Download He fusion

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.
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).
Let’s start with 2 hydrogen nuclei - protons
As they fuse, one of the protons emits a
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Let’s look at the equation…
Enter two hydrogen nuclei (protons)…
1
1
H +
1
2
H  H +
1
1
0

+

1
… which fuse to become a deuterium nucleus (a
heavier isotope of hydrogen with a mass of 2)
plus a positron and a neutrino
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).
And the equation for this?
2
1
H +
1
3
H

He
+

2
1
Stage 3
Finally two of these helium-3’s collide and fuse
into a stable helium-4, and shedding the two
spare protons…
And this equation is …
3
He
+
2
3
4
1
He

He
+
2
H
2
2
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?
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
having a mass of 12u
12
C nucleus
6
which means 1u = 1.660 540 x 10-27 kg …ish
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 =
Oh! Some of the mass
seems to have disappeared!
-4.002603
u
________
0.0265… u
= 4.40 x 10-29 kg
We have a missing mass ( a “mass deficit”)
= 4.40 x 10-29 kg
Here comes that equation…
E =
where
2
mc
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
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?
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?
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!
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.
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