The stable nucleus that has a radius `1//3` that of `Os^(189)` is-

Options

A. `._(3)Li^(7)`
B. `._(2)He^(4)`
C. `._(5)B^(10)`
D. `._(6)C^(12)`

Explanation

To find the stable nucleus that has a radius \( \frac{1}{3} \) that of \( \text{Os}^{189} \), we will follow these steps:

### Step 1: Determine the radius of \( \text{Os}^{189} \)
The radius of a nucleus can be approximated 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.

For \( \text{Os}^{189} \):
\[ R_{\text{Os}} = R_0 \cdot (189)^{1/3} \]

### Step 2: Calculate the radius of the nucleus we are looking for
We need to find a nucleus whose radius is \( \frac{1}{3} \) that of \( \text{Os}^{189} \):
\[ R = \frac{1}{3} R_{\text{Os}} = \frac{1}{3} R_0 \cdot (189)^{1/3} \]

### Step 3: Relate the radius to the mass number of the new nucleus
Using the formula for the radius, we can set up the equation for the new nucleus:
\[ R = R_0 A^{1/3} \]
Setting the two expressions for \( R \) equal gives:
\[ \frac{1}{3} R_0 (189)^{1/3} = R_0 A^{1/3} \]

### Step 4: Simplify the equation
We can cancel \( R_0 \) from both sides (assuming \( R_0 \neq 0 \)):
\[ \frac{1}{3} (189)^{1/3} = A^{1/3} \]

### Step 5: Solve for \( A \)
Cubing both sides to eliminate the cube root:
\[ \left(\frac{1}{3}\right)^3 (189) = A \]
\[ A = \frac{189}{27} = 7 \]

### Step 6: Identify the stable nucleus with mass number 7
The stable nucleus with mass number 7 is \( \text{Li}^{7} \) (Lithium-7).

### Final Answer:
The stable nucleus that has a radius \( \frac{1}{3} \) that of \( \text{Os}^{189} \) is \( \text{Li}^{7} \).

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