Astrophysics and Cosmology

[logistic_guy]

In the later stages of stellar evolution, a star (if massive enough) will begin fusing carbon nuclei to form, for example, magnesium:

\(\displaystyle {}^{12}_{\ \ 6}\text{C} + {}^{12}_{\ \ 6}\text{C} \rightarrow {}^{24}_{12}\text{Mg} + \gamma\)

\(\displaystyle \bold{(a)}\) How much energy is released in this reaction? \(\displaystyle \bold{(b)}\) How much kinetic energy must each carbon nucleus have (assume equal) in a head-on collision if they are just to touch so that the strong force can come into play? \(\displaystyle \bold{(c)}\) What temperature does this kinetic energy correspond to?

 

[logistic_guy]

\(\displaystyle \bold{(a)}\)

The released energy is just the \(\displaystyle Q\)-value.

\(\displaystyle Q = (\) masses before reaction \(\displaystyle -\) masses after reaction \(\displaystyle )c^2\)

\(\displaystyle = (12 \ \text{u} + 12 \ \text{u} - 23.985042 \ \text{u} - 0 \ \text{u})c^2 = (0.014958 \ \text{u})c^2\)

We need to get this energy in \(\displaystyle \text{eV}\). Therefore, let us calculate the rest energy for \(\displaystyle 1\) atomic mass unit (\(\displaystyle \text{u}\)).

\(\displaystyle \text{u} = 1.6605 \times 10^{-27} \ \text{kg}\)

Then,

\(\displaystyle E = mc^2 = 1.6605 \times 10^{-27} \times (299792458)^2 = 1.492382974 \times 10^{-10} \ \text{J}\)

Let us convert it to \(\displaystyle \text{eV}\).

\(\displaystyle E = \frac{1.492382974 \times 10^{-10}}{1.602 \times 10^{-19}} = 9.3157 \times 10^{8} \ \text{eV} = 931.57 \ \text{MeV}\)

This gives:

\(\displaystyle \text{u}c^2 = 931.57 \ \text{MeV}\)

Then,

\(\displaystyle Q = (0.014958 \ \text{u})c^2 \ \frac{931.57 \ \text{MeV}}{\text{u}c^2} = 13.93 \ \text{MeV}\) of energy is released

 

[logistic_guy]

How much kinetic energy must each carbon nucleus have (assume equal) in a head-on collision if they are just to touch so that the strong force can come into play?

When two nuclei collide their kinetic energies transform to potential energy. That is:
\(\displaystyle 2K = U\)

From accumulated knowledge we know that the potential energy \(\displaystyle U = qV\)

where \(\displaystyle V\) is the potential.

We know that \(\displaystyle qV = q\frac{q}{4\pi \epsilon_0 r}\)

But since we have two radii, we have to double the distance.

\(\displaystyle U = qV = q\frac{q}{4\pi \epsilon_0 2r}\)

So, we have two nuclei colliding and each one holding 6 protons (carbon). The formula becomes:

\(\displaystyle U = \frac{6 \times e \times 6 \times e}{4\pi \epsilon_0 2r}\)

\(\displaystyle \frac{e^2}{4\pi \epsilon_0} = 1.44 \ \text{MeV} \cdot \text{fm} \ \ \ \ \) (I will derive this result in future Epsiodes.)

Then, we have:

\(\displaystyle U = \frac{36 \times 1.44}{2r}\)

Or

\(\displaystyle U = \frac{18 \times 1.44}{r}\)

The radius of carbon nucleus can be approximated by:

\(\displaystyle r = (1.2 \ \text{fm})A^{1/3}\)

where \(\displaystyle A\) is the mass number.

Then,

\(\displaystyle r = (1.2 \ \text{fm})12^{1/3}\)

The energy becomes:

\(\displaystyle U = \frac{18 \times 1.44 \ \text{MeV} \cdot \text{fm}}{(1.2 \ \text{fm})12^{1/3}}\)

Or

\(\displaystyle 2K = \frac{18 \times 1.44 \ \text{MeV}}{(1.2)12^{1/3}}\)

Or

\(\displaystyle K = \frac{9 \times 1.44 \ \text{MeV}}{(1.2)12^{1/3}} = \textcolor{blue}{4.717 \ \text{MeV}}\)

 

[logistic_guy]

\(\displaystyle \bold{(c)}\) What temperature does this kinetic energy correspond to?

We have a beautiful formula that relates kinetic energy and temperature. That is:

\(\displaystyle K = \frac{3}{2}kT\)

where \(\displaystyle k\) is the Boltzmann constant.

Plug in numbers.

\(\displaystyle 4.717 \ \text{MeV} = \frac{3}{2}\left(1.38 \times 10^{-23}\ \frac{\text{J}}{\text{K}}\right)T\)

We have problem in the units above as they don't match. So we need to get rid of \(\displaystyle \text{MeV}\).

\(\displaystyle 4.717 \ \text{MeV} \times \frac{1.602 \times 10^{-13} \ \text{J}}{\text{MeV}}= \frac{3}{2}\left(1.38 \times 10^{-23}\ \frac{\text{J}}{\text{K}}\right)T\)

This gives:

\(\displaystyle T = \color{blue} 3.65 \times 10^{10} \ \text{K}\)