Understanding spectral line of hyperfine transition

Question

I was looking at the contents of the Pioneer Plaque and I pretty much understood everything that was there, but not the so-called "Hyperfine transition of neutral hydrogen".

Looking mostly at Wikipedia articles, I could understand that what is depicted there is the frequency (in centimeters) of the electromagnetic radiation of the spin up-spin down transition, and, crudely, this is what 'hyperfine transition' means.

But, what I don't understand is how do the spectral line enters in all this. Do the photons emitted generate both emission and absorption lines? Or certain particles generate emission lines and other generate absorption? And how is this related to the 'hyperfine transition'? Does the spectral line has a period/frequency?

Answer 1

(Hmpf I was almost finished when John Rennie posted his anwer .... you can see it as a supplement.)

What generates absorption and emission lines are the differences in energy levels. As an example take sodium: when you heat the sodium vapor you push electrons in excited energy states, when those electrons jump from the excited in the ground state they emit photons with a specific energy. What you then see are emission lines, in the image below this is the part on the top.

On the other hand you could send white light (composed of all wavelengths) through a non-heated sodium vapor. In this vapor photons with a certain frequency,i.e. energy, push electrons from the ground in the excited state. In this case you see absorption lines, in the picture below this is the part on the bottom.

So the photons emitted make emission lines and the photons which are absorbed are responsible for absorption lines.

The same principle takes place with the hyperfine transition: you have two states with different energies. When the electrons jumps from the higher (spin up) state to the lower (spin down) it emits a photon and conversely it has to absorb a photon if it wants to jump from the lower to the higher state.

So the energy is in both cases the same, but one time the photon is absorbed and the other time it is emitted.

Yes the spectral lines have a frequency. Either you measure the frequency directly or you measure the wavelength of the photons. When you know the wavelength you can calculate the frequency via $$\lambda f = c, $$ where $\lambda$ is the wavelength and $c$ is the speed of light. (Indeed, when you put the two numbers from the Wikipedia article in you get this result.) This in turn gives you the energy the photons have and consequently the difference of the two energy levels: $$E=\hbar\omega\quad\text{with}\;\omega=2\pi f $$

Answer 2

If you take an isolated hydrogen atom it will settle into the ground state with the spins opposed and stay there for ever. Not very exciting.

However if you take a hydrogen gas and you heat it, e.g. by shining starlight on it, then the hydrogen atoms will start buzzing around and colliding with each other, and every now and then when two hydrogen atoms collide one or both of them will be jolted into the excited state with the spins aligned. The excited hydrogen atom will then relax to the ground state and emit a photon with a wavelength of 21cm.

So a hot hydrogen gas emits light at the 21cm wavelength. If you look at the light from the hot gas with a spectrometer you'll see a bright line at 21cm (well, it's not an optical wavelength so your detector does the "seeing").

Now suppose you take a cold hydrogen gas. This won't be emitting light with the 21cm wavelength but it will absorb 21cm light by exciting ground state hydrogen atoms to the excited state. The excited atoms will relax, but the re-emitted light may be in a different direction, or the excited atoms may lose energy by colliding with other hydrogen atoms and not re-emit the light at all.

So if a broad range of wavelengths passes through a cold hydrogen gas the 21cm light will be absorbed. If you look at the light from the cold gas with a spectrometer you'll see a dark line at 21cm.

It's not my area, but I think in practice the temperature required to produce the emission is so low that we only ever see emission at 21cm and never absorption.

The 21cm is the wavelength, $\lambda$, of the electromagnetic radiation. The frequency, $\nu$, is given by:

$$ \nu = \frac{c}{\lambda} \approx 1.42\text{GHz} $$

For comparison, microwave ovens produce electromagnetic radiation with a frequency of 2.45GHz.

Answer 3

John Rennie's description of where the energy difference comes from that produces the line is of course correct.

What he omitted was that it is a "forbidden transition", that is radiatively facilitated only by an extremely unlikely magnetic dipole interaction (unlike the more likely electric dipole mode which is forbidden by quantum mechanical selection rules). The radiative lifetime is about 10 million years.

This means that in anything but the sparsest of gases, it is overwhelmingly collisionally dexcited. However, if it does occur radiatively, the photon's mean free path in the interstellar medium is very long - it is hard to absorb for exactly the same reason it is hard to emit. Its penetrative power through the ISM is quite similar to X-rays. Because the radiative transition has a long lifetime, the spectral line emission is also extremely narrow, making it a brilliant diagnostic of motion.

Emission is usually seen from clouds of cold, neutral hydrogen gas. Absorption at 21cm requires a very thick cloud of hydrogen gas to be in front of a source of radio continuum, such that there is some chance of a 21cm photon exciting the transition between the energy levels and being absorbed.