# Both "^18O and "^16O are found in nature. However, "^16O Is the most common. Why?

Oct 29, 2016

$\text{_8^16"O}$ is the most common isotope because it is one of the main products formed in stars.

#### Explanation:

Both $\text{_8^16"O}$ and $\text{_8^18"O}$ are stable isotopes, and both were originally formed in stars.

Formation of $\text{_8^16"O}$

When a star runs out of hydrogen to fuse in its core, it begins to collapse until the central temperature rises to ${10}^{8} \textcolor{w h i t e}{l} \text{K}$.

At this temperature, the α particles can fuse rapidly, and $\text{_8^16"O}$ is formed in a "triple alpha" process.

$\text{_2^4"He" + ""_2^4"He" → ""_4^8"Be}$

$\text{_4^8"Be" + ""_2^4"He" → ""_6^12"C}$

$\text{_6^12"C" + ""_2^4"He" → ""_8^16"O}$

Formation of $\text{_8^18"O}$

In contrast, $\text{_8^18"O}$ is formed mostly in massive stars.

The processes involved are

$\text{_6^12"C" → ""_7^13"N" → ""_6^13"C" → ""_7^14"N" → ""_8^15"O" → ""_7^15"N"→ ""_8^16"O" → ""_9^17"F" → ""_8^17"O"→ ""_9^18"F" → ""_8^18"O}$

Thus, $\text{_8^18"O}$ is much less abundant than $\text{_8^16"O}$.