Chapter 30, Problem 1OQ

To determine

The emitted particle in the decay process of ${}_{15}^{32}\text{P}$ to ${}_{16}^{32}\text{S}$ from the given options.

Answer to Problem 1OQ

Option (c).

Explanation of Solution

In the decay of radioactive nucleus, the daughter nucleus has the same number of nucleon as the parent nucleon, but the atomic number is changed. Such type of radioactive decay is called beta decay. There are two types of beta decay exist, beta minus decay and beta plus decay. A neutron in the nucleus is transformed in to proton and electron in beta minus decay, and a proton is transformed into a neutron and a positron in beta plus decay.

Here the parent nuclei is ${}_{15}^{32}\text{P}$, and the daughter nuclei is ${}_{16}^{32}\text{S}$. The atomic number is increased by one. So this decay process is the example of beta minus decay. The nucleon number and charge should be conserved in a decay process. So the emitted particle is an electron. This process can be written as,

    ${}_{15}^{32}\text{P}{\to }_{16}^{32}\text{S}{+}_{-1}^0\text{e}+{\overline{v}}_{\text{e}}$

Conclusion:

Since emitted particle is an electron, option (c) is correct.

Emitted particle is an electron. Thus, option (a) is incorrect.

Emitted particle is an electron. Thus, option (b) is incorrect.

Emitted particle is an electron. Thus, option (d) is incorrect.

Emitted particle is an electron. Thus, option (e) is incorrect.