The satble nucleus that has a radius half that of `Fe^(56)` is

Options

A. `Li^(7)`
B. `Na^(21)`
C. `S^(16)`
D. `Ca^(40)`

Answer

To solve the problem of finding the stable nucleus that has a radius half that of \( \text{Fe}^{56} \), we can follow these steps:

### Step 1: Understand the formula for nuclear radius
The radius \( r \) of a nucleus can be expressed using the formula:
\[ r = r_0 A^{1/3} \]
where \( r_0 \) is a constant (approximately \( 1.2 \, \text{fm} \)), and \( A \) is the mass number of the nucleus.

### Step 2: Calculate the radius of \( \text{Fe}^{56} \)
For \( \text{Fe}^{56} \), the mass number \( A \) is 56. Therefore, we can write the radius \( r \) as:
\[ r_{\text{Fe}} = r_0 (56)^{1/3} \]

### Step 3: Determine the radius of the nucleus we are looking for
We need to find a nucleus whose radius is half that of \( \text{Fe}^{56} \). Thus, we can express this as:
\[ r_1 = \frac{1}{2} r_{\text{Fe}} = \frac{1}{2} r_0 (56)^{1/3} \]

### Step 4: Simplify the expression for \( r_1 \)
Substituting the expression for \( r_{\text{Fe}} \):
\[ r_1 = \frac{1}{2} r_0 (56)^{1/3} = r_0 \frac{(56)^{1/3}}{2} \]

### Step 5: Relate \( r_1 \) to a new mass number \( A_1 \)
Using the formula for nuclear radius again, we can express \( r_1 \) in terms of a new mass number \( A_1 \):
\[ r_1 = r_0 (A_1)^{1/3} \]
Setting the two expressions for \( r_1 \) equal gives:
\[ r_0 (A_1)^{1/3} = r_0 \frac{(56)^{1/3}}{2} \]

### Step 6: Cancel \( r_0 \) and solve for \( A_1 \)
We can cancel \( r_0 \) from both sides (assuming \( r_0 \neq 0 \)):
\[ (A_1)^{1/3} = \frac{(56)^{1/3}}{2} \]
Cubing both sides results in:
\[ A_1 = \left(\frac{(56)^{1/3}}{2}\right)^3 = \frac{56}{8} = 7 \]

### Step 7: Identify the stable nucleus with mass number 7
The mass number \( A_1 = 7 \) corresponds to the stable nucleus of lithium, \( \text{Li}^{7} \).

### Final Answer
The stable nucleus that has a radius half that of \( \text{Fe}^{56} \) is \( \text{Li}^{7} \).

---