Answer:
(a) ![[Y_{p} ]_{max} = \frac{2f}{1+f}](https://tex.z-dn.net/?f=%5BY_%7Bp%7D%20%5D_%7Bmax%7D%20%3D%20%5Cfrac%7B2f%7D%7B1%2Bf%7D)
(b)
;
= 0.026
Explanation:
Since the neutron-to-proton ratio at the time of nucleosynthesis is given:

Therefore:

Then, to determine the maximum ⁴He fraction if all the available
neutrons bind to all the protons. Since, there are 2 protons and 2 neutrons in a ⁴He nucleus, it shows that there would be
nuclei of ⁴He.
In addition, a ⁴He nucleus has a mass of
, where
is the mass of one proton. Thus,
nuclei of such nuclei will have a mass of
*
.
Assuming that
, there would be a total of
protons and neutrons with a total mass of
.
Thus:![[Y_{p} ]_{max} = \frac{2f}{1+f}](https://tex.z-dn.net/?f=%5BY_%7Bp%7D%20%5D_%7Bmax%7D%20%3D%20%5Cfrac%7B2f%7D%7B1%2Bf%7D)
(b) Given:
; τ
= 3*60s = 180 s
![f_{new} = \frac{n_{nf} }{n_{pf} } = \frac{exp (-200/180)}{5 +[1- exp(-200/180)]} =\frac{0.077}{5.923} = 0.013](https://tex.z-dn.net/?f=f_%7Bnew%7D%20%3D%20%5Cfrac%7Bn_%7Bnf%7D%20%7D%7Bn_%7Bpf%7D%20%7D%20%3D%20%5Cfrac%7Bexp%20%28-200%2F180%29%7D%7B5%20%2B%5B1-%20exp%28-200%2F180%29%5D%7D%20%3D%5Cfrac%7B0.077%7D%7B5.923%7D%20%3D%200.013)
= (2*0.013)/(1+0.013) = 0.026