Question

What type of decay is evident in the following nuclear reaction: 64C + e + 44N?

a) Alpha decay
b) Beta decay
c) Gamma emission
d) None of the above

Answer 1

Step-by-step explanation:

The nuclear reaction given is: 64C + e + 44N.

In this reaction, an electron (e) is emitted. This indicates that a beta particle is being released.

Beta decay is a type of radioactive decay in which a beta particle, either an electron or a positron, is emitted from the nucleus of an atom. It occurs when there is an imbalance between the number of protons and neutrons in the nucleus, leading to the conversion of a neutron into a proton or a proton into a neutron.

Therefore, the type of decay evident in the given nuclear reaction is b) Beta decay.

Answer 2

The nuclear reaction provided demonstrates beta decay, where a neutron is converted into a proton and an electron, changing carbon-14 into nitrogen-14.

Step-by-step explanation:

The nuclear reaction in question is beta decay. This can be determined by examining the change in atomic numbers between the reactants and products. In the given reaction, carbon transforms into nitrogen, with the emission of an electron (e). During beta decay, a neutron in the nucleus is transformed into a proton and an electron is emitted. The atomic number increases by one, changing the element from carbon to nitrogen, but the mass number remains the same, indicating that this is a beta decay process, which is consistent with our equation where C-14 becomes N-14 plus an electron (e). Hence, the correct type of decay is evident in the nuclear reaction 64C + e + 44N.