An artifical radioactive isotope gave `""_(7)^(14)N` after two successive `beta`-particle emissions. The number of neutrons in the parent nucleus must be

Options

A. 9
B. 14
C. 5
D. 7

Answer

To solve the problem, we need to determine the number of neutrons in the parent nucleus of an artificial radioactive isotope that decays to produce nitrogen-14 (\( _{7}^{14}N \)) after two successive beta particle emissions.

### Step-by-Step Solution:

1. **Understand Beta Decay**:
- A beta particle emission involves the transformation of a neutron into a proton, which increases the atomic number by 1 while the mass number remains unchanged.

2. **Identify the Product**:
- The product of the decay is nitrogen-14, which has an atomic number (Z) of 7 and a mass number (A) of 14.

3. **Establish Parent Nucleus**:
- Let the parent nucleus be represented as \( _{Z}^{A}X \), where Z is the atomic number and A is the mass number of the parent nucleus.

4. **Determine Changes in Atomic Number**:
- Since two beta particles are emitted, the atomic number of the parent nucleus will decrease by 2:
\[ Z - 2 = 7 \implies Z = 7 + 2 = 9 \]

5. **Determine Changes in Mass Number**:
- The mass number remains unchanged during beta decay:
\[ A = 14 \]

6. **Identify the Parent Nucleus**:
- Therefore, the parent nucleus can be represented as \( _{9}^{14}X \).

7. **Calculate the Number of Protons and Neutrons**:
- The number of protons (P) in the parent nucleus is equal to its atomic number, which is 9.
- The mass number (A) is the sum of protons (P) and neutrons (N):
\[ A = P + N \implies 14 = 9 + N \]
- Rearranging gives:
\[ N = 14 - 9 = 5 \]

8. **Conclusion**:
- The number of neutrons in the parent nucleus is 5.

### Final Answer:
The number of neutrons in the parent nucleus must be **5**.

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