In the reaction `Np_(93)^239 to Pu_94^239 + (?)`, the missing particle is

Options

A. Proton
B. Positron
C. Electron
D. Neutron

Answer

To solve the reaction \( \text{Np}_{93}^{239} \rightarrow \text{Pu}_{94}^{239} + (?) \), we need to identify the missing particle in this nuclear reaction.

### Step-by-Step Solution:

1. **Identify the Reactants and Products**:
- The reactant is Neptunium (Np) with atomic number 93 and mass number 239.
- The product is Plutonium (Pu) with atomic number 94 and mass number 239.

2. **Analyze the Change in Atomic Number**:
- The atomic number of Neptunium (Np) is 93, and it changes to Plutonium (Pu) which has an atomic number of 94.
- This indicates that the atomic number has increased by 1.

3. **Analyze the Change in Mass Number**:
- The mass number of both Neptunium (Np) and Plutonium (Pu) is 239.
- This indicates that the mass number remains unchanged during this reaction.

4. **Determine the Type of Decay**:
- Since the atomic number increases by 1 while the mass number remains the same, this reaction is indicative of beta decay.
- In beta decay, a neutron in the nucleus is converted into a proton, and a beta particle (which is an electron) is emitted.

5. **Identify the Missing Particle**:
- In beta decay, the emitted particle is a beta particle, which can be represented as \( \beta^- \) or an electron \( e^- \).
- Therefore, the missing particle in the reaction is a beta particle.

### Final Reaction:
The complete reaction can be written as:
\[ \text{Np}_{93}^{239} \rightarrow \text{Pu}_{94}^{239} + \beta^- \]

### Conclusion:
The missing particle in the reaction is a beta particle (electron).

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