Question

For reasons having to do with quantum mechanics. a given kind of atom can emit only certain wavelengths of light. These spectral lines serve as a " fingerprint." For instance, hydrogen's only visible spectral lines are$\mathbf{656}\mathbf{,}$ $\mathbf{486}\mathbf{,}$$\mathbf{434}\mathbf{,}$and$\mathbf{410}\mathit{n}\mathit{m}$ . If spectra/ lines were ofabsolutely precise wavelength. they would be very difficult to discern. Fortunately, two factors broaden them: the uncertainty principle (discussed in Chapter 4) and Doppler broadening. Atoms in a gas are in motion, so some light will arrive that was emitted by atoms moving toward the observer and some from atoms moving away. Thus. the light reaching the observer will Cover a range ofwavelengths. (a) Making the assumption that atoms move no foster than their rms speed-given by ,${\mathit{v}}_{\mathbf{r}\mathbf{m}\mathbf{s}}\mathbf{=}\sqrt{\mathbf{2}{\mathbf{K}}_{\mathbf{B}}\mathbf{T}\mathbf{/}\mathbf{m}}$ where${\mathit{K}}_{\mathbf{B}}$ is the Boltzmann constant. Obtain a formula for the range of wavelengths in terms of the wavelength$\mathit{\lambda }$ of the spectral line, the atomic mass$\mathit{m}$ , and the temperature$\mathit{T}$. (Note: .${\mathit{v}}_{\mathbf{r}\mathbf{m}\mathbf{s}}\mathbf{<}\mathbf{<}\mathit{c}$) (b) Evaluate this range for the$\mathbf{656}\mathit{n}\mathit{m}$ hydrogen spectral line, assuming a temperature of$\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathit{K}$ .

Short answer

(a) The expression for the range of the wavelength is $\frac{2a}{c}\sqrt{\frac{3K_B}{T}}m$.

(b) The range of wavelength of hydrogen line is$0.15\text{\hspace{0.17em}nm}$ .

Step-by-step solution

Write the given data from the question

The root mean square of value of velocity,'$v_{rms}=\sqrt{\frac{3K_{B}T}{m}}$.

The wavelength is .$\lambda =656\text{\hspace{0.17em}nm}$

Temperature is $T=5\times {10}^4\text{\hspace{0.17em}K}$.

The atomic mass is$\text{m}$ .

 Step 2: Determine the formulas to derive the expression for the range of the wavelength and value of wavelength.

The expression to calculate the observer frequency in terms of source frequency, speed and speed of light is given by Doppler effect as follows.

${\mathit{f}}_{\mathbf{o}\mathbf{b}\mathbf{s}}\mathbf{=}{\mathit{f}}_{\mathbf{s}\mathbf{o}\mathbf{u}\mathbf{r}\mathbf{c}\mathbf{e}}\frac{\sqrt{\mathbf{1}\mathbf{-}{\mathbf{\left(}\frac{\mathbf{v}}{\mathbf{c}}\mathbf{\right)}}^{\mathbf{2}}}}{\mathbf{1}\mathbf{+}\frac{\mathbf{v}}{\mathbf{c}}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\theta }}$

Here, $\mathit{c}$is the speed of light, $\mathit{\theta }$is the angle between the source of light and observer and$\mathit{v}$is the velocity of the source.

The expression between the frequency, wavelength and speed of light is given as follows.

$\mathit{\lambda }\mathbf{=}\frac{\mathbf{c}}{\mathbf{f}}$

Here, $\mathit{\lambda }$is the wavelength,$\mathit{c}$ is the speed of light and$\mathit{f}$ is the frequency.

(a) Derive the expression for the range of the wavelength

Determine the Doppler equation in terms of wavelength.

Substitute$\frac{c}{{\lambda }_{obs}}$for$f_{obs}$and $\frac{c}{{\lambda }_{source}}$for $f_{source}$into equation (i).

