oxygen + oxygen [imath]\bold{gives}[/imath] silicon + helium

[logistic_guy]

Repeat Problem \(\displaystyle 16\) for the reaction

\(\displaystyle {}^{16}_{\ \ 8}\text{O} + {}^{16}_{\ \ 8}\text{O} \rightarrow {}^{28}_{14}\text{Si} + {}^{4}_{2}\text{He}\)

 

[khansaheb]

Repeat Problem 16\displaystyle 1616 for the reaction

Go ahead and repeat......

 

[logistic_guy]

Go ahead and repeat......

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

\(\displaystyle Q = (15.994915 \ \text{u} + 15.994915 \ \text{u} - 27.976927 \ \text{u} - 4.002603 \ \text{u})c^2\)


\(\displaystyle = (0.0103 \ \text{u})c^2 = (0.0103 \ \text{u})c^2 \ \frac{931.57 \ \text{MeV}}{\text{u}c^2} = 9.595 \ \text{MeV}\)

 

[logistic_guy]

\(\displaystyle \bold{(b)}\)

Now we have two oxygen nuclei collide each other. From the previous exercise we found that the formula to find the kindetic energy is:

\(\displaystyle 2K = \frac{Z_1Z_2e^2}{4\pi \epsilon_0 2r}\)

where \(\displaystyle Z_1, Z_2\) are the atomic numbers of each nucleus respectively. And \(\displaystyle r\) is the radius of one nucleus and it is defined as \(\displaystyle r = (1.2 \ \text{fm})A^{1/3}\) where \(\displaystyle A\) is the atomic mass number.

Here is a full explanation how we got this formula.

Astrophysics and Cosmology

In the later stages of stellar evolution, a star (if massive enough) will begin fusing carbon nuclei to form, for example, magnesium: {}^{12}_{\ \ 6}\text{C} + {}^{12}_{\ \ 6}\text{C} \rightarrow {}^{24}_{12}\text{Mg} + \gamma \bold{(a)} How much energy is released in this reaction? \bold{(b)}...

www.freemathhelp.com

\(\displaystyle K = \frac{Z_1Z_2e^2}{4\pi \epsilon_0 4r} = \frac{8*8*1.44 \ \text{MeV} \cdot \text{fm}}{4(1.2 \ \text{fm})16^{1/3}} = \color{blue} 7.62 \ \text{MeV}\)

 

[logistic_guy]

\(\displaystyle \bold{(c)}\)

Repeating what I did in the Cosmology thread.

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.

\(\displaystyle 7.62 \ \text{MeV} = \frac{3}{2}\left(1.38 \times 10^{-23}\ \frac{\text{J}}{\text{K}}\right)T\)

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

\(\displaystyle 7.62 \ \text{MeV} \times \frac{1.602 \times 10^{-13} \ \text{J}}{\text{MeV}}= \frac{3}{2}\left(1.38 \times 10^{-23}\ \frac{\text{J}}{\text{K}}\right)T\)

This gives:

\(\displaystyle T = \color{blue} 5.897 \times 10^{10} \ \text{K}\)