Consider the temperature T = 9.283620004×10° °K at which neutrons freeze-out in the early universe. The difference in ma

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Consider The Temperature T 9 283620004 10 K At Which Neutrons Freeze Out In The Early Universe The Difference In Ma 1 (164.74 KiB) Viewed 14 times

Consider the temperature T = 9.283620004×10° °K at which neutrons freeze-out in the early universe. The difference in mass between a neutron and proton is known to be (equation 9.6) mn - mp -m₂ = 2.299634877771×10-³⁰ kg. Let us now suppose that the difference in mass is 10 times smaller, i.e., m - mp = 2.299634877771×10-³¹ kg: ● ● Р What would be the equilibrium neutron to proton ratio n/n at the freeze-out temperature? Given that the electron rest mass is m = 9.109406376286×10-³¹ kg, explain why it would be energetically impossible for a neutron to decay via the weak interaction n→ p* + e¯ +V₂. Although impossible to decay via the above reaction, neutrons could decay via positron capture n+e¹ → p¹ +√¸, but even this would not happen after t»4 sec when all positrons would + e' + e' have annihilated with electrons, and so we can assume that the neutron to proton ratio remains at the number you found in the first part of this problem. Now to make a He nucleus two protons must fuse with two neutrons. Given the neutron to proton ratio you have found 4 4 He Η estimate the Helium mass fraction Y₁ = N₁ H/NH (with Hydrogen nucleus consisting of 1 proton). Use MeV for units of mass and temperature.