The stable nucleus which has a radius half of `Zn​^(64)`​ is -

Options

A. `Ca​^(40 )`
B. `S^(16)`
C. `Na^(21)`
D. `Be^(8)`

Explanation

To solve the problem of finding the stable nucleus that has a radius half of that of \( \text{Zn}^{64} \), we can follow these steps:

### Step 1: Understand the relationship between the radius of a nucleus and its mass number.
The radius \( R \) of a nucleus is given by the formula:
\[ R = R_0 A^{1/3} \]
where \( R_0 \) is a constant and \( A \) is the mass number of the nucleus.

### Step 2: Calculate the radius of \( \text{Zn}^{64} \).
For \( \text{Zn}^{64} \), the mass number \( A = 64 \). Thus, the radius \( R \) can be calculated as:
\[ R_{\text{Zn}} = R_0 (64)^{1/3} \]

### Step 3: Calculate \( (64)^{1/3} \).
Calculating the cube root of 64:
\[ (64)^{1/3} = 4 \]
So, we have:
\[ R_{\text{Zn}} = R_0 \cdot 4 \]

### Step 4: Determine the radius of the nucleus that is half of \( R_{\text{Zn}} \).
We want to find a nucleus whose radius \( R_x \) is half of \( R_{\text{Zn}} \):
\[ R_x = \frac{1}{2} R_{\text{Zn}} = \frac{1}{2} (R_0 \cdot 4) = 2 R_0 \]

### Step 5: Relate the radius of nucleus \( x \) to its mass number.
Using the formula for the radius again:
\[ R_x = R_0 A_x^{1/3} \]
Setting this equal to \( 2 R_0 \):
\[ R_0 A_x^{1/3} = 2 R_0 \]

### Step 6: Simplify the equation.
Dividing both sides by \( R_0 \):
\[ A_x^{1/3} = 2 \]

### Step 7: Cube both sides to find \( A_x \).
Cubing both sides gives:
\[ A_x = 2^3 = 8 \]

### Step 8: Identify the element with mass number 8.
The element with a mass number of 8 is Beryllium (Be).

### Conclusion:
The stable nucleus which has a radius half of \( \text{Zn}^{64} \) is \( \text{Be} \) (Beryllium).

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