The radius of `._32X` nuclide is measured to be twice the radius of `._4^9Be`. The number of neutrons in X is

Options

A. 40
B. 72
C. 32
D. 36

Answer

To solve the problem, we need to find the number of neutrons in the nuclide `._32X`, given that its radius is twice that of the `._4^9Be` nuclide.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a nuclide `._32X` where 32 is the atomic number (Z).
- We also have beryllium `._4^9Be` where 4 is the atomic number and 9 is the mass number (A).
- The radius of `._32X` is twice the radius of `._4^9Be`.

2. **Using the Formula for Nuclear Radius**:
- The radius of a nucleus can be expressed as:
\[ R = r_0 A^{1/3} \]
- Here, \( r_0 \) is a constant, and \( A \) is the mass number.

3. **Setting Up the Equation**:
- Let \( R_X \) be the radius of `._32X` and \( R_{Be} \) be the radius of `._4^9Be`.
- According to the problem:
\[ R_X = 2 R_{Be} \]
- Using the formula for the radius:
\[ r_0 A_X^{1/3} = 2 (r_0 A_{Be}^{1/3}) \]
- The \( r_0 \) cancels out:
\[ A_X^{1/3} = 2 A_{Be}^{1/3} \]

4. **Substituting the Mass Number of Beryllium**:
- The mass number of beryllium \( A_{Be} = 9 \):
\[ A_X^{1/3} = 2 (9^{1/3}) \]

5. **Cubing Both Sides**:
- To eliminate the cube root, we cube both sides:
\[ A_X = (2 \cdot 9^{1/3})^3 \]
- This simplifies to:
\[ A_X = 8 \cdot 9 = 72 \]

6. **Finding the Number of Neutrons**:
- The number of neutrons \( N \) in a nuclide is given by:
\[ N = A - Z \]
- Substituting the values we found:
\[ N = 72 - 32 = 40 \]

7. **Conclusion**:
- The number of neutrons in the nuclide `._32X` is **40**.

### Final Answer:
The number of neutrons in `._32X` is **40**.