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Hydrogen Burning and the Suns Nuclear Fusion

  1. Oct 9, 2008 #1
    hey guys, im new here and thought you might like to help me out. im a first year astronomy student at Curtin University in Western Australia.

    my current assignment has asked me to calculate the mass of hydrogen converted in the sun to helium and energy over 10 billion years (10 billion years being its lifespan on the HR diagram). (PS this is Question 5 of 8 haha)

    now, i understand peoples reluctance to help others who dont prove they have done any work themselves. which is lovely, considering youll now be reading all my own calculations and the result im getting!

    my problem is that my amount of hydrogen burnt in 10 billion years seems to be only 14% of the mass of hydrogen in the sun now. shouldnt it be MUCH more than 14%? stars are supposed to burn of all of their hydrogen by the time they hit red giant phase.

    without further ado, and recognising my use of the letter e in figures is scientific notation (10 to the power of) and nothing to do with natural logs.

    FIRSTLY, the amount of seconds in 10 billion years.

    60 x 60 = 3600 seconds per hour

    3600 x 24 = 86400 seconds per day

    86400 x 365 = 31536000 seconds per non leap year
    86400 x 366 = 31622400 seconds per leap year.

    since 10 billion years will consist of 75% non leap years and 25% leap years:

    0.75 x 10 billion years = 7.5 e 9 years
    0.25 x 10 billion years = 2.5 e 9 years


    (2.5e9 x 31622400) + (7.5e9 x 31536000) = 3.16224 e 17 seconds in 10 billion years

    SECONDLY since the sun emits 3.84e26 joules per second of energy, the total power emitted over 10 billion years would be:

    3.84 e 26 x 3.16224 e 17 = 1.214 e 44 watts (or joules per second) of power in 10 billion years we will need this figure later.

    THIRDLY now that we have the total power output of the sun over 10 billion years we can use e=mc^2 to find out how much mass has been converted into energy in that time.

    1.214 e 44 = m (3 e 8)^2

    rearranging for m, gives m = 1.349 e 27 kilograms of mass converted to energy in the sun over 10 billion years

    FOURTH so now we get onto the heavy stuff. knowing the masses of the hydrogen atom and helium atoms, and the reaction in the sun which converts the former to the latter + energy, i should be able to get a figure for the total hydrogen burnt. lets give it ago!

    The fusion reaction summed up is 4 Hydrogen + 2 electrons ----> 1 Helium + 2 neutrinos + 6 photons. Now, remember this is a summary of the 3 steps, but they arent important. only need to know the input and final output.

    the mass of a hydrogen atom is 1.673 e -27
    the mass of a helium atom is 6.647 e -27

    now there are 4 hydrogens going into the reaction so:

    4 x 1.673 e -27 = 6.692 e -27 the mass of hydrogen going into a single full reaction


    6.692 e -27 = (6.647 e -27 + X) with X being the amount of mass converted to energy per reaction.

    rearranging for X gives X = 4.5 e -29 kg mass converted into energy per reaction.

    FIFTH so if the total mass converted to energy over 10 billion years is 1.349 e 27, and we know now that the mass being converted to energy per full reaction is 4.5 e -29 then:

    1.349 e 27 / 4.5 e -27 = 2.998 e 55 total reactions in 10 billion years

    so if 6.692 e -27 kg of hydrogen goes into a single reaction, and there is 2.998 e 55 total reactions, then:

    6.692 e -27 x 2.998 e 55 = 2.006 e 29 kg of hydrogen converted into helium and energy in 10 billion years

    so now you see my problem. if sun right now has 70% hydrogen, making that total hydrogen to be 1.393 e 30 kg of hydrogen in the sun right now, then how can the hydrogen burnt over 10 billion years be only one seventh (14%) of the mass of hydrogen right now?
    Last edited: Oct 9, 2008
  2. jcsd
  3. Oct 9, 2008 #2
    It's been a while since I did this stuff, but I seem to remember calculations like this indicating that hydrogen fusion is not the sole energy source. Seems like I remember something about a Carbon-Nitrogen cycle as well? In other words, perhaps the inconsistency you mention was evidence that there was something else going on; i.e. that the power output of the sun was not due solely to H fusion?

    If that's not the answer then I'm as confused as you, because I read somewhere the sun is only expected to last another 5 B years.
  4. Oct 9, 2008 #3
    merryjman: youve got me thinking now :D (or :(( im not sure)

    i reread my question, and it told me to assume hydrogen->helium is the only nuclear reaction occuring in the sun. there is another reaction which involved helium, beryllium, boron and back to helium, and there is also the CNO cycle which you mentioned. the problem is, even if i was to consider these, which im told not to, i would still expect my figure for the burning of hydrogen over 10 billion years from the main reaction to equate to something much more than 14% of the suns current mass of hydrogen.

    consider that the sun is approximately 5 billion years old, and currently has a total mass of 1.99 e 30 kg. hydrogen is 70% of this, so thats 1.393 e 30 kg of hydrogen. double this, and youve got the amount of hydrogen the sun started with at time=0 years.

    thats 2.786 e 30 kg of hydrogen. now, if stars are supposed to burn all of that hydrogen in their lifetime (in this case 10 billion years for the sun), my sun is apparently burning only 7.2% of that amount in 10 billion years.

    now even considering the CNO and beryllium/boron/helium reactions included, youd still expect the % of hydrogen burnt from the primary reaction to be exceedingly larger than 7.2%
  5. Oct 9, 2008 #4


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  6. Oct 9, 2008 #5
    so...after thinking that stars burn most if not all of their hydrogen over that 10 billion year period, im actually right in calculating that the sun will only burn 7% of its hydrogen?

    many MANY thanks to both merryjman and mgb_phys for putting me back on the right track, and possibly saving me the enbarressment of my lecturer telling me im actually correct
  7. Oct 9, 2008 #6


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    It's a few years since I last slept through stellar evolution but I seem to remember about 10% H used

    Your PM Question.
    Because of the radiation pressure the material in the outer layers doesn't fall into the core to be burnt. The burn spreads out into outer layers leving behind the end-products of earlier fusion. As the fuel in the inner layers is used up, the radiation pressure decreases, the star begins to collapse, the pressure and temperature in the core rises and a new reaction kicks off. Ultimately you end up with an onion structure with different layers burning different material.

    (burning of course means nuclear fusion in stellar terms)
  8. Oct 10, 2008 #7
    mgb to the rescue! So now, I've got a side question - if sun-like stars only burn ~10% of their H fuel, then why do we think our sun will become a red giant? I was always under the impression that stars became giants because their primary H->He reaction was done. Does this have to do with the onion thing you mentioned; i.e. the H in the star's center is used up and a new reaction begins from that center? What would that reaction be?
  9. Oct 10, 2008 #8


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    When a star's core is depleted of hydrogen, the core contracts (due to gravity), and the star begins fusing hydrogen in a thin shell outside the core. As it turns out, there's virtually no mixing between the outer layers and the core -- hydrogen that begins near the surface of the star stays there forever and never gets fused.

    The star becomes a red giant because its enormous luminosity creates enormous radiation pressure, which expands the outer layers of the star. The average surface temperature a red giant is low, but the total luminosity is immense.

    - Warren
  10. Oct 11, 2008 #9
    thanks, all, useful and interesting information.
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