Energy generated by proton-proton cycle

  • Thread starter lovelyrwwr
  • Start date
  • #1
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My attempt

Q Reaction 1:
Q1 = 2(1.007825) - (2.014102 + 0.000549) = 9.99E-4 u

Q Reaction 2:
Q2 = 1.007825 + 2.014102 - 4.002603 = -0.980676 u

Q Reaction 3:
Q3 = 3.016030 - (4.002603 + 2(1.007825)) = 0.013807 u
---------
2(Q1) + 2(Q2) + Q3 = -1.945547 u

(-1.945547 u)(931.5 MeV) = -1812.28 MeV

But it is not supposed to be negative? Thanks for any help.
 

Answers and Replies

  • #2
6,054
390
The second equation given in the problem is wrong. See if you can figure out why and how to correct it.
 
  • #3
346
48
One of the isotopes in equation 2 is wrong.

Also note that ion's fuse not neutral atoms! You should really be using the rest mass of the ion, not the rest mass of the atom. You can approximate this by subtracting the mass of the neutral atom by the mass of an electron.

To see why this important we should really write equation 1 as
[itex]\stackrel{1}{1}{H^{+}}+e^- +\stackrel{1}{1}H^{+}+e^-=\stackrel{2}{1}H^{+}+2e^-+e^+ +\nu[/itex]

From this we can calculate Q1
[itex]Q1=2mass(\stackrel{1}{1}{H^{+}}) +2mass(e^-) -mass(\stackrel{2}{1}H^{+}) - 2 mass(e^-) - mass(e^+)[/itex]

If we approximate the mass of the ion as the sum of the masses of the electron and the neutral atom we then get
[itex]Q1=2mass(\stackrel{1}{1}{H} ) -mass(\stackrel{2}{1}H) - mass(e^-) - mass(e^+)[/itex]
 
  • #4
34,467
10,587
In the core of our sun, nearly all hydrogen and helium atoms are ionized.
 

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