The radius of