# HI cloud and 21cm line

1. May 11, 2009

### TFM

1. The problem statement, all variables and given/known data

Consider an HI cloud of temperature T ≥ 3 K.

a)

Calculate the energy level difference between the hydrogen atom states leading to 21cm line emission. Show that the ratio of hydrogen atoms in the upper to the lower state of the 21 cm line is approximately 3:1.

b)

Assuming the cloud to be optically thin, find a relation between the total 21 cm Luminosity (in Watts) and the mass of HI (in solar masses). The decay rate is $$A_{21} ≈ (1.1 × 10^7 yr)^{−1}$$.

2. Relevant equations

3. The attempt at a solution

I have done the first part of a), finding the Energy Difference. I got this to be

$$9.43*10^{-25}$$Joules.

I thought the formula for the second part opf a, showing the ratio is approx. 3, required this formula:

$$\frac{n_2}{n_1} = exp(\frac{-(e_2 - E_1)}{kT})$$

However, this doesn't give me a reasonable value, the value comes a=out to be about 1.

I am also slight uncertain about what to do on part b.

Any ideas would be most appreciated,

TFM

2. May 12, 2009

### Redbelly98

Staff Emeritus
(a) You did the calculation correctly, but have left out a key factor.

Hint:
For any temperature, a state of higher energy will always have fewer atoms than a state of lower energy. This is because the expression

n2/n1 = e-ΔE/kT

is always less than 1, for any temperature.

How then might it be possible for the higher energy to have 3 times as many atoms as the lower energy?

3. May 12, 2009

### TFM

It would seem to be impossible then, if for any temperature there will be always be more atoms in a lower state of energy.

4. May 12, 2009

### Redbelly98

Staff Emeritus
If there were really just one upper state and just one lower state, that would be true.

5. May 12, 2009

### TFM

Well, wouldn't there be two upper and lower states, since the proton/electron pair can be either parallel up or parallel down (higher energy) or anti-parallel, with either the electron up/proton down or proton up/electron down?

6. May 12, 2009

### Redbelly98

Staff Emeritus
While that appears to be a reasonable way of thinking about it ... it's incorrect :-(

Instead, think of: what is the total spin S when both spins are parallel? How many mS values are there for this value of S?

7. May 12, 2009

### TFM

Well, when they are parallel, the spin will be either positive or negative 1, since they will be either both plus half, or both minus half. When they are anti-parallel, the total spin will be 0, since they will always be one positive half, one negative half.

8. May 12, 2009

### Redbelly98

Staff Emeritus
Okay, good. (Actually the parallel case is "spin = 1", since we are just talking about the magnitudes of the spin).

How many states exist for spin=1? And for spin=0? Hint: this is the m quantum number.

9. May 12, 2009

### TFM

Looking through my noted, but all I am getting is m_s = plus/minus one half, which doesn't seem to work for this.

would it be m = 0 for s = 0, and m = -1, 0, 1 for s = 1?

10. May 12, 2009

### Redbelly98

Staff Emeritus
Yes, that's right.

11. May 13, 2009

### TFM

So there is three upper states, and only one lower state, would this how the 3:1 ratio?

12. May 13, 2009

### Redbelly98

Staff Emeritus
Yes, exactly.

13. May 13, 2009

### TFM

Excellent.

So for the next part, what is the best way to show the relationship between H21 luminoscity and the mass. So would the right thing to be to use the fomrula:

$$L = 4\pi r^2 \sigma T^4$$

and the fact that:

$$v = \frac{m}{\rho}$$

and for a sphere:

$$v = \frac{4}{3}\pi r^3$$

which can be rearranged to give:

$$L = 4\pi \left( ^3\sqrt{\frac{3}{4\pi}\frac{m}{\rho}} \right)^2 \sigma T^4$$

where it is cube root,

however, this doesn't use the decay rate that is given...?

14. May 14, 2009

### Redbelly98

Staff Emeritus
That approach doesn't work, since the H atoms are radiating at a single wavelength, and not a blackbody spectrum.

Probably best to think first what would be the luminosity of one single H atom. Then figure out how many H atoms there are.

15. May 15, 2009

### TFM

well the luminosity is the energy radiated over time, now we know the energy difference. Would that be useful?

16. May 15, 2009

### Redbelly98

Staff Emeritus
Yes. And, the average time it takes to radiate that energy is useful as well.

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