Try rubbing a plastic pen through your hair, and you’ll find that you can pick up a tiny scrap of paper when the pen is about one centimeter above the paper. From this simple experiment you can estimate how much an atom in the paper is polarized by the pen! You will need to make several assumptions and approximations. Hints may be found at the end of the chapter. (a) Suppose that the center of the outer electron cloud ($\mathit{q}\mathbf{=}\mathbf{-}\mathbf{4}\mathit{e}$) of a carbon atom shifts a distance s when the atom is polarized by the pen. Calculate s algebraically in terms of the charge Q on the pen. (b) Assume that the pen carries about as much charge Q as we typically find on a piece of charged invisible tape. Evaluate s numerically. How does this compare with the size of an atom or a nucleus? (c) Calculate the polarizability $\mathit{\alpha }$of a carbon atom. Compare your answer to the measured value of $\mathbf{1}\mathbf{.}\mathbf{96}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{40}}\mathbf{}\mathit{C}\mathbf{.}\mathit{m}\mathbf{/}\mathbf{(}\mathbf{N}\mathbf{/}\mathbf{C}\mathbf{)}$(T. M. Miller and B. Bederson, “Atomic and molecular polarizabilities: a review of recent advances,” Advances in Atomic and Molecular Physics, 13, 1–55, 1977).(d) Carefully list all assumptions and approximations.

The given data is listed below as:

• The distance of the pen from the paper scrap is, $\mathit{d}\mathbf{=}\mathbf{1}\mathbf{}\mathit{c}\mathit{m}$.

• The charge of the center of the cloud’s outer electron is, $\mathit{q}\mathbf{=}\mathbf{-}\mathbf{4}\mathit{e}$.

The electromagnetism is referred to as the interaction of the fields or the electric currents and also the magnetic fields. Moreover, the electromagnetism is helpful for calculating both the electricity and the magnetism in order to find the electromagnetism of an object.

The equation of the magnitude of the electric field of the carbon atom is expressed as:

$\mathit{E}\mathbf{=}\left(\frac{1}{4\pi {\epsilon }_0}\right)\mathbf{·}\left(\frac{2p}{d^3}\right)$

Here, $\mathbf{\left(}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathbf{\right)}$is the electric field constant, p is the dipole and d is the distance between the outer electron to the carbon atom.

The above equation can be expressed as:

$\mathit{E}\mathbf{=}\mathbf{\left(}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathbf{\right)}\mathbf{·}\mathbf{\left(}\frac{\mathbf{2}\mathbf{qs}}{{\mathbf{d}}^{\mathbf{3}}}\mathbf{\right)}$

Here, q is the charge of the carbon atom and s is the distance moved by the atom when it is polarized by a plastic pen.

Substitute the values in the above equation.

$\mathit{E}\mathbf{=}\mathbf{\left(}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathbf{\right)}\mathbf{·}\mathbf{\left(}\frac{\mathbf{2}\mathbf{\left(}\mathbf{4}\mathbf{e}\mathbf{\right)}}{{\mathbf{d}}^{\mathbf{3}}}\mathbf{\right)}$

The equation of the force exerted on the carbon atom by the pen is expressed as:

$\mathit{F}\mathbf{=}\mathit{Q}\mathit{E}$

Here, Q is the charge of the pen and E is the magnitude of the electric field of the carbon atom.

Substitute the values in the above equation.

$\mathit{F}\mathbf{=}\mathbf{-}\mathit{Q}\mathbf{\left(}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathbf{\right)}\mathbf{·}\mathbf{\left(}\frac{\mathbf{2}\mathbf{\left(}\mathbf{4}\mathbf{e}\mathbf{\right)}\mathbf{s}}{{\mathbf{d}}^{\mathbf{3}}}\mathbf{\right)}$

As the piece of the paper is being moved due to the charge of the pen, then the equation of the force due to gravity is expressed as:

${\mathit{F}}_{\mathbf{1}}\mathbf{=}\mathit{m}\mathit{g}$

Here, m is the mass of the carbon atom and g is the acceleration due to gravity.

