The end product of the `beta^-` decay of `._(15)^(32)P` is

Options

A. `._(16)^(32)S`
B. `._(16)^(35)S`
C. `._(14)^(32)N`
D. `._(17)^(32)Cl`

Answer

To solve the problem of determining the end product of the beta-minus decay of phosphorus-32 (represented as \(_{15}^{32}P\)), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial Element**:
- We have phosphorus with atomic number \(Z = 15\) and atomic mass \(A = 32\).
- This can be represented as \(_{15}^{32}P\).

2. **Understand Beta Minus Decay**:
- In beta-minus decay, a neutron in the nucleus is converted into a proton, and an electron (beta particle) is emitted.
- The reaction can be represented as:
\[ n \rightarrow p + e^- + \bar{\nu} \]
- Here, \(n\) is a neutron, \(p\) is a proton, \(e^-\) is the emitted electron, and \(\bar{\nu}\) is an antineutrino.

3. **Determine Changes in Atomic Number and Mass**:
- The atomic mass \(A\) remains unchanged because a neutron is converted into a proton (total nucleon count remains the same).
- The atomic number \(Z\) increases by 1 because we gain a proton:
\[ Z' = Z + 1 \]
- Thus, for phosphorus:
\[ Z' = 15 + 1 = 16 \]
- The atomic mass remains:
\[ A' = A = 32 \]

4. **Identify the New Element**:
- The new element after beta decay will have atomic number 16 and atomic mass 32, which corresponds to sulfur (\(_{16}^{32}S\)).

5. **Conclusion**:
- Therefore, the end product of the beta-minus decay of \(_{15}^{32}P\) is \(_{16}^{32}S\) (sulfur).

### Final Answer:
The end product of the beta-minus decay of \(_{15}^{32}P\) is \(_{16}^{32}S\) (sulfur).

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