Question

in the equation, \\( ^{14}_{6}c \
ightarrow ^{14}_{7}n + ^{0}_{-1}b \\), the ________ decay of radioactive carbon - 14 results in the creation of a new nitrogen - 14 atom.

Explanation:

Beta decay involves the emission of a beta particle (in this case, \(_{-1}^0B\)). In beta decay, a neutron in the nucleus is converted into a proton and an electron (beta particle). Here, carbon - 14 (\(_{6}^{14}C\)) undergoes a transformation where a neutron changes to a proton (increasing the atomic number from 6 to 7, forming nitrogen - 14 \(_{7}^{14}N\)) and emitting a beta particle (\(_{-1}^0B\)). Alpha decay would involve the emission of \(_{2}^{4}He\) particles, and gamma decay involves the emission of high - energy photons (\(\gamma\)) with no change in atomic or mass number (except for energy state).

Answer:

Beta