Question

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?

Short answer

(a) Given situations show that, Doppler broadening is more important.

(b) The other effect (uncertainty principle) predominates when it has a longer wavelength of lower temperature.

Step-by-step solution

(a) Variation of wavelength due to Doppler effect:

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.

(a) Variation in wavelength due to uncertainty principle:

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.

Conclusion:

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

(b) The other effect (uncertainty principle) predominates:

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.