The ratio of radii of nuclei `._13 A1^27` and `._52 X^A` is `3 : 5`. The number of neutrons in the nuclei of `X` will be

Options

A. 52
B. 73
C. 125
D. 13

Answer

To solve the problem, we need to find the number of neutrons in the nucleus of element \( X \) given the ratio of the radii of the nuclei of \( _{13}^{27}Al \) and \( _{52}^{A}X \) is \( 3:5 \).

### Step-by-Step Solution:

1. **Understanding the Relationship of Nuclear Radius**:
The radius of a nucleus is given by the formula:
\[ R = R_0 A^{1/3} \]
where \( R_0 \) is a constant and \( A \) is the atomic mass number.

2. **Setting Up the Radii for Both Nuclei**:
Let \( R_1 \) be the radius of the aluminum nucleus and \( R_2 \) be the radius of the nucleus of element \( X \).
- For aluminum \( ( _{13}^{27}Al ) \):
\[ R_1 = R_0 (27)^{1/3} \]
- For element \( X \):
\[ R_2 = R_0 A^{1/3} \]

3. **Taking the Ratio of the Radii**:
According to the problem, the ratio of the radii is given as:
\[ \frac{R_1}{R_2} = \frac{3}{5} \]
Substituting the expressions for \( R_1 \) and \( R_2 \):
\[ \frac{R_0 (27)^{1/3}}{R_0 A^{1/3}} = \frac{3}{5} \]
The \( R_0 \) cancels out:
\[ \frac{(27)^{1/3}}{A^{1/3}} = \frac{3}{5} \]

4. **Cubing Both Sides**:
To eliminate the cube root, we cube both sides:
\[ \frac{27}{A} = \left(\frac{3}{5}\right)^3 \]
Calculating \( \left(\frac{3}{5}\right)^3 \):
\[ \left(\frac{3}{5}\right)^3 = \frac{27}{125} \]
Thus, we have:
\[ \frac{27}{A} = \frac{27}{125} \]

5. **Cross-Multiplying to Solve for \( A \)**:
Cross-multiplying gives:
\[ 27 \times 125 = 27 \times A \]
Dividing both sides by 27:
\[ A = 125 \]

6. **Finding the Number of Neutrons**:
The atomic mass \( A \) is the sum of the number of protons and neutrons:
\[ A = Z + N \]
where \( Z \) is the number of protons and \( N \) is the number of neutrons. For element \( X \):
- \( Z = 52 \) (given)
- \( A = 125 \) (calculated)

Thus:
\[ N = A - Z = 125 - 52 = 73 \]

### Final Answer:
The number of neutrons in the nucleus of element \( X \) is \( 73 \).

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