Chapter 24, Problem 56P

(a)

To determine

The number of components and the wavelength of each components.

Answer to Problem 56P

The numbers of components are $1110101,1110102,1110100,1110103\text{ and so on}$.

The wavelengths of each component are $632.80914\text{nm , 632}\text{.80857 nm , 632}\text{.80971nm}$.

Explanation of Solution

Write the expression for the separation distance.

$d=\frac{N\lambda }{2}$        (I)

Here, $d$ is the separation distance, $N$ is the number and $\lambda$ is the wavelength.

Write the expression for the used wavelength.

${\lambda }_{\text{used}}=\frac{{\lambda }_1+{\lambda }_2}{2}$        (II)

Here, ${\lambda }_1,{\lambda }_2$ are the wavelengths.

Rewrite the expression (I) in terms of the number of components by using (II).

$N_{\text{used}}=\frac{2d}{\left(\frac{{\lambda }_1+{\lambda }_2}{2}\right)}$        (III)

Rewrite the equation (I) in terms of wavelength.

  $\lambda =\frac{2d}{N}$        (IV)

Conclusion:

Substitute, $35.124103\text{cm}$ for $d$, $632.80840\text{nm}$ for ${\lambda }_1$, $632.80980\text{nm}$ for ${\lambda }_2$ in Equation (II) to find $N_{\text{used}}$.

  $\begin{array}{c}{N}_{\text{used}}=\frac{2\left[\left(35.124103\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right]}{\left(\frac{\left[\left(632.80840\text{nm}\right)\left(\frac{1\times {10}^{-9}\text{m}}{1\text{nm}}\right)\right]+\left[\left(632.80980\text{nm}\right)\left(\frac{1\times {10}^{-9}\text{m}}{1\text{nm}}\right)\right]}{2}\right)}\\ =1110101.07\end{array}$

Hence, the number of components will be 1110100, 1110101, 1110102, 1110103 and so on.

For the first wavelength:

Substitute, $35.124103\text{cm}$ for $d$, 1110100 for $N$ in Equation (IV) to find $\lambda$.

$\begin{array}{c}\lambda =\frac{2\left[\left(35.124103\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right]}{\left(1110100\right)}\\ =632.80971\times {10}^{-9}\text{m}\\ \text{=}632.80971\text{nm}\end{array}$

For the second wavelength:

Substitute, $35.124103\text{cm}$ for $d$, 1110101 for $N$ in Equation (IV) to find $\lambda$.

$\begin{array}{c}\lambda =\frac{2\left[\left(35.124103\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right]}{\left(1110101\right)}\\ =632.80914\times {10}^{-9}\text{m}\\ \text{=}632.80914\text{nm}\end{array}$

For the third wavelength:

Substitute, $35.124103\text{cm}$ for $d$, 1110102 for $N$ in Equation (IV) to find $\lambda$.

$\begin{array}{c}\lambda =\frac{2\left[\left(35.124103\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right]}{\left(1110102\right)}\\ =632.80857\times {10}^{-9}\text{m}\\ \text{=}632.80857\text{nm}\end{array}$

For the fourth wavelength:

Substitute, $35.124103\text{cm}$ for $d$, 1110103 for $N$ in Equation (IV) to find $\lambda$.

$\begin{array}{c}\lambda =\frac{2\left[\left(35.124103\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right]}{\left(1110103\right)}\\ =632.80800\times {10}^{-9}\text{m}\\ \text{=}632.80800\text{nm}\end{array}$

Hence only first three wavelengths are possible.

Thus, the numbers of components are only three and the wavelengths of each components are $632.80914\text{nm , 632}\text{.80857 nm , 632}\text{.80971nm}$.

(b)

To determine

The root mean square speed for neon atom.

Answer to Problem 56P

The root mean square speed for neon atom is $\underset{\_}{697\text{m/s}}$.

Explanation of Solution

Write the expression for the root mean square speed by using conservation of energy.

$\begin{array}{l}\frac{1}{2}{m}_0{v}^2=\frac{3}{2}kT\\ v=\sqrt{\frac{3kT}{m_0}}\end{array}$        (V)

Here, $m_0$ is the mass, $v$ is the rms speed, $k$ is the Boltzmann constant and $T$ is the temperature.

Conclusion:

Substitute, $120°\text{C}$ for $T$, $1.38\times {10}^{-23}\text{J/K}$ for $k$, $20.18\text{u}$ for $m_0$ in Equation (V) to find $v$.

$\begin{array}{c}v=\sqrt{\frac{3\left(1.38\times {10}^{-23}\text{J/K}\right)\left(120°\text{C}+273\text{K}\right)}{\left[\left(20.18\text{u}\right)\left(\frac{1.66\times {10}^{-27}\text{kg}}{1\text{u}}\right)\right]}}\\ =697\text{m/s}\end{array}$

Thus, the root mean square speed for neon atom is $\underset{\_}{697\text{m/s}}$.

(c)

To determine

Whether the Doppler Effect for light emission by moving neon atoms should realistically make the bandwidth of the light amplifier larger than the 0.00140nm at $120°\text{C}$.

Answer to Problem 56P

The Doppler Effect for light emission by moving neon atoms should realistically make the bandwidth of the light amplifier larger than the 0.00140nm at $120°\text{C}$.

Explanation of Solution

Write the expression for the frequency from Doppler Effect.

$f^{\prime }=f\sqrt{\frac{c+v}{c-v}}$        (VI)

Here, $f^{\prime }$ is the apparent frequency, $f$ is the original frequency and $c$ is the light of speed.

Since, the frequency and the wavelength is inversely proportional, the apparent wavelength equation can be calculated by using equation (VI).

Write the expression for the apparent wavelength by using (II).

${\lambda }^{\prime }=\left(\frac{{\lambda }_1+{\lambda }_2}{2}\right)\sqrt{\frac{c-v}{c+v}}$        (VII)

Here, ${\lambda }^{\prime }$ is the apparent wavelength and $\lambda$ is the wavelength.

Conclusion:

Substitute, $697\text{m/s}$ for $v$, $3\times {10}^8\text{m/s}$ for $c$, $632.80840\text{nm}$ for ${\lambda }_1$, $632.80980\text{nm}$ for ${\lambda }_2$ in Equation (VII) to find ${\lambda }^{\prime }$.

$\begin{array}{c}{\lambda }^{\prime }=\left(\frac{\left[\left(632.80840\text{nm}\right)\left(\frac{1\times {10}^{-9}\text{m}}{1\text{nm}}\right)\right]+\left[\left(632.80980\text{nm}\right)\left(\frac{1\times {10}^{-9}\text{m}}{1\text{nm}}\right)\right]}{2}\right)\sqrt{\frac{\left(3\times {10}^8\text{m/s}\right)-\left(697\text{m/s}\right)}{\left(3\times {10}^8\text{m/s}\right)+\left(697\text{m/s}\right)}}\\ =632.80763\text{nm}\end{array}$

The calculated wavelength is beyond limit. There will be many atoms which are moving more than rms speed. Hence, it can be expected that there will be more Doppler broadening of the resonance amplification peak.