# I Why are Beryllium-7 neutrinos monoenergetic?

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1. Mar 23, 2017

### Smalde

In one step of the solar ppII chain the following reaction (may) take place:

⁷Be + e⁻ --> ⁷Li + ν_e

This reaction was used in the famous homestake experiment to detect solar neutrinos when the famous solar neutrino problem came into play.

In http://cds.cern.ch/record/1733288/files/vol42-issue10-p015-e.pdf Nobel Laureate Raymond Davis Jr describes that his "[...]first experiment was a study of the recoil energy of a lithium-7 nucleus resulting from the electron capture decay of beryllium-7. In a beryllium-7 decay, a single monoenergetic neutrino is emitted with an energy of 0.862 MeV[...]"

My question is: Why are these neutrinos monoenergetic?

I understand that most of the Lithium atoms are generated in their ground state. But wouldn't the kinetic energies of the electron and Beryllium matter?

The only reason I can come to think is that the reactants, electron and beryllium, have to be slow (i.e. their kinetic energies negligible in comparison to their rest mass energies) for the whole energy of the neutrino to be determined by the mass difference between product and educt. This can be the case if we think that these reactions take (mostly) only place in the core and the sun. One can think that due to the extreme densities the particles do not move extremely fast.
Another option would be that the kinetic energy of the neutrinos is in no way determined by the original kinetic energies.

So, what is it?

Thanks!

Pol

2. Mar 23, 2017

### Staff: Mentor

Yes, but they are small (few keV) compared to the neutrino energy (1 MeV), and usually smaller than the energy resolution of the detector.
Compare this to three-body processes where the neutrino can have every energy from zero to some maximum.

3. Mar 23, 2017

### Smalde

Ok, thanks, that solves it.
That's what I thought, but I couldn't really find anything on the internet about this. It was an interesting search anyway, since I learned more about both the homestake and the borexino experiments. :)

4. Mar 23, 2017

### Orodruin

Staff Emeritus
Many times you will see things like this where "monoenergetic" really means "to a very good approximation monoenergetic". Take neutrinoless double beta decay for example - the summed energy of the electrons is essentially the same regardless of the kinematic configuration. I once put the computation of the natural width of this distribution as a problem in relativistic kinematics on an exam. The result is ... small.

5. Mar 23, 2017

### Smalde

So... a new problem arises when trying to calculate the momentum of the neutrino.

Assuming negligible momentum of both the electron and the beryllium and negligible neutrino mass:

$$P_e = (m_e c,\vec{0})$$
$$P_{⁷Be} = (m_{⁷Be} c,\vec{0})$$
$$P_{⁷Li} = (E/c,\vec{p}_{⁷Li})$$
$$P_{\nu} = (|\vec{p}_{\nu}|,\vec{p}_{\nu})$$

This means that $$\vec{p}_{\nu} = -\vec{p}_{⁷Li}$$

Thus

$$m_e c² + m_{⁷Be} c² = E + |\vec{p}_{\nu}|c = \sqrt{m²_{⁷Li}c⁴ + |\vec{p}_{⁷Li}|²c²} + |\vec{p}_{\nu}|c= \sqrt{m²_{⁷Li}c⁴ + |\vec{p}_{\nu}|²c²}+ |\vec{p}_{\nu}|c$$

And solving for $$|\vec{p}_{\nu}|$$ one has $$|\vec{p}_{\nu}|c=\frac{m_ec²}{2} \frac{m_e+2m_{⁷Be}}{m_e+m_{⁷Be}} \approx m_ec² = 0.51MeV$$

This differs from the 0.862MeV Raymond Davis Jr speaks about.

Where am I wrong?

6. Mar 23, 2017

### Staff: Mentor

Something went wrong in the "and solving for pv" step. The lithium mass should not disappear, and things shouldn't get proportional to the electron mass. Electron and beryllium mass only appear as sum, that shouldn't change.

