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:

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