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 They’ll have to be travelling really fast to get close before their mutual repulsion forces them apart. 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.

4. Let’s look at the equation… Enter two hydrogen nuclei (protons)… H + H 1111 1111 … which fuse to become a deuterium nucleus (a heavier isotope of hydrogen with a mass of 2)  H 2121 plus a positron +  0101 and a neutrino +

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

6. And the equation for this? H + H  He +  2121 1111 3232

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 … He + He  He + 2 H 3232 3232 4242 1111 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 This is based on the nice, stable C nucleus having a mass of 12u 12 6 which means 1u = 1.660 540 x 10 -27 kg…ish

10. Doing the sums (with a few less sig figs…) Oh! Some of the mass seems to have disappeared! 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 = 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 = mc 2 whereE = energy released m = mass deficit c = speed of light  E= 4.40 x 10 -29 x (3.0 x 10 8 ) 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 10 26 W (and remember, 1W = 1 Js -1 ) so how many of these fusions are taking place per second?

13. Energy released per fusion = 3.96 x 10 -12 J  no. fusions = 3.8 x 10 26 3.96 x 10 -12 The luminosity of the Sun is 3.8 x 10 26 W = 3.8 x 10 26 J s -1 = 9.6 x 10 37 per sec At this rate, how long will the Sun last?

14. 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 10 37 = 4.22 x 10 9 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 10 9 kg mass of the Sun= 1.99 x 10 30 kg  lifetime of Sun= 1.99 x 10 30 4.22 x 10 9 = 4.71 x 10 30 s = 1.49 x 10 13 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 10 13 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 10 13 = 4.5 x 10 9 years