<blockquote>
<p>The elements will give an average of 27/56 protons which will turn to neutrons. I have multiplied this with number of neutrons (n=1.4*solar mass/mass of neutron) but that doesn't seem to give any results.</p>
</blockquote>
<p>That should give you a result. I get <a href="https://www.wolframalpha.com/input?i=27%2F56*%281.4*solar+mass%29%2F%28mass+of+neutron%29" rel="nofollow noreferrer"><span class="math-container">$8.01\times10^{56}$</span></a>, which is very close to the given answer.</p>

CC BY-SA 4.0
2024-10-30T14:37:49.507Z
I got the same answer as well, but there are other available options which are closer approximations to 8.01*10^56.  Hence I was wondering if my method is incorrect or if there is another way to approach the problem.

For reference, these are the options:
a) 7.77*10^56
b) 1.74*10^57
c)8.68*10^56
d)8.38*10^57
Anupama Mukherjee
CC BY-SA 4.0
2024-10-30T14:45:57.930Z
@AnupamaMukherjee  How did you get 27/56? That said, this was a multiple guess question and only two are close to this rough estimate. I would have chosen (a) as being the closest to my rough estimate.
David Hammen
CC BY-SA 4.0
2024-10-30T14:47:56.910Z
We have Fe,Co, Ni are the given elements and each of them has a mass number 56, hence the average mass is 56. Next, they give 26,27,28 protons and the average comes out to be 27.
Anupama Mukherjee
CC BY-SA 4.0
2024-10-30T14:58:27.717Z
That the three elements in question are split evenly is not a good assumption.
David Hammen
CC BY-SA 4.0
2024-10-30T15:05:09.297Z
Is there any other approach to it, then? I have assumed that they split equally since it is not mentioned in the problem statement. 

Whereas if we assume the ratio to be 26/56 the answer is a very close 7.84*10^56. However, I don't know the reasoning for the ratio to be 26/56.
Anupama Mukherjee
CC BY-SA 4.0
2024-10-30T16:13:04.580Z
@AnupamaMukherjee It would be better to assume almost all iron. Iron is far more abundant than nickel, which in turn is far more abundant than cobalt. Also, the second most abundant isotope of iron is $^{54}\text{Fe}$, which is going to lower the ratio to a bit less than 26/56.
David Hammen
CC BY-SA 4.0
2024-10-31T06:39:51.653Z
26/56 because it calls it an "iron core" in the question.
ProfRob
CC BY-SA 4.0
2024-10-30T12:29:25.840Z
2024-10-30T12:29:25.840Z
David Hammen
<p>I have come across the following problem in an Astronomy Olympiad:</p>
<blockquote>
<p>Formation of the iron peak elements Fe,Co and Ni by nuclear fusion marks the end of energy production in a star. If the mass of the iron core is greater than the Chandrasekhar mass (1.4 solar masses) the core cannot be supported by the degeneracy pressure of the electrons and will collapse to form a neutron star. Matter is crushed to nuclear densities with protons combining with electrons as p+e→ n + ve. Neutron stars whose masses are not too high are supported by neutron degeneracy pressure against gravitational collapse. The number of neutrinos released, when a neutron star of 1.4 solar masses forms, is approximately?</p>
</blockquote>
<p>My attempt:
The elements will give an average of 27/56 protons which will turn to neutrons. I have multiplied this with number of neutrons (n=1.4*solar mass/mass of neutron) but that doesn't seem to give any results.</p>
<p>The correct answer is given: 7.77*10^56</p>

CC BY-SA 4.0
2024-10-30T11:22:49.953Z
The info in the problem statement appears to neglect the huge numbers of thermal neutrinos (and antineutrinos), and only considers the neutrinos released when protons convert to neutrons. So it's a rather artificial problem. For info on the thermal neutrinos, please see https://astronomy.stackexchange.com/q/57832/16685 & links therein.
PM 2Ring
CC BY-SA 4.0
2024-10-30T12:32:31.743Z
@PM2Ring Yes, the question is a bit artificial. So? Lots of introductory level questions in the sciences might as well start with "assume a spherical cow ..."
David Hammen
CC BY-SA 4.0
2024-10-30T12:36:54.417Z
@David Sure. My comment is just a warning to the OP (& other interested readers) that there are other neutrinos to consider in a supernova.
PM 2Ring
<neutron-star>
How to calculate the number of neutrinos during neutron star formation?
Anupama Mukherjee
# How to calculate the number of neutrinos during neutron star formation?

