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 find the end product of the beta minus decay of phosphorus-32 \((^{32}_{15}P)\), we can follow these steps:

### Step 1: Understand Beta Minus Decay
In beta minus decay, a neutron in the nucleus is converted into a proton. This process increases the atomic number of the element by 1 while keeping the mass number unchanged.

**Hint:** Remember that in beta minus decay, the transformation involves a neutron becoming a proton.

### Step 2: Identify the Initial Nucleus
The initial nucleus is phosphorus-32, which has:
- Atomic number \(Z = 15\) (indicating it is phosphorus)
- Mass number \(A = 32\)

**Hint:** The atomic number tells you which element you are dealing with.

### Step 3: Apply the Beta Minus Decay Process
When phosphorus-32 undergoes beta minus decay:
- The atomic number increases by 1: \(Z = 15 + 1 = 16\)
- The mass number remains the same: \(A = 32\)

**Hint:** The mass number is conserved in this decay process.

### Step 4: Identify the New Element
The element with atomic number 16 is sulfur (S). Therefore, the product of the decay is sulfur-32.

**Hint:** Use the periodic table to find the element corresponding to the new atomic number.

### Step 5: Write the Decay Equation
The complete decay can be represented as:
\[ ^{32}_{15}P \rightarrow ^{32}_{16}S + e^- + \bar{\nu} \]
where \(e^-\) is the emitted beta particle (electron) and \(\bar{\nu}\) is the emitted antineutrino.

**Hint:** The decay equation shows the transformation and conservation of charge.

### Conclusion
The end product of the beta minus decay of phosphorus-32 is sulfur-32.

**Final Answer:** \(^{32}_{16}S\)

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