Answer:
(a) [tex][Y_{p} ]_{max} = \frac{2f}{1+f}[/tex]
(b) [tex]f_{new} = 0.013[/tex]; [tex][Y_{p} ]_{max} [/tex] = 0.026
Explanation:
Since the neutron-to-proton ratio at the time of nucleosynthesis is given:
[tex]f = \frac{n_{n} }{n_{p} }[/tex]
Therefore:
[tex]n_{n} = f*n_{p}[/tex]
Then, to determine the maximum ⁴He fraction if all the available [tex]n_{n}[/tex] neutrons bind to all the protons. Since, there are 2 protons and 2 neutrons in a ⁴He nucleus, it shows that there would be [tex]n_{n}/2[/tex] nuclei of ⁴He.
In addition, a ⁴He nucleus has a mass of [tex]4m_{p}[/tex], where [tex]m_{p}[/tex] is the mass of one proton. Thus, [tex]n_{n}/2[/tex] nuclei of such nuclei will have a mass of [tex]n_{n}/2[/tex]*[tex]4m_{p}[/tex].
Assuming that [tex]m_{p}=m_{n}[/tex], there would be a total of [tex](n_{n}+n_{p})[/tex] protons and neutrons with a total mass of [tex](n_{n}+n_{p})*m_{p}[/tex].
Thus:[tex][Y_{p} ]_{max} = \frac{2f}{1+f}[/tex]
(b) Given:
[tex]t_{nuc} = 200 s[/tex]; τ[tex]_{n}[/tex] = 3*60s = 180 s
[tex]f_{new} = \frac{n_{nf} }{n_{pf} } = \frac{exp (-200/180)}{5 +[1- exp(-200/180)]} =\frac{0.077}{5.923} = 0.013[/tex]
[tex][Y_{p} ]_{max} = \frac{2f}{1+f}[/tex] = (2*0.013)/(1+0.013) = 0.026
