Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 74 / 4 / i

Table of contents
- Solution i

Solution

In the hydrogen–helium approximation, take helium to be helium-4, and assume essentially all neutrons remaining at the onset of efficient nuclear burning enter it. Each helium-4 nucleus contains two neutrons and two protons, hence

n_{He} = \frac{n _{n}}{2}, n_{H} = n_{p} - n_{n}, n_{B} = n_{p} + n_{n} .

Here

n_{p}, n_{n}

count constituent nucleons, not just free particles after burning. Neglecting the small nucleon mass difference and nuclear binding-energy correction, the helium mass is

4 m_{u} n_{He}

and the total baryon mass is

m_{u} n_{B}

. Thus neutron-limited helium synthesis gives

Y_{p} = \frac{4 n _{He}}{n _{B}} = \frac{2 n _{n}}{n _{p} + n _{n}} = 2 (1 + \frac{n _{p}}{n _{n}})^{- 1} .

For

n_{n} / n_{p} ≃ 1/7

, this yields

Y_{p} ≃ 1/4

and the hydrogen mass fraction is

X_{p} ≃ 3/4

. The assumptions require

n_{p} \geq n_{n}

and negligible free neutrons, deuterium and helium-3: the phrase hydrogen and helium must here mean predominantly hydrogen-1 and helium-4, not an arbitrary mixture of helium isotopes.

 Ancestors (10)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version