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High temperatures, nucleons, early universe

  1. May 27, 2013 #1
    1. The problem statement, all variables and given/known data
    At very high temperatures (as in the very early universe), the proton ad neutron can be thought of as two different states of the same particle, called the "nucleon".
    (The reactions that convert a proton to a neutron or vice versa require the absorption of an electron or a positron or a neutrino, but all of these particles tend to be very abundant at sufficiently high temperatures)
    Since the neutron's mass is higher than the proton's by ## 2.3 x 10^{-30} ## kg, it's energy is higher by this amount by ## c^2 ##.
    Suppose, then, that at some very early time, the nucleons were in thermal equilibrium with the rest of the universe at ## 10^{11} ## K.
    What fraction of the nucleons at that time were protons and what fraction were neutrons?

    2. Relevant equations

    ## E = mc^2 ##

    ## \dfrac{\rho (s_2)}{\rho (s_1)} = e^{-[E(s_2) - E(s_1)]/k T} ##

    3. The attempt at a solution

    So it's given that the difference in energy is:

    ## E_2 - E_1 = 2.3 x 10^{-30} \cdot 300000000^2 ##
    ## E_2 - E_1 = 2.07 \cdot 10^{-13} ##

    And the ratio of probabilities is given by:

    ## \dfrac{\rho (s_2)}{\rho (s_1)} = e^{-[E(s_2) - E(s_1)]/k T} ##

    ## \dfrac{\rho (s_2)}{\rho (s_1)} = e^{-[2.07 \cdot 10^{-13}]/[1.381 \cdot 10^-23 \cdot 10^{11}]} ##

    ## \dfrac{\rho (s_2)}{\rho (s_1)} = \dfrac{0.860801...}{1} ##

    Does that seem right?

    Converting that to fractions, I get:
    Fraction of Neutrons: 0.462597
    Fraction of Protons: 0.537402
     
  2. jcsd
  3. May 27, 2013 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Seems about right. But you should put units on numbers that have dimensions.
     
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