…(ii)

$\begin{array}{l}\frac{c}{{\lambda }_{obs}}=\frac{c}{{\lambda }_{source}}\frac{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}{1+\frac{v}{c}\cos \theta }\\ {\lambda }_{obs}={\lambda }_{source}\frac{1+\frac{v}{c}\cos \theta }{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}\end{array}$

When the source receding the observer then the angle between the observer and source is equal to$\theta =0°$.

Substitute$0°$for $\theta$into equation (ii),

$\begin{array}{l}{\lambda }_{obs}^1={\lambda }_{source}\frac{1+\frac{v}{c}\cos 0°}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}\\ {\lambda }_{obs}^1={\lambda }_{source}\frac{1+\frac{v}{c}}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}\end{array}$

When the source approaching the observer then the angle between the observer and source is equal to $\theta =180°$.

Substitute $180°$for$\theta$into equation (ii),

$\begin{array}{l}{\lambda }_{obs}^2={\lambda }_{source}\frac{1+\frac{v}{c}\cos 180°}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}\\ {\lambda }_{obs}^2={\lambda }_{source}\frac{1-\frac{v}{c}}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}\end{array}$

The range of the wavelength is given by,

$\Delta \lambda ={\lambda }_{obs}^1-{\lambda }_{obs}^2$

Substitute ${\lambda }_{source}\frac{1+\frac{v}{c}}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}$for ${\lambda }_{obs}^1$and ${\lambda }_{source}\frac{1-\frac{v}{c}}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}$for into above equation.

$\begin{array}{l}\Delta \lambda ={\lambda }_{source}\frac{1+\frac{v}{c}}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}-{\lambda }_{source}\frac{1-\frac{v}{c}}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}\\ \Delta \lambda =\frac{{\lambda }_{source}}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}[(1+\frac{v}{c})-(1-\frac{v}{c})]\\ \Delta \lambda =\frac{{\lambda }_{source}}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}\frac{2v}{c}\end{array}$

Since , $v<<c$which means${(v/c)}^2$is very low, so,$\sqrt{1-{(\text{v}/c)}^2}\approx 1$.

$\Delta \lambda =\frac{2v{\lambda }_{source}}{c}$

Substitute $v_{rms}$for$v$ and$\lambda$ for ${\lambda }_{source}$into above equation.

$\Delta \lambda =\frac{2\lambda v_{rms}}{c}$

Substitute $\sqrt{\frac{3K_{B}T}{m}}$for$v_{rms}$ into above equation.

$\Delta \lambda =\frac{2\lambda }{c}\sqrt{\frac{3K_{B}T}{m}}$ …(iii)

Hence the expression for the range of the wavelength is $\frac{2a}{c}\sqrt{\frac{3K_B}{T}}m$.

(b) Determine the value of the range of the wavelength.

Calculate the value of the range of wavelength.

Substitute $1.38\times {10}^{-23}\text{\hspace{0.17em}J}/\text{K}$for$K_B$ , $656\text{\hspace{0.17em}nm}$for$\lambda$ , $3\times {10}^8\text{\hspace{0.17em}m}/\text{s}$, $5\times {10}^4\text{\hspace{0.17em}K}$for$T$ and$1.67\times {10}^{-27}\text{\hspace{0.17em}kg}$ into equation (iii).

$\begin{array}{l}\Delta \lambda =\frac{2\times 656\text{\hspace{0.17em}nm}}{3\times {10}^8\text{\hspace{0.17em}m}/\text{s}}\sqrt{\frac{3\times 1.38\times {10}^{-23}\text{\hspace{0.17em}J}/\text{K}\times 5\times {10}^4\text{\hspace{0.17em}K}}{1.67\times {10}^{-27}\text{kg}}}\\ \Delta \lambda =437.33\sqrt{12.395\times {10}^8}\text{nm}\\ \Delta \lambda =\frac{437.33}{{10}^8}\times 35206.83\text{nm}\\ \Delta \lambda =0.15\text{\hspace{0.17em}nm}\end{array}$

Hence the range of wavelength of$656\text{\hspace{0.17em}nm}$ hydrogen line is$0.15\text{\hspace{0.17em}nm}$ .