For maintaining equilibrium, the sum of the force due to gravity and the force exerted on the carbon atom by the pen must be equal to zero. Hence, the equation can be expressed as:

${\mathit{F}}_{\mathbf{1}}\mathbf{+}\mathit{F}\mathbf{=}\mathbf{0}$

Here, F is the force exerted on the carbon atom by the pen and ${\mathit{F}}_{\mathbf{1}}$is the force due to gravity.

Substitute the values in the above equation.

$$\begin{gathered} \mathit{m}\mathit{g}\mathbf{-}\mathit{Q}\mathbf{\left(}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathbf{\right)}\mathbf{·}\mathbf{\left(}\frac{\mathbf{2}\mathbf{\left(}\mathbf{4}\mathbf{e}\mathbf{\right)}\mathbf{s}}{{\mathbf{d}}^{\mathbf{3}}}\mathbf{\right)}\mathbf{=}\mathbf{0} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{m}\mathit{g}\mathbf{=}\mathbf{\left(}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathbf{\right)}\mathbf{·}\mathbf{\left(}\frac{\mathbf{2}\mathbf{\left(}\mathbf{4}\mathbf{e}\mathbf{\right)}\mathbf{s}}{{\mathbf{d}}^{\mathbf{3}}}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{2}\mathbf{\left(}\mathbf{4}\mathbf{e}\mathbf{\right)}\mathit{s}\mathbf{=}\frac{\mathbf{mg}\mathbf{\left(}\mathbf{4}{\mathbf{\epsilon }}_{\mathbf{0}}\mathbf{\right)}{\mathbf{d}}^{\mathbf{3}}}{\mathbf{Q}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{s}\mathbf{=}\frac{\mathbf{mg}\mathbf{\left(}{\mathbf{\pi \epsilon }}_{\mathbf{0}}\mathbf{\right)}{\mathbf{d}}^{\mathbf{3}}}{\mathbf{2}\mathbf{eQ}} \end{gathered}$$ …(i)

Thus, the value of s algebraically in terms of the charge Q on the pen is $\frac{\mathbf{mg}\mathbf{\left(}{\mathbf{\pi \epsilon }}_{\mathbf{0}}\mathbf{\right)}{\mathbf{d}}^{\mathbf{3}}}{\mathbf{2}\mathbf{eQ}}$.

From the equation (i), the equation of the distance moved by the pen while being polarized by the plastic pen is expressed as:

$\mathit{s}\mathbf{=}\frac{\mathbf{mg}\mathbf{\left(}{\mathbf{\pi \epsilon }}_{\mathbf{0}}\mathbf{\right)}{\mathbf{d}}^{\mathbf{3}}}{\mathbf{2}\mathbf{eQ}}$

Here, m is the mass of one carbon atom, gis the acceleration due to gravity, ${\mathit{\epsilon }}_{\mathbf{0}}$ is the permittivity, d is the distance between the outer electron to the carbon atom, eis the charge of an electron and Q is the charge of the pen.

It can be observed that the distance between the outer electron to the carbon atom is being taken as the Bohr’s radius. Moreover, the charge of the pen is the charge of the tape that is the charge of an electron.

Let us assume that the mass of one carbon atom that is $\mathbf{1}\mathbf{.}\mathbf{99}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}\mathbf{}\mathit{g}$and the value of permittivity is $\mathbf{8}\mathbf{.}\mathbf{59}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{}{\mathit{C}}^{\mathbf{2}}\mathbf{·}{\mathit{m}}^{\mathbf{-}\mathbf{2}}\mathbf{·}{\mathit{g}}^{\mathbf{-}\mathbf{1}}$. Another assumption is that the charge of one electron is $\mathbf{1}\mathbf{.}\mathbf{602}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathbf{}\mathit{C}$which is the charge of the pen. The final assumption is that the distance between the outer electron to the carbon atom is being taken as the Bohr’s radius that is $\mathbf{(}\mathbf{5}\mathbf{.}\mathbf{29}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{11}}\mathbf{}\mathbf{m}\mathbf{)}$.

Substitute the values in the above equation.

$$\begin{gathered} \mathit{s}\mathbf{=}\frac{\left(1.99\times {10}^{-23}g\right)\left(3.14\times 8.59\times {10}^{-7}{C}^2·m^{-2}·g^{-1}\right){\left(5.29\times {10}^{-11}m\right)}^{\mathbf{3}}}{\mathbf{2}\left(1.602\times {10}^{-19}C\right)\left(1.602\times {10}^{-19}C\right)} \\ \mathbf{}\mathbf{}\mathbf{=}\frac{\left(1.99\times {10}^{-23}g\right)\left(2.69\times {10}^{-6}{C}^2·m^{-2}·g^{-1}\right)\left(1.48\times {10}^{-31}{m}^3\right)}{\mathbf{2}\left(2.58\times {10}^{-38}{C}^2\right)} \\ \mathbf{}\mathbf{}\mathbf{=}\frac{\left(7.92\times {10}^{-60}{C}^2·m\right)}{\left(5.17\times {10}^{-38}{C}^2\right)} \\ \mathbf{}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{53}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{22}}\mathbf{}\mathit{m} \end{gathered}$$

Thus, the numerical value of s is $\mathbf{1}\mathbf{.}\mathbf{53}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{22}}\mathbf{}\mathit{m}$.

The size of an atom is $\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathbf{}\mathit{m}$. The equation of the difference with the distance moved by the pen while being polarized by the plastic pen and the size of an atom is expressed as:

a = b - s

Here, a is the difference with the distance moved by the pen while being polarized by the plastic pen and the size of an atom, bis the size of an atom and s is the distance moved by the pen while being polarized by the plastic pen.

Substitute the values in the above equation.

$$\begin{gathered} \mathit{a}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathbf{}\mathit{m}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{53}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{22}}\mathbf{}\mathit{m} \\ \mathbf{}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathbf{}\mathit{m} \end{gathered}$$

Thus, the size of an atom is bigger by a distance of $\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathbf{}\mathit{m}$than the distance moved by the pen while being polarized by the plastic pen.

The equation of the electric field for the plastic pen is expressed as:

$\mathit{E}\mathbf{=}\mathit{k}\frac{\mathbf{Q}}{{\mathbf{d}}^{\mathbf{2}}}$

Here, k is the electric field constant, Qis the charge of the pen and d is the distance between the outer electron to the carbon atom.

The equation of the polarizability of the carbon atom is expressed as:

$$\begin{gathered} \mathit{q}\mathit{s}\mathbf{=}\mathit{\alpha }\mathit{E} \\ \mathbf{}\mathbf{}\mathit{\alpha }\mathbf{=}\frac{\mathbf{q}\mathbf{s}}{\mathbf{E}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{q}\mathbf{s}{\mathbf{d}}^{\mathbf{2}}}{\mathbf{k}\mathbf{Q}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\left(4e\right)\mathbf{s}{\mathbf{d}}^{\mathbf{2}}}{\mathbf{k}\mathbf{Q}} \end{gathered}$$

Here, $\mathit{\alpha }$ is the polarizability of the carbon atom, E is the magnitude of the electric field of the plastic pen, qis the charge of the carbon atom and s is the distance moved by the atom when it is polarized by a plastic pen.

Substitute the values in the above equation.

$$\begin{gathered} \mathit{\alpha }\mathbf{=}\frac{\mathbf{4}\left(1.6\times {10}^{-19}C\right)\left(1.53\times {10}^{-22}m\right){\left(5.29\times {10}^{-11}m\right)}^{\mathbf{2}}}{\left(9\times {10}^{9}N·m^2/C^2\right)\left(1.6\times {10}^{-19}C\right)} \\ \mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{4}\left(1.53\times {10}^{-22}m\right)\left(2.79\times {10}^{-21}{m}^2\right)}{\left(9\times {10}^{9}N·m^2/C^2\right)} \\ \mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{4}\left(1.71\times {10}^{-22}{m}^3\right)}{\left(9\times {10}^{9}N·m^2/C^2\right)} \\ \mathbf{}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{902}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{52}}\mathbf{}{\mathit{C}}^{\mathbf{2}}\mathbf{·}\mathit{m}\mathbf{/}\mathit{N} \end{gathered}$$

Thus, the polarizability of the carbon atom is $\mathbf{1}\mathbf{.}\mathbf{902}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{52}}\mathbf{}{\mathit{C}}^{\mathbf{2}}\mathbf{·}\mathit{m}\mathbf{/}\mathit{N}$.

The measured value is given as $\mathbf{1}\mathbf{.}\mathbf{96}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{40}}\mathbf{}{\mathit{C}}^{\mathbf{2}}\mathbf{·}\mathit{m}\mathbf{/}\mathit{N}$.

The equation of the difference between the measured value and the polarizability of the carbon atom is expressed as:

$\mathit{d}\mathbf{=}\mathit{x}\mathbf{-}\mathit{\alpha }$

Here, d is the difference between the measured value and the polarizability of the carbon atom, xis the measured value and $\mathit{\alpha }$is the polarizability of the carbon atom.

Substitute the values in the above equation.

$$\begin{gathered} \mathit{d}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{96}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{40}}\mathbf{}{\mathit{C}}^{\mathbf{2}}\mathbf{·}\mathit{m}\mathbf{/}\mathit{N}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{902}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{52}}\mathbf{}{\mathit{C}}^{\mathbf{2}}\mathbf{·}\mathit{m}\mathbf{/}\mathit{N} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{96}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{40}}\mathbf{}{\mathit{C}}^{\mathbf{2}}\mathbf{·}\mathit{m}\mathbf{/}\mathit{N} \end{gathered}$$

Thus, the measured value is bigger by $\mathbf{1}\mathbf{.}\mathbf{96}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{40}}\mathbf{}{\mathit{C}}^{\mathbf{2}}\mathbf{·}\mathit{m}\mathbf{/}\mathit{N}$than the polarizability of the carbon atom.

The first assumption took in this question is that for maintaining equilibrium, the sum of the force due to gravity and the force exerted on the carbon atom by the pen must be equal to zero. The second assumption is that the distance between the outer electron to the carbon atom is being taken as the Bohr’s radius. Moreover, the third assumption made in this question is the charge of the pen is the charge of the tape that is the charge of an electron.

The fourth assumption made is that the size of an atom is $\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathbf{}\mathit{m}$. The fifth assumption is the mass of one carbon atom that is $\mathbf{1}\mathbf{.}\mathbf{99}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}}\mathit{g}$and the acceleration due to gravity is $\mathbf{9}\mathbf{.}\mathbf{8}\mathbf{}\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{2}}$. The sixth assumption is the permittivity that is $\mathbf{8}\mathbf{.}\mathbf{59}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{}{\mathit{C}}^{\mathbf{2}}\mathbf{·}{\mathit{m}}^{\mathbf{-}\mathbf{2}}\mathbf{·}{\mathit{g}}^{\mathbf{-}\mathbf{1}}$. The seventh assumption is the charge of one electron and the value of pi that are $\mathbf{1}\mathbf{.}\mathbf{602}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathbf{}\mathit{C}$and $\mathbf{3}\mathbf{.}\mathbf{14}$respectively. The eighth and the final assumption is that the distance between the outer electron to the carbon atom is being taken as the Bohr’s radius that is $\mathbf{(}\mathbf{5}\mathbf{.}\mathbf{29}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{11}}\mathbf{}\mathbf{m}\mathbf{)}$.

Thus, the assumptions made in this question are, the sum of the force due to gravity and the force exerted on the carbon atom by the pen must be equal to zero. The distance between the outer electron to the carbon atom is being taken as the Bohr’s radius. The charge of the pen is the charge of the tape that is the charge of an electron. The size of an atom is $\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathbf{}\mathit{m}$. The mass of one carbon atom that is $\mathbf{1}\mathbf{.}\mathbf{99}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{23}}\mathbf{}\mathit{g}$and the acceleration due to gravity is $\mathbf{9}\mathbf{.}\mathbf{8}\mathbf{}\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{2}}$. The permittivity that is $\mathbf{8}\mathbf{.}\mathbf{59}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{}{\mathit{C}}^{\mathbf{2}}\mathbf{·}{\mathit{m}}^{\mathbf{-}\mathbf{2}}\mathbf{·}{\mathit{g}}^{\mathbf{-}\mathbf{g}}$. The charge of one electron and the value of pi that are $\mathbf{1}\mathbf{.}\mathbf{60}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathbf{}\mathit{C}$and $\mathbf{3}\mathbf{.}\mathbf{14}$respectively. The distance between the outer electron to the carbon atom is being taken as the Bohr’s radius that is $\mathbf{(}\mathbf{5}\mathbf{.}\mathbf{29}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathbf{}\mathbf{m}\mathbf{)}$.