Question

How does carbon \[14\] decay?

Answer 1

Hint:‌As we know that the Carbon \[14\] is a highly unstable and radioactive isotope of carbon and it is normally written as $_6^{14}C$, created by the collision of a neutron with the nuclei of nitrogen in air and it possess six protons and eight neutrons. It generally emits a beta particle and nitrogen \[14\].


Complete answer:
As we already know that the carbon actually exists in three forms. Two of them are stable and non-radioactive isotopes which are ${}^{12}C$ and ${}^{13}C$ and one is an isotope of carbon which is carbon \[14\] represented as $_6^{14}C$ and it is highly unstable and a radioactive isotope which is generally formed as a result of reaction between a neutron and nitrogen \[14\] resulting into the formation of carbon \[14\] and a proton which can be shown as below:
$n + {}_7^{14}N \to {}_6^{14}C + p$
As it is created, it can be destroyed also. The carbon \[14\] decays by emitting beta particles and gives nitrogen which is shown by the help of a chemical equation as below:
${}_6^{14}C \to {}_7^{14}N + \beta - particle({e^ - })$
When the beta particle is emitted that mean an electron is released and an electron antineutrino which is one of the neutrons in the carbon \[14\] nucleus is changed into a proton thereby resulting in the stable and non-radioactive isotope of nitrogen \[14\] from the carbon \[14\] nucleus.
Therefore, the correct answer is that the carbon \[14\] decays by emitting beta particles.

Note:always remember that the carbon \[14\] which is found in nature can be regenerated by cosmic rays hitting the atmosphere resulting in creation of atoms of carbon \[14\] while colliding with the nuclei in the upper atmosphere and liberates neutrons which interacts with the nitrogen present in air. Similarly when it decays it given off the antineutron and nitrogen \[14\]. Carbon \[14\] can be used as a radioactive marker.