(Asked by: username_0 on 2024-10-30)

I have come across the following problem in an Astronomy Olympiad:

> Formation of the iron peak elements Fe,Co and Ni by nuclear fusion marks the end of energy production in a star. If the mass of the iron core is greater than the Chandrasekhar mass (1.4 solar masses) the core cannot be supported by the degeneracy pressure of the electrons and will collapse to form a neutron star. Matter is crushed to nuclear densities with protons combining with electrons as p+e→ n + ve. Neutron stars whose masses are not too high are supported by neutron degeneracy pressure against gravitational collapse. The number of neutrinos released, when a neutron star of 1.4 solar masses forms, is approximately?

My attempt:
The elements will give an average of 27/56 protons which will turn to neutrons. I have multiplied this with number of neutrons (n=1.4\*solar mass/mass of neutron) but that doesn't seem to give any results.

The correct answer is given: 7.77\*10^56

**Tags:** neutron-star
> **Comments:**
> **username_1**: The info in the problem statement appears to neglect the huge numbers of thermal neutrinos (and antineutrinos), and only considers the neutrinos released when protons convert to neutrons. So it's a rather artificial problem. For info on the thermal neutrinos, please see https://astronomy.stackexchange.com/q/57832/16685 & links therein.  
> **username_2**: @PM2Ring Yes, the question is a bit artificial. So? Lots of introductory level questions in the sciences might as well start with "assume a spherical cow ..."  
> **username_1**: @David Sure. My comment is just a warning to the OP (& other interested readers) that there are other neutrinos to consider in a supernova.  

---

## 1 answers

### Answer by username_2 on 2024-10-30. Score: 3 | Accepted answer

> The elements will give an average of 27/56 protons which will turn to neutrons. I have multiplied this with number of neutrons (n=1.4\*solar mass/mass of neutron) but that doesn't seem to give any results.

That should give you a result. I get [$8.01\times10^{56}$](https://www.wolframalpha.com/input?i=27%2F56*%281.4*solar+mass%29%2F%28mass+of+neutron%29), which is very close to the given answer.

> **Comments:**
> **username_0**: I got the same answer as well, but there are other available options which are closer approximations to 8.01*10^56.  Hence I was wondering if my method is incorrect or if there is another way to approach the problem.  For reference, these are the options: a) 7.77*10^56 b) 1.74*10^57 c)8.68*10^56 d)8.38*10^57  
> **username_2**: @AnupamaMukherjee  How did you get 27/56? That said, this was a multiple guess question and only two are close to this rough estimate. I would have chosen (a) as being the closest to my rough estimate.  
> **username_0**: We have Fe,Co, Ni are the given elements and each of them has a mass number 56, hence the average mass is 56. Next, they give 26,27,28 protons and the average comes out to be 27.  
> **username_2**: That the three elements in question are split evenly is not a good assumption.  
> **username_0**: Is there any other approach to it, then? I have assumed that they split equally since it is not mentioned in the problem statement.   Whereas if we assume the ratio to be 26/56 the answer is a very close 7.84*10^56. However, I don't know the reasoning for the ratio to be 26/56.  
> **username_2**: @AnupamaMukherjee It would be better to assume almost all iron. Iron is far more abundant than nickel, which in turn is far more abundant than cobalt. Also, the second most abundant isotope of iron is $^{54}\text{Fe}$, which is going to lower the ratio to a bit less than 26/56.  
> **username_3**: 26/56 because it calls it an "iron core" in the question.  

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