Calculating Energy Released Per Proton & Mass Converted to Energy/sec

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The discussion focuses on calculating energy released per proton during nuclear fusion in the sun, using specific fusion reactions involving hydrogen and helium isotopes. The energy released per proton is determined to be 6.55 MeV, and the sun's luminosity is given as 3.826 x 10^26 W. Participants discuss how to calculate the number of protons converted to helium each second and the corresponding mass converted to energy, arriving at a mass conversion of approximately 4.3 x 10^9 kg per second. The calculations involve using the equation E = mc² to relate energy output to mass. Overall, the thread emphasizes the importance of precise unit conversions and the steps needed to derive these values from the fusion equations.
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H1 means hydrogen with mass 1, H2 means hydrogen with mass 2 (has a neutron), etc


1) H1 + H1 -> H2 + positron + neutrino + 1.18 Mev (+0.26MeV)
2) H1 + H2 -> He3 + photon + 5.49MeV
3) He3 + He3 -> He4 + 2*H1 + 12.86 MeV

(The neutrino produced in 1 for all practical purposes do not interact with matter, so their energy can be ignored (neutrino energy 0.26MeV)

Using the above 3 equations, calculate the energy released per proton. Note that the equations must be properly combined in order to find the overall reaction! Given the sun's luminosity calculate the amount of hydrogen converted to helium every second. How much mass per second is converted directly to energy?


Homework Equations


sun's luminosity: 3.826*10^26 W

answers: 6.55 MeV, 6.2*10^11 kg, 4.3x10^9 kg.

The Attempt at a Solution



I have shown each proton has released 6.55 MeV. but how do i calculate the amount of hydrogen converted to helium every second?
 
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(I removed part of my reply, since I missed at first where you said you'd gotten the 6.55 MeV per proton.)

Having gotten that answer, you have the energy release (in MeV) from one pp-reaction. Given what the luminosity (power output) of the Sun is, how many of these reactions would be needed per second? How many protons cease to exist as individuals and become groups of four in nuclei every second? How much mass is that?

Last question -- Since you have the Sun's luminosity, what mass is that energy equivalent to (that will take you from, say, Joules per second to kg/sec)? What famous equation might you use?
 
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part c I've got, directly using E = mc² gives 4.3*10^9 kg

but i still have trouble with part b :(

Luminosity of sun is 3.826*10^26 W which is 3.826*10^26 J/s
6.55 MeV = 1.049 * 10^-12 protons/J .
 
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Qyzren said:
i still have trouble with part b :(

Luminosity of sun is 3.826*10^26 W which is 3.826*10^26 J/s
6.55 MeV = 1.049 * 10^-12 J/proton .
so if you divide it you can get 3.64*10^38 J/proton...
Watch your units: that division gives you 3.64·10^38 protons/sec. What mass does that many protons have?

just using E = mc2 to convert J to kg won't get me the answer of 6.2*10^11 kg :(

That equation is for finding the mass-equivalent of the 3.826·10^26 J/sec, which is the answer for the third question.
 
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