In nuclear reaction `._2He^4 + ._ZX^A to ._(Z+2)Y^(A+3)+R`,R denotes

Options

A. electron
B. positron
C. proton
D. neutron

Answer

To solve the nuclear reaction problem given, we need to identify what R represents in the reaction:

\[ \text{He}^4 + \text{X} \rightarrow \text{Y} + R \]

where \(\text{He}^4\) is Helium-4, \(\text{X}\) has atomic number \(Z\) and mass number \(A\), and \(\text{Y}\) has atomic number \(Z + 2\) and mass number \(A + 3\).

### Step 1: Write down the reaction
We start with the given reaction:

\[ \text{He}^4 + \text{X} \rightarrow \text{Y} + R \]

### Step 2: Apply charge conservation
For charge conservation, we need to ensure that the total charge on the left side equals the total charge on the right side.

- The charge of \(\text{He}^4\) is 2 (since Helium has an atomic number of 2).
- The charge of \(\text{X}\) is \(Z\).
- The charge of \(\text{Y}\) is \(Z + 2\).
- Let the charge of \(R\) be \(n\).

Setting up the equation for charge conservation:

\[ 2 + Z = (Z + 2) + n \]

### Step 3: Solve for \(n\)
Rearranging the equation:

\[ 2 + Z = Z + 2 + n \]

Cancelling \(Z\) and \(2\) from both sides gives:

\[ 0 = n \]

This means that \(R\) has a charge of 0.

### Step 4: Apply mass conservation
Now, we apply mass conservation. The total mass on the left side must equal the total mass on the right side.

- The mass of \(\text{He}^4\) is 4.
- The mass of \(\text{X}\) is \(A\).
- The mass of \(\text{Y}\) is \(A + 3\).
- Let the mass of \(R\) be \(m\).

Setting up the equation for mass conservation:

\[ 4 + A = (A + 3) + m \]

### Step 5: Solve for \(m\)
Rearranging the equation:

\[ 4 + A = A + 3 + m \]

Cancelling \(A\) from both sides gives:

\[ 4 = 3 + m \]

Thus,

\[ m = 1 \]

### Step 6: Identify \(R\)
Now we know that \(R\) has a charge of 0 and a mass of 1. The only particle that fits this description is a neutron, which has no charge and a mass number of 1.

### Conclusion
Therefore, \(R\) denotes a neutron.

### Final Answer
R denotes a neutron.

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