# Calculating the energy of anti neutrino

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1. Apr 28, 2017

### Rahulrj

1. The problem statement, all variables and given/known data
A free neutron is unstable and decays into proton electron and an anti neutrino. The rest masses of these particles are $m_n = 939.6 MeV, m_p = 938.3 MeV m_e = 0.51 MeV$and $m_{\nu} = 0$ so that the change in the total rest mass in the decay is 0.79 MeV. If in a particular decay, the neutron and proton created are at rest, the energy of the anti neutrino is?

2. Relevant equations
$Q = m_i*c^2 - \sum m_f*c^2$

3. The attempt at a solution
I was solving a question paper that has got random topics and I do not have much idea on this problem using the above equation that I only know. I would appreciate if someone could provide a link to the associated theory or examples that is pretty much self sufficient to do similar problems.

2. Apr 28, 2017

### phyzguy

Assuming the neutron and proton are at rest, both the electron and the anti-neutrino carry away energy and momentum. If you assume that energy and momentum are conserved, you should be able to solve for the energies of the outgoing electron and anti-neutrino.

3. Apr 29, 2017

### Rahulrj

Okay, I tried doing it but I am not able to get the right answer. This is how I did it:
$E_n = E_p +E_e+E_{\nu}$
Since neutron and proton are at rest, The E values for them would be the rest mass energy therefore
$1.3 c^2 = E_e+ E_{\nu}$ I am not quite sure about the next step but that's the only way I found
so now $1.3 c^2 = T_{e}+E_{0e}+T_{\nu} + E_{0\nu}$ and I can substitute the following $E_{0\nu} = 0$and $T_{\nu} = 0$(since $T = (\gamma -1)m_{0\nu}c^2$ $where m_{0\nu} = 0$)
I can now find $\gamma$
$0.79 c^2 = T_e$ and $\gamma = 2.54$
using this I can substitute values in the main equation to get
$1.3c^2-2.54*.51 c^2 = E_{\nu}$
and the answer I get is $4.6*10^{-3} MeV$ which is no where close to any of the options given.
So where have I gone wrong?

4. Apr 29, 2017

### kuruman

The change in rest mass energy is 939.6 - (938.6 + 0.511) MeV = 0.79 MeV. You forgot to include the rest mass of the electron. Also, you state that the neutrino has zero energy. That can't be. If you assume that it is massless, then its energy is related to momentum by E = pc. You cannot solve this problem by energy conservation alone; you also have to conserve momentum as @physguy suggested.

5. Apr 29, 2017

### Rahulrj

Since its been given that neutron and proton are at rest, why is $939.6-938.6 = E_e+E_{\nu}$ incorrect?

6. Apr 29, 2017

### kuruman

$n \rightarrow p+e+\nu$
What's the rest mass difference before minus after?

7. Apr 29, 2017

### Rahulrj

I did not get what you meant.

8. Apr 29, 2017

### kuruman

The total rest mass before the decay is 939.6 MeV (neutron)
The total rest mass after the decay is 938.3 MeV + 0.511 MeV = 938.81 MeV (proton + electron)

What do you get when you subtract the second number from the first? Answer: 0.79 MeV as given by the problem.

9. Apr 29, 2017

### Rahulrj

Yes Indeed but then isn't the statement that proton and neutron at rest relevant to be considered? if its not to be considered why then is it even mentioned saying that in a 'particular decay' it is so?
What I meant by writing like this $939.6-938.3 = E_e+E_{\nu}$is that the difference in rest mass energy of proton and neutron equals the sum of total energy (KE+rest mass) of electron and anti neutrino and this makes sense to me as per the question. The energy and momentum as phyzguy said are carried away by electron and anti neutrino. Please point out if I am wrong in this, I cannot think it other than this way.

10. Apr 29, 2017

### Rahulrj

I got an answer and I am not sure if its correct so I'll post it here:
$p_e = -p_{\nu}$ so I shall refer to them both as p (equal magnitude)
$1.3c^2 = E_e+E_{\nu}$
$E_e = \sqrt {(pc)^2+(m_ec^2)^2}$
substituting it into the energy equation
$1.3c^2=\sqrt {(pc)^2+(m_ec^2)^2}+pc$
bringing pc to the LHS
$pc = \sqrt {(pc)^2+(m_ec^2)^2}-1.3c^2$
and after some alegbra
I find $p = 0.55c$ I sense an inconsistency in the units here however in the question masses are given in MeV.
and since the p's are same for electron and anti neutrino
$E_{\nu} = 0.55 c^2$ if masses were given in the units of MeV/c^2 as I think it should be the answer would then be 0.55 MeV.
It took me a while to figure out why the units were inconsistent until I saw it in the question.

11. Apr 30, 2017

### kuruman

Actually you should look at it as $E_{\nu}=pc=0.55c^2$. I got the same answer. Note that the numbers, such as 1.3, have units of MeV/c2, so 1.3 c2 translates to 1.3 MeV.

It is interesting to note that the 0.79 MeV given by the problem is not used. I think the problem expects one to use the 0.79 MeV as the sum of the kinetic energy of the electron and the neutrino energy in a non-relativistic calculation. However the relativistic calculation shows that the energy of the electron is 1.3 - 0.55 = 0.75 MeV. This implies γ = 1.5 which should be in the relativistic regime.

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