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kogti [31]
3 years ago
6

Imagine an alternative universe where the characteristic decay time of neutrons is 3 min instead of 15 min. All other properties

remain unchanged from what was discussed in class.
(a) Show that the maximum possible value of the primordial helium fraction is: Ymax = 2f 1 + f (1) where f = nn/np ≥ 1 is the neutron-to-proton ratio at the time of nucleosynthesis. Prove that this assertion is true.
(b) Estimate the maximum fraction of the baryonic matter in the form of helium today. You may assume all He nuclei exist in the form of helium 4
Physics
1 answer:
FrozenT [24]3 years ago
6 0

Answer:

(a) [Y_{p} ]_{max} = \frac{2f}{1+f}

(b) f_{new} = 0.013; [Y_{p} ]_{max} = 0.026

Explanation:

Since the neutron-to-proton ratio at the time of nucleosynthesis is given:

f = \frac{n_{n} }{n_{p} }

Therefore:

n_{n} = f*n_{p}

Then, to determine the maximum ⁴He fraction if all the available n_{n} neutrons bind to all the protons. Since, there are 2 protons and 2 neutrons in a ⁴He nucleus, it shows that there would be n_{n}/2 nuclei of ⁴He.

In addition, a ⁴He nucleus has a mass of 4m_{p}, where m_{p} is the mass of one proton. Thus, n_{n}/2 nuclei of such nuclei will have a mass of n_{n}/2*4m_{p}.

Assuming that m_{p}=m_{n}, there would be a total of (n_{n}+n_{p}) protons and neutrons with a total mass of (n_{n}+n_{p})*m_{p}.

Thus:[Y_{p} ]_{max} = \frac{2f}{1+f}

(b) Given:

t_{nuc} = 200 s;   τ_{n} = 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

[Y_{p} ]_{max} = \frac{2f}{1+f} = (2*0.013)/(1+0.013) = 0.026

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Answer:

The answer to your question is:

a)  t1 = 2.99 s ≈ 3 s

b)  vf = 39.43 m/s

Explanation:

Data

vo = 10 m/s

h = 74 m

g = 9.81 m/s

t = ?   time to reach the ground

vf = ?   final speed

a)    h = vot + (1/2)gt²

     74 = 10t + (1/2)9.81t²

     4.9t² + 10t -74 = 0                  solve by using quadratic formula

   

   t = (-b ± √ (b² -4ac) / 2a

   t = (-10 ± √ (10² -4(4.9(-74) / 2(4.9)

   t = (-10 ± √ 1550.4 ) / 9.81

  t1 = (-10 + √ 1550.4 ) / 9.81               t2 = (-10 - √ 1550.4 ) / 9.81

  t1 = (-10 ± 39.38 ) / 9.81                    t2 = (-10 - 39.38) / 9.81

   t1 = 2.99 s ≈ 3 s                               t2 = is negative then is wrong there are

                                                                   no negative times.

b) Formula vf = vo + gt

                  vf = 10 + (9.81)(3)

                  vf = 10 + 29.43

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