Radius of `._2^4 He` nucleus is `3` Fermi. The radius of `._82^206 Pb` nucleus will be.

Options

A. 5 Fermi
B. 6 Fermi
C. 11.16 Fermi
D. 8 Fermi

Answer

To find the radius of the \( _{82}^{206}Pb \) nucleus given that the radius of the \( _{2}^{4}He \) nucleus is 3 Fermi, we can use the formula for the radius of a nucleus:

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

where:
- \( R \) is the radius of the nucleus,
- \( R_0 \) is a constant (approximately \( 1.2 \, \text{Fermi} \)),
- \( A \) is the mass number of the nucleus.

### Step 1: Calculate the radius of the \( _{2}^{4}He \) nucleus.

Given:
- Mass number \( A \) of \( _{2}^{4}He \) is 4.
- Radius \( R \) of \( _{2}^{4}He \) is 3 Fermi.

Using the formula:

\[ R_{He} = R_0 \cdot 4^{1/3} \]

### Step 2: Set up the equation for the \( _{82}^{206}Pb \) nucleus.

For the \( _{82}^{206}Pb \) nucleus:
- Mass number \( A \) is 206.

Using the same formula:

\[ R_{Pb} = R_0 \cdot 206^{1/3} \]

### Step 3: Relate the two radii.

From the two equations, we can set up a ratio:

\[ \frac{R_{He}}{R_{Pb}} = \frac{R_0 \cdot 4^{1/3}}{R_0 \cdot 206^{1/3}} \]

The \( R_0 \) cancels out:

\[ \frac{R_{He}}{R_{Pb}} = \frac{4^{1/3}}{206^{1/3}} \]

### Step 4: Substitute the known value of \( R_{He} \).

We know \( R_{He} = 3 \, \text{Fermi} \):

\[ \frac{3}{R_{Pb}} = \frac{4^{1/3}}{206^{1/3}} \]

### Step 5: Solve for \( R_{Pb} \).

Cross-multiplying gives:

\[ 3 \cdot 206^{1/3} = R_{Pb} \cdot 4^{1/3} \]

Now, solving for \( R_{Pb} \):

\[ R_{Pb} = \frac{3 \cdot 206^{1/3}}{4^{1/3}} \]

### Step 6: Calculate \( R_{Pb} \).

Calculating the cube roots and substituting:

\[ R_{Pb} = 3 \cdot \frac{206^{1/3}}{4^{1/3}} = 3 \cdot \left(\frac{206}{4}\right)^{1/3} \]

Calculating \( \frac{206}{4} = 51.5 \):

\[ R_{Pb} = 3 \cdot (51.5)^{1/3} \]

Using a calculator, \( (51.5)^{1/3} \approx 3.68 \):

\[ R_{Pb} \approx 3 \cdot 3.68 \approx 11.16 \, \text{Fermi} \]

### Final Result:

The radius of the \( _{82}^{206}Pb \) nucleus is approximately \( 11.16 \, \text{Fermi} \).

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