7. Mar 23, 2017

### Orodruin

Staff Emeritus
The easier way is to use conservation of 4-momentum directly. You know that $p_{e}+p_{Be} - p_\nu = p_{Li}$. Squaring this results in
$$(m_e + m_{Be})^2 - 2 (m_e + m_{Be}) E = m_{Li}^2 \quad \Longrightarrow \quad E = \frac{(m_e + m_{Be})^2 - m_{Li}^2}{2(m_e + m_{Be})} \simeq 0.86~\mbox{MeV}$$
(we are working in units where $c = 1$). I don't see how you got rid of the lithium mass when solving for the momentum. Note that for massless neutrinos the momentum is equal to the energy.

Edit: Be careful with which beryllium and lithium masses you put in ...

8. Mar 23, 2017

### Smalde

Oh, right, reviewing my notes something did clearly go wrong in the solving for $|\vec{p}_\nu|$ step... I wasn't careful enough and crossed a beryllium term on the left hand and a lithium term on the right as if they were the same. I did not check the result because it was the same result I got with a wrong ansatz before. hehe

Ok, everything clear now.

Davis may keep his Nobel :P

9. Mar 23, 2017

### Orodruin

Staff Emeritus
So the perhaps easier way to put it in terms of things you will find tabulated is to rewrite the entire expression as
$$E \simeq \frac{(m_e + m_{Be} + m_{Li})(m_e + m_{Be} - m_{Li})}{2(m_e + m_{Be})} \simeq \frac{Q_{EC}2m_{Be}}{2m_{Be}} = Q_{EC}.$$
You will find the Q-value for beryllium electron capture in any table.

10. Mar 23, 2017

### Staff: Mentor

For those decays, a first order approximation works as well: The Q-value is 0.86 MeV, therefore the neutrino will have an energy of nearly 0.86 MeV and a momentum of nearly 0.86 MeV/c. The lithium nucleus will have the same (absolute) momentum, which means it has an energy of p2/(2m) = 0.0004 MeV. We could subtract this from the 0.86 MeV, but it is irrelevant with two significant figures. In addition, the correction is just 400 eV, well below the width of the actual neutrino spectrum from thermal motion.
It is not an exact solution - as the neutrino energy is lower, the momentum will be lower as well, and so on - but it leads to a very good approximation very fast.
Even thermal effects are tens of orders of magnitude larger.

11. Mar 23, 2017

### Orodruin

Staff Emeritus
Even if there was no thermal motion, the exercise would quickly become pointless as the Q-value itself comes associated with an error of the order of 70 eV (source). While the lithium momentum might lead to corrections slightly larger than the uncertainty, any corrections beyond this will not affect precision anyway.

12. Mar 23, 2017

### Staff: Mentor

That uncertainty is surprisingly large. Penning traps can measure nuclear masses with about 1 eV precision. Apparently no one bothered measuring Be-7 with the best traps yet.

13. Mar 23, 2017

### Orodruin

Staff Emeritus
Perhaps not the most "sexy" of research topics. Compare "you will be working with a team of 3000 researchers to uncover the unknown properties of elementary particles at new energy scales" with "you will be measuring the Q-value of beryllium 7 to decrease the relative uncertainty from 1e-4 to 1e-5". Which PhD project would you choose?

14. Mar 23, 2017

### Staff: Mentor

More like "you will be working with a team of 3000 researchers, your project is to make the jet energy measurement a little bit more precise, and work on the background contributions to improve SUSY exclusion limits in some specific region of phase space using this specific decay channel".
Sure, there are a few analyses measuring something for the first time, and even some analyses finding unexpected things (pentaquarks, ...), but even those are the result of larger analysis groups where everyone contributes a little bit.

All the thousands of branching fractions of hadrons known today had to be measured at some point. That is not very sexy, but it is necessary to understand all the background contributions in other analyses.

15. Mar 23, 2017

Staff Emeritus
That's probably not the choice. The choice is more likely, of the ~5000 nuclei out there, why measure this one right now as opposed to some other one. Somewhere out there is the lab that is the best at this sort of thing. And that PI has a list of nuclei he's weighing, and that list is driven by several things: the ease of the measurement, the interest in the measurement, what the competition is likely to do, etc. If he's weighing (say) Ca-48 now, why switch to Be-7?

16. Mar 24, 2017

### Smalde

Because of this thread, obviously :P