energy from fusion - “that” equation
DESCRIPTION
Energy from fusion - “that” equation. 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. - PowerPoint PPT PresentationTRANSCRIPT
![Page 1: Energy from fusion - “that” equation](https://reader031.vdocuments.us/reader031/viewer/2022032106/56812ce2550346895d91a5fa/html5/thumbnails/1.jpg)
Energy from fusion - Energy from fusion - “that” equation“that” equation
![Page 2: Energy from fusion - “that” equation](https://reader031.vdocuments.us/reader031/viewer/2022032106/56812ce2550346895d91a5fa/html5/thumbnails/2.jpg)
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).
![Page 3: Energy from fusion - “that” equation](https://reader031.vdocuments.us/reader031/viewer/2022032106/56812ce2550346895d91a5fa/html5/thumbnails/3.jpg)
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.
![Page 4: Energy from fusion - “that” equation](https://reader031.vdocuments.us/reader031/viewer/2022032106/56812ce2550346895d91a5fa/html5/thumbnails/4.jpg)
Let’s look at the equation…
Enter two hydrogen nuclei (protons)…
H + H
1
1
1
1
… which fuse to become a deuterium nucleus (a heavier isotope of hydrogen with a mass of 2)
H2
1
plus a positron
+ 0
1
and a neutrino
+
![Page 5: Energy from fusion - “that” equation](https://reader031.vdocuments.us/reader031/viewer/2022032106/56812ce2550346895d91a5fa/html5/thumbnails/5.jpg)
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 …
![Page 6: Energy from fusion - “that” equation](https://reader031.vdocuments.us/reader031/viewer/2022032106/56812ce2550346895d91a5fa/html5/thumbnails/6.jpg)
And the equation for this?
H + H He +
2
1
1
1
3
2
![Page 7: Energy from fusion - “that” equation](https://reader031.vdocuments.us/reader031/viewer/2022032106/56812ce2550346895d91a5fa/html5/thumbnails/7.jpg)
Stage 3
Finally two of these helium-3’s collide and fuse into a stable helium-4, and shedding the two spare protons…
![Page 8: Energy from fusion - “that” equation](https://reader031.vdocuments.us/reader031/viewer/2022032106/56812ce2550346895d91a5fa/html5/thumbnails/8.jpg)
And this equation is …
He + He He + 2 H
3
2
3
2
4
2
1
1
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?
![Page 9: Energy from fusion - “that” equation](https://reader031.vdocuments.us/reader031/viewer/2022032106/56812ce2550346895d91a5fa/html5/thumbnails/9.jpg)
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
![Page 10: Energy from fusion - “that” equation](https://reader031.vdocuments.us/reader031/viewer/2022032106/56812ce2550346895d91a5fa/html5/thumbnails/10.jpg)
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
![Page 11: Energy from fusion - “that” equation](https://reader031.vdocuments.us/reader031/viewer/2022032106/56812ce2550346895d91a5fa/html5/thumbnails/11.jpg)
We have a missing mass ( a “mass deficit”)
= 4.40 x 10-29 kg
Here comes that equation…
E = mcE = mc22
where E = energy releasedm = mass deficitc = speed of light
E = 4.40 x 10-29 x (3.0 x 108)2 = 3.96 x 10-12 J
![Page 12: Energy from fusion - “that” equation](https://reader031.vdocuments.us/reader031/viewer/2022032106/56812ce2550346895d91a5fa/html5/thumbnails/12.jpg)
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?
![Page 13: Energy from fusion - “that” equation](https://reader031.vdocuments.us/reader031/viewer/2022032106/56812ce2550346895d91a5fa/html5/thumbnails/13.jpg)
Energy released per fusion = 3.96 x 10-12 J
no. fusions = 3.8 x 1026 3.96 x 10-12
The luminosity of the Sun is 3.8 x 1026 W
= 3.8 x 1026 J s-1
= 9.6 x 1037 per sec
At this rate, how long will the Sun last?
![Page 14: Energy from fusion - “that” equation](https://reader031.vdocuments.us/reader031/viewer/2022032106/56812ce2550346895d91a5fa/html5/thumbnails/14.jpg)
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!
![Page 15: Energy from fusion - “that” equation](https://reader031.vdocuments.us/reader031/viewer/2022032106/56812ce2550346895d91a5fa/html5/thumbnails/15.jpg)
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.
![Page 16: Energy from fusion - “that” equation](https://reader031.vdocuments.us/reader031/viewer/2022032106/56812ce2550346895d91a5fa/html5/thumbnails/16.jpg)
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