Spectral lines are fuzzy due to two effects: Doppler broadening and the uncertainty principle. The relative variation in wavelength due to the first effect (see Exercise 2.57) is given by

$\frac{\mathbf{∆}\mathbf{\lambda }}{\mathbf{\lambda }}\mathbf{=}\frac{\sqrt{\mathbf{3}{\mathbf{k}}_{\mathbf{B}}\mathbf{T}\mathbf{/}\mathbf{m}}}{\mathbf{c}}$

Where T is the temperature of the sample and m is the mass of the particles emitting the light. The variation due to the second effect (see Exercise 4.72) is given by

$\frac{\mathbf{∆}\mathbf{\lambda }}{\mathbf{\lambda }}\mathbf{=}\frac{\mathbf{\lambda }}{\mathbf{4}\mathbf{\pi c}}$

Where, $\mathbf{∆}\mathbf{t}$ is the typical transition time

(a) Suppose the hydrogen in a star has a temperature of $\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{}\mathbf{K}$. Compare the broadening of these two effects for the first line in the Balmer series (i.e.,${\mathbf{n}}_{\mathbf{i}}\mathbf{=}\mathbf{3}\mathbf{\to }{\mathbf{n}}_{\mathbf{f}}\mathbf{=}\mathbf{2}$ ). Assume a transition time of 10^-8s. Which effect is more important?

(b) Under what condition(s) might the other effect predominate?

According to the Doppler effect or Doppler shift, the frequency of a source changes if the observer is in relative motion w.r.t source.

Given, the expression of relative variation of wavelength due to the Doppler effect is given by,

$\frac{\mathbf{∆}\mathbf{\lambda }}{\mathbf{\lambda }}\mathbf{=}\frac{\sqrt{\mathbf{3}{\mathbf{k}}_{\mathbf{B}}\mathbf{T}\mathbf{/}\mathbf{m}}}{\mathbf{c}}$

Where,${\mathbf{k}}_{\mathbf{B}}$is theBoltzmann constant,$\mathbf{\lambda }$is the Wavelength of the light, is the temperature (in K), is the mass of hydrogen atom,and c is the speed of light.

Now,

$$\begin{gathered} \frac{∆\mathrm{\lambda }}{\mathrm{\lambda }}=\frac{\sqrt{3{\mathrm{k}}_{\mathrm{B}}\mathrm{T}/\mathrm{m}}}{\mathrm{c}} \\ =\frac{\sqrt{3\left(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\right)\left(5\times {10}^4\mathrm{K}\right)/\left(1.67\times {10}^{-27}\mathrm{kg}\right)}}{3\times {10}^8\mathrm{m}/\mathrm{s}} \\ =1.74\times {10}^{-8} \end{gathered}$$

Hence, wavelength is varied by a fraction of $1.74\times {10}^{-8}$for Doppler.

Given,variation in wavelength due to uncertainty principle

$$\begin{gathered} \frac{∆\mathrm{\lambda }}{\mathrm{\lambda }}=\frac{\mathrm{\lambda }}{4\mathrm{\pi c}∆\mathrm{t}} \\ =\frac{656\times {10}^{-9}\mathrm{m}}{4\mathrm{\pi }\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)\left({10}^{-8}\mathrm{s}\right)} \\ =1.74\times {10}^{-8} \end{gathered}$$

Hence, wavelength is varied by a fraction of $1.74\times {10}^{-8}$for uncertainty principle.

It can be clearly observed from Step 1 and 2 that, the Doppler broadening is more important. ([Doppler:$1.17\times {10}^4$] [Uncertainty:$1.17\times {10}^{-8}$ ])

As you know that, variation in wavelength due to uncertainty principle is given by,

$\frac{∆\lambda }{\lambda }=\frac{\lambda }{4\mathrm{\pi c}∆\mathrm{t}}$

It increases when wavelength $\lambda$is higher. And according to Doppler effect,

$\frac{∆\mathrm{\lambda }}{\mathrm{\lambda }}=\frac{\sqrt{3{\mathrm{k}}_{\mathrm{B}}\mathrm{T}/\mathrm{m}}}{\mathrm{c}}$

It lowers when the temperature lowers.

Hence, the Uncertainty principle broadening can predominate the Doppler effect if wavelength increases or temperature decreases.