If you take the model in Ex. 4.1 at face value, what natural frequency do you get? Put in the actual numbers. Where, in the electromagnetic spectrum, does this lie, assuming the radius of the atom is 0.5 Å? Find the coefficients of refraction and dispersion, and compare them with the measured values for hydrogen at $\mathbf{0}\mathbf{°}\mathbf{C}$and atmospheric pressure:$\mathbf{A}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{36}\mathbf{\times }{\mathbf{10}}^{-4}$,$\mathbf{B}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{7}\mathbf{\times }{\mathbf{10}}^{-15}{\mathbf{m}}^2$ .

Write the expression forelectric field.

$E=\frac{1}{4\pi {\epsilon }_0}\frac{qd}{a^3}$ …… (1)

Here, ${\epsilon }_0$is the permittivity of free space, q is the charge, d is the distance and a is the radius of an atom.

Write the expression for binding force using equation (1) as,

$\begin{array}{c}{F}_{binding}=-qE\\ =-\left(\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{a^3}\right)x\\ =-k_{spring}x\end{array}$ …… (2)

Here,$k_{\text{spring}}$ is the spring constant.

Substitute $F_{\text{binding}}=\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{a^3}$and $k_{\text{spring}}=m{\omega }_0^2$in equation (2).

$\begin{array}{c}\left(\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{a^3}\right)x=-m{\omega }_0^{2}x\\ {\omega }_0=\sqrt{\frac{q^2}{4\pi {\epsilon }_{0}m{a}^3}}\end{array}$ …… (3)

Write the expression for the natural frequency.

${\nu }_0=\frac{{\omega }_0}{2\pi }$

Substitute${\omega }_0=\sqrt{\frac{q^2}{4\pi {\epsilon }_{0}m{a}^3}}$ in the above expression.

${\nu }_0=\frac{1}{2\pi }\sqrt{\frac{q^2}{4\pi {\epsilon }_{0}m{a}^3}}$ …… (4)

Here, $q$is charge of electron,$m$ is mass of electron, ${\epsilon }_0$is permittivity of space, and$a$ is distance.

Substitute ,$q=1.6\times {10}^{-19}\text{\hspace{0.17em}C}$ ,${\epsilon }_0=8.85\times {10}^{-12}{\text{\hspace{0.17em}C}}^2/\text{N}\cdot {\text{m}}^2$,$m=9.11\times {10}^{-31}\text{\hspace{0.17em}kg}$and $a=0.5\stackrel{\text{0}}{\text{A}}$in equation (4).

$\begin{array}{c}{\nu }_0=\frac{1}{2\pi }\sqrt{\frac{{(1.6\times {10}^{-19}\text{\hspace{0.17em}C})}^2}{4\pi (8.85\times {10}^{-12}{\text{\hspace{0.17em}C}}^2/\text{N}\cdot {\text{m}}^2)(9.11\times {10}^{-31}\text{\hspace{0.17em}kg}){(0.5\stackrel{0}{\text{A}}\times \frac{{10}^{-10}\text{\hspace{0.17em}m}}{1\stackrel{0}{\text{A}}})}^3}}\\ {\nu }_0=7.16\times {10}^{15}\text{\hspace{0.17em}Hz}\end{array}$

Therefore, the natural frequency is $7.16\times {10}^{15}\text{\hspace{0.17em}Hz}$, and it lies in ultraviolet range.

Write the formula for the index of refraction.

$\begin{array}{c}n=1+\left(\frac{N{q}^2}{2m{\epsilon }_0}\sum _j\frac{f_j}{{\omega }_j^2}\right)+{\omega }^2\left(\frac{N{q}^2}{2m{\epsilon }_0}\sum _j\frac{f_j}{{\omega }_j^4}\right)\\ n=1+A(1+\frac{B}{{\lambda }^2})\end{array}$

Here,$\lambda$ is the wavelength.

Here,$\lambda =\frac{2\pi c}{{\omega }_0}$

Write the formula for coefficient of refraction (A).

