Radius of `._2^4He`The nucleus is 3 Fermi. The radius of `._16^32S` nucleus will be

Options

A. 6 fermi
B. 4 fermi
C. 5 fermi
D. 8 fermi

Answer

To find the radius of the sulfur-32 nucleus given that the radius of the helium-4 nucleus is 3 Fermi, we can use the formula for the radius of a nucleus:

\[ R = R_0 \cdot A^{1/3} \]

where \( R_0 \) is a constant (approximately 1.2 to 1.3 Fermi) and \( A \) is the mass number of the nucleus.

### Step-by-Step Solution:

1. **Identify the mass numbers**:
- For helium-4 (\( _2^4He \)), the mass number \( A_{He} = 4 \).
- For sulfur-32 (\( _{16}^{32}S \)), the mass number \( A_{S} = 32 \).

2. **Write the formula for the radius of each nucleus**:
- Radius of helium:
\[ R_{He} = R_0 \cdot (A_{He})^{1/3} = R_0 \cdot (4)^{1/3} \]
- Radius of sulfur:
\[ R_{S} = R_0 \cdot (A_{S})^{1/3} = R_0 \cdot (32)^{1/3} \]

3. **Set up the ratio of the radii**:
- The ratio of the radius of helium to the radius of sulfur is:
\[ \frac{R_{He}}{R_{S}} = \frac{(4)^{1/3}}{(32)^{1/3}} = \left(\frac{4}{32}\right)^{1/3} = \left(\frac{1}{8}\right)^{1/3} \]

4. **Calculate the cube root**:
- The cube root of \( \frac{1}{8} \) is:
\[ \left(\frac{1}{8}\right)^{1/3} = \frac{1}{2} \]

5. **Express the radius of sulfur in terms of the radius of helium**:
- Rearranging gives:
\[ R_{S} = R_{He} \cdot 2 \]

6. **Substitute the known radius of helium**:
- Given \( R_{He} = 3 \) Fermi:
\[ R_{S} = 2 \cdot 3 = 6 \text{ Fermi} \]

### Final Answer:
The radius of the sulfur-32 nucleus is **6 Fermi**.

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