Chapter 19, Problem 16Q

To determine

The way the universe would have been different if the half-life of

${}^{8}Be$ was zero and if it was stable.

Answer to Problem 16Q

If the half- life of ${}^{8}Be$ was not zero the universe would be richer with atoms other than Hydrogen and Helium, but zero half- life of ${}^{8}Be$ prevents the process. If the ${}^{8}Be$ nucleus was stable, ${}^{12}C$ and organic molecules would form life in other planets as well.

Explanation of Solution

Given:

The half- life of ${}^{8}Be$ is $2.6\times {10}^{-16}\text{seconds}$.

${}^{8}Be$ decays in to two $\alpha$ particles.

Formula used:

The half- life equation,

$\begin{array}{l}A=A_0\times {\left[\frac{1}{2}\right]}^{\frac{t}{h}}{\text{Where, A= Final amount of particles, A}}_0=\text{Initial amount of partclies, }\\ \text{t= time and h= half- life}\end{array}$

Calculation:

Analyzing the two situations where the half-life of ${}^{8}Be$ was zero and ${}^{8}Be$ nucleus was stable.

For zero half- life, $h=0$. Then,

$\begin{array}{l}A=A_0\times {\left[\frac{1}{2}\right]}^{\frac{t}{0}}\\ A=A_0\times {\left[\frac{1}{2}\right]}^{\infty }\\ A=A_0\times 0\\ A=0\end{array}$

This means ${}^{8}Be$ decays extremely faster, even faster than $2.6\times {10}^{-16}\text{seconds}$.

If ${}^{8}Be$ nucleus was stable and did not decay, then the half- life is $\infty$.

For $\infty$ half- life, h= $\infty$. Then,

$\begin{array}{l}A=A_0\times {\left[\frac{1}{2}\right]}^{\frac{t}{\infty }}\\ A=A_0\times {\left[\frac{1}{2}\right]}^0\\ A=A_0\times 1\\ A=A_0\end{array}$

The final amount of particles is the same as the initial amount. The result implies that ${}^{8}Be$ nucleus is extremely stable and exists.

$Be$ is the fourth element in the periodic table and a solid at room temperature. It has four protons and five neutrons in its nucleus. $Be$ has many known isotopes and ${}^{8}Be$ is one of them. All those isotopes have different decaying paths while ${}^{8}Be$ decays into two $\alpha$ particles.

Most of the elements having a similar number of protons and neutrons are stable especially if the number of protons is even. ${}^{8}Be$ also has four protons and four neutrons. According to the above description ${}^{8}Be$ should be stable. However, the half-life of ${}^{8}Be$ isotope is extremely low ( $2.6\times {10}^{-16}\text{seconds}$ ) leaving it heavily unstable. Due to this reason ${}^{8}Be$ is non- existent.

An $\alpha$ particle is equivalent to a Helium- 4 nuclei. Inside a star after the Hydrogen fusion, Helium fusion begins, producing elements heavier than Helium which means Helium fuses in to ${}^{8}Be$. But due to the very low half-life, ${}^{8}Be$ decays instantly back into two $\alpha$ particles.

In a core of a star with enough pressure and mass, two $\alpha$ particles can collide to form ${}^{8}Be$ and within its very short lifespan another $\alpha$ particle can hit it forming a ${}^{12}C$ which is stable. If this happens future stars might consist $Be$ and heavier elements. Then the universe would be richer with other atoms as well instead of only Hydrogen and Helium. Having intensively low half-life would stop this process leaving Hydrogen and Helium be the only elements left in the cosmos.

On the other hand, if ${}^{8}Be$ was extremely stable and did not decay which means having an extremely high half- life, even smaller stars could produce heavier elements including ${}^{8}Be$ .Since $Be$ is solid at room temperature, most of the planets would consist denser rocks. Stable ${}^{8}Be$ would lead to form ${}^{12}C$ easier which then leads to form organic molecules as well. Thus life could be formed in many planets in the universe.