Energy from fusion - “that” equation

1. Energy from fusion - “that” equation

2. The energy from stars comes from nuclear fusion in the core. Light nuclei fuse together & release energy - it takes less “binding energy” to hold the slightly bigger nucleus together than it did to hold the separate pieces together. For hydrogen fusing into helium it’s a three stage process, called the p-p process because it starts with a couple of protons (aka 2 hydrogen nuclei).

3. Let’s start with 2 hydrogen nuclei - protons As they fuse, one of the protons emits a positron (a sort of anti-electron - exactly like an electron but positive). Having lost it’s positive charge it’s now neutral - it’s become a neutron. Did you notice the other little particle fly out at the same time? That was a neutrino, an almost massless particle with no charge. It just carries away a bit of excess energy. They’ll have to be travelling really fast to get close before their mutual repulsion forces them apart.

4. Let’s look at the equation… Enter two hydrogen nuclei (protons)… 1 1 1 1 2 1 0 1 H + H  H +  +  … which fuse to become a deuterium nucleus (a heavier isotope of hydrogen with a mass of 2) plus a positron and a neutrino

5. Stage 2 - the deuterium fuses with another hydrogen… … to make the isotope of helium with a mass of 3 … … and this releases a bit more energy in a little photon of light (or more likely gamma waves).

6. And the equation for this? 2 1 3 2 1 1 H + H  He + 

7. Stage 3 Finally two of these helium-3’s collide and fuse into a stable helium-4, and shedding the two spare protons…

8. And this equation is … 3 2 3 2 4 2 1 1 He + He  He + 2 H So over the three stages, we’ve effectively had… 4 hydrogen nuclei  1 helium nucleus + energy But how do we work out the energy released?

9. To measure the masses of things in the nucleus, we don’t use kg because that’s far too big a unit. Instead we use the atomic mass unit, u 12 6 This is based on the nice, stable C nucleus having a mass of 12u which means 1u = 1.660 540 x 10-27 kg …ish

10. Doing the sums (with a few less sig figs…) mass of hydrogen nucleus = 1.007276 u  mass of 4 hydrogen nuclei = 4.029104 u mass of helium & 2 positrons = 4.002603 u - ________ 0.0265… u Oh! Some of the mass seems to have disappeared! = 4.40 x 10-29 kg

11. We have a missing mass ( a “mass deficit”) = 4.40 x 10-29 kg Here comes that equation… E = mc2 where E = energy released m = mass deficit c = speed of light  E = 4.40 x 10-29 x (3.0 x 108)2 = 3.96 x 10-12 J

12. Energy released = 3.96 x 10-12 J That doesn’t seem very big, but remember - that’s the energy released by just 4 hydrogen nuclei (protons) fusing into 1 helium. The luminosity of the Sun is 3.8 x 1026 W (and remember, 1W = 1 Js-1) so how many of these fusions are taking place per second?

13. The luminosity of the Sun is 3.8 x 1026 W = 3.8 x 1026 J s-1 Energy released per fusion = 3.96 x 10-12 J  no. fusions = 3.8 x 1026 3.96 x 10-12 = 9.6 x 1037 per sec At this rate, how long will the Sun last?

14. At this rate, how long will the Sun last? Remember, the mass deficit every time one of these processes happens was 4.40 x 10-29 kg  mass loss per sec = mass deficit x no. per sec. = 4.40 x 10-29 x 9.6 x 1037 = 4.22 x 109 kg - that’s 4¼ billion tonnes disappearing per second!

15. At this rate, how long will the Sun last?  mass loss per sec = 4.22 x 109 kg mass of the Sun = 1.99 x 1030 kg  lifetime of Sun = 1.99 x 1030 4.22 x 109 = 4.71 x 1030 s = 1.49 x 1013 years But a deeper understanding of astrophysics suggests that fusions at the core will die out when just over 0.0003 of its mass has been lost.

16. At this rate, how long will the Sun last?  lifetime of Sun = 1.49 x 1013 years But a deeper understanding of astrophysics suggests that fusions at the core will die out when just over 0.0003 of its mass has been lost.  lifetime of Sun as a star doing fusion = 0.0003 x 1.49 x 1013 = 4.5 x 109 years