Question #a7e21

Question

Question #a7e21

1 Answer

Impact of this question

2032 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-22T00:41:40

Carbon-12 is formed in dying stars via the Triple-alpha Process ; it is actually a two step process that must occur in less time than it takes for the intermediate product, Berillium-8 ( $_{4}^{8} B e$ $""_4^8Be$ ), to decay ( $t < 6.7 \times 10^{- 7} s$ $t<6.7xx10^-7" s"$ ).

Here are the equations for this two step process:

$_{2}^{4} H e +_{2}^{4} H e \to_{4}^{8} B e$ $""_2^4He+""_2^4Herarr""_4^8Be$
$_{4}^{8} B e +_{2}^{4} H e \to_{6}^{12} C + 2 γ$ $""_4^8Be+""_2^4Herarr""_6^12C+ 2gamma$

At the same time, some of the carbon-12 is fused into oxygen-16 using an additional alpha particle:

$_{6}^{12} C +_{2}^{4} H e \to_{8}^{16} O + γ$ $""_6^12C+""_2^4Herarr""_8^16O + gamma$

Please read the above reference for more details.

Science

Math

Humanities

... and beyond

Question #a7e21

1 Answer

Related questions

Impact of this question