Let us solve the question by analyzing the beta decay process and determining the identity of the daughter nuclide M.
We start with Phosphorus-30, which is represented as follows:
[tex]\[
{}_{15}^{30}P
\][/tex]
Here, [tex]\(15\)[/tex] is the atomic number (number of protons), and [tex]\(30\)[/tex] is the mass number (sum of protons and neutrons).
The process we are dealing with involves beta decay, which is indicated by the emission of a beta particle:
[tex]\[
{}_{-1}^0 e
\][/tex]
A beta particle is essentially an electron that is emitted during the transformation of a neutron into a proton. Therefore, the atomic number of the original atom will increase by 1, while the mass number remains unchanged because an electron has negligible mass compared to protons and neutrons.
In other words, during beta decay:
- The atomic number (Z) increases by 1.
- The mass number (A) remains the same.
However, it appears in our specific reaction, the beta particle is defined a bit differently with a negative atomic number contribution, possibly due to different notation conventions.
Here, let us analyze the given reaction to derive the identity of M.
Given:
[tex]\[
{}_{15}^{30} P \rightarrow M +- {}_{-1}^0 e
\][/tex]
1. Mass number:
The mass number before and after the decay remains the same because beta particles have a mass number of zero.
2. Atomic number:
Since in this equation, emitting a [tex]\( \beta \)[/tex]-particle (a particle with atomic number [tex]\(-1\)[/tex]) seems to subtract from the atomic number, let’s calculate the atomic number of M:
[tex]\[
Z_\text{M} = Z_\text{P} + Z_\text{e}
\][/tex]
Where:
[tex]\[
Z_\text{P} = 15
\][/tex]
[tex]\[
Z_\text{e} = -1
\][/tex]
Therefore:
[tex]\[
Z_\text{M} = 15 + (-1) = 14
\][/tex]
Thus, the atomic number of M is [tex]\(14\)[/tex]. The mass number remains [tex]\(30\)[/tex] as before:
[tex]\[
A_\text{M} = A_\text{P} + A_\text{e} = 30 + 0 = 30
\][/tex]
So, M is represented as:
[tex]\[
{}_{14}^{30}M
\][/tex]
Given that the atomic number [tex]\(Z = 14\)[/tex], we can identify the element M as Silicon (Si) because Silicon is the element with atomic number 14.
Therefore, M is:
[tex]\[
{}_{14}^{30}Si
\][/tex]
To summarize, M stands for Silicon-30 (Si-30), which results from the beta decay of Phosphorus-30.
