# Neutrinos from the Sun

1. Nov 15, 2007

### Nakalay

Hi. I am curious about the following nuclear reaction:

$$\nu_{e} + ^{71}_{31}Ga \rightarrow ^{71}_{32}Ge + e^{-}$$.

Can anyone explain to me why this reaction has an energy threshold of approximatly 0.23 MeV? How does one calculate the minimum energy of the neutrino for this reaction to take place?

2. Nov 15, 2007

### malawi_glenn

Just calculate the Q-value of the reaction.

Is this homework?

If you do the calculation, you will see that the threshould energy is approx 0.25MeV (depending on how many/good numbers you have from your table, I just did)

If you have no idea what Q-value is, then I suggest you look it up and present your attempt to a solution before we proceed. This is very straightforward and these kinds of reactions are covered in all introductory nuclear physics books.

Last edited: Nov 15, 2007
3. Nov 15, 2007

### Nakalay

All right, that was surprisingly easy. I found the energy threshold to be approximatly 0.236 MeV:

$$Q = -\Delta m\times c^{2} = (70.924954 u - 70.924701 u)\times931.494 \frac{MeV}{u} \approx 0.236 MeV$$

Thanks for the help.

4. Nov 15, 2007

### malawi_glenn

I think this is wrong.

Tabulated are the ATOMIC masses, i.e all electrons are included.'

Before the reaction you have 31 electrons, and after you have 32 + 1 = 33; so you must remove 2electron masses.

Now i calculate with the nuclear masses.

$$M( ^{71}_{31}Ga ) = 70.924701 - 31*0.0005486 = 70.9076944 u$$

$$M( ^{71}_{32}Ge ) = 70.924954 - 32*0.0005486 = 70.9073988 u$$

$$Q = M( ^{71}_{32}Ge ) + m_e - M( ^{71}_{32}Ge ) = 70.9073988 + 0.0005486 - 70.9076944 = 2.53*10^{-4} u = 0.235667 MeV$$

Might not differ so much in this case, but the principle is imortant. In another reaction than this, it might make big difference.

5. Nov 15, 2007

### Nakalay

Great, thanks again. I was actually wondering why I hadn't accounted for the electrons.
I've lost my physics book and I haven't been able to find much information elsewhere on the web so your help is much appreciated.

6. Nov 16, 2007

### malawi_glenn

But have you looked at your library? There should be plenty of books about introductory nuclear physics. Remeber that even quite old books from the 1950-1960 is good for this.

7. Nov 16, 2007

### Nakalay

I've found my book again so no worries. There is one thing I have not yet fully understood, though. Why is it that you have to use the nuclear masses instead of the atomic masses in the calculation?

8. Nov 16, 2007

### malawi_glenn

Only the nucleus takes part in the reaction. The atomic electrons dont parcipitate.

This is reflected in the Q-value, which is defined for the nucleus mass.

Last edited: Nov 16, 2007