}\text{J}\times \frac{1\text{eV}}{1.6\times {10}^{-19}\text{J}}\\ =7.66\times {10}^{-5}\text{eV}\end{array}$

Therefore, the energy needed to flip the dipole is $\underset{\_}{7.66\times {10}^{-5}\text{eV}}$.

(c)

To determine

The implication average kinetic energy of a molecule for the orientation of water molecule in the relatively strong field of ${10}^6\text{V}\cdot {\text{m}}^{-1}$.

Answer to Problem 86E

The value of the $\cos \theta <1$ since, at this field and kinetic energy orientation of water molecule is not possible.

Explanation of Solution

Given that the average kinetic energy of the molecule is $0.04\text{eV}$.

Let the angle $\theta$ the potential energy of water molecule is equal to kinetic energy.

  $0.04\text{eV}=pE\cos \theta$        (IV)

Conclusion:

Substitute $6.13\times {10}^{-30}\text{C}\cdot \text{m}$ for $p$ and ${10}^6\text{V}\cdot {\text{m}}^{-1}$ for $E$ in Equation (IV) to find the implication average kinetic energy of a molecule for the orientation of water molecule in the relatively strong field of ${10}^6\text{V}\cdot {\text{m}}^{-1}$.

  $\begin{array}{c}0.04\text{eV}=\left(6.13\times {10}^{-30}\text{C}\cdot \text{m}\right)\left({10}^6\text{V}\cdot {\text{m}}^{-1}\right)\cos \theta \\ \cos \theta =\frac{0.04\text{eV}\times 1.6\times {10}^{-19}\text{J}}{\left(6.13\times {10}^{-30}\text{C}\cdot \text{m}\right)\left({10}^6\text{V}\cdot {\text{m}}^{-1}\right)}\\ =1044\end{array}$

Therefore, the value of the $\cos \theta <1$ since, at this field and kinetic energy orientation of water molecule is not possible.