$A=\frac{N{q}^2}{2m{\epsilon }_0}\frac{f}{{\omega }_0^2}$ …… (5)

Here, $N$is number of molecules per unit volume, and $f$number of electrons per molecules.

Here,$f=2\left({\text{for\hspace{0.17em}H}}_{\text{2}}\right)$ .

Write the formula for coefficient of dispersion (B).

$B={\left(\frac{2\pi c}{{\omega }_0}\right)}^2$ …… (6)

Here, c is the speed of light.

Substitute$q=1.6\times {10}^{-19}\text{\hspace{0.17em}C}$,${\epsilon }_0=8.85\times {10}^{-12}{\text{\hspace{0.17em}C}}^2/\text{N}\cdot {\text{m}}^2$, $m=9.11\times {10}^{-31}\text{\hspace{0.17em}kg}$and $a=0.5\stackrel{\text{0}}{\text{A}}$in equation (3).

$\begin{array}{c}{\omega }_0=\sqrt{\frac{q^2}{4\pi {\epsilon }_{0}m{a}^3}}\\ {\omega }_0=\sqrt{\frac{{(1.6\times {10}^{-19}\text{\hspace{0.17em}C})}^2}{4\pi (8.85\times {10}^{-12})(9.11\times {10}^{-31}\text{\hspace{0.17em}kg}){(0.5\stackrel{0}{\text{A}}\times \frac{{10}^{-10}\text{\hspace{0.17em}m}}{1\stackrel{0}{\text{A}}})}^3}}\\ {\omega }_0=4.5\times {10}^{16}\text{\hspace{0.17em}Hz}\end{array}$

Calculate the value of $N$as follows.

$\begin{array}{c}N=\frac{\text{Avogadro’s\hspace{0.17em}number}}{22.4\text{\hspace{0.17em}L}}\\ N=\frac{6.02\times {10}^{23}}{22.4\times {10}^{-3}}\\ N=2.69\times {10}^{25}\end{array}$

Substitute$N=2.69\times {10}^{25}$, $q=1.6\times {10}^{-19}\text{\hspace{0.17em}C}$,$f=2$,$m=9.11\times {10}^{-31}\text{\hspace{0.17em}kg}$, ${\epsilon }_0=8.85\times {10}^{-12}{\text{\hspace{0.17em}C}}^2/\text{N}\cdot {\text{m}}^2$and${\omega }_0=4.5\times {10}^{16}\text{\hspace{0.17em}Hz}$in equation (5).

$\begin{array}{c}A=\frac{(2.69\times {10}^{25}){(1.6\times {10}^{-19}\text{\hspace{0.17em}C})}^2\left(2\right)}{2(9.11\times {10}^{-31}\text{\hspace{0.17em}kg})(8.85\times {10}^{-12}{\text{\hspace{0.17em}C}}^2/\text{N}\cdot {\text{m}}^2){({\omega }_0=4.5\times {10}^{16}\text{\hspace{0.17em}Hz})}^2}\\ A=4.2\times {10}^{-5}\end{array}$

This value is about$\frac{1}{3}$of the actual value.

Substitute$c=3\times {10}^8\text{\hspace{0.17em}m}/\text{s}$and${\omega }_0=4.5\times {10}^{16}\text{\hspace{0.17em}Hz}$in equation (6).

$\begin{array}{c}B={\left(\frac{2\pi \times 3\times {10}^8}{4.5\times {10}^{16}}\right)}^2\\ B=1.8\times {10}^{-15}{\text{\hspace{0.17em}m}}^2\end{array}$

This value is about $\frac{1}{4}$of the actual value.

Therefore, the coefficient of refraction is $4.2\times {10}^{-5}$and it is about $\frac{1}{3}$of the actual value. The coefficient of dispersionis $1.8\times {10}^{-15}{\text{\hspace{0.17em}m}}^2$, and it is about$\frac{1}{4}$ of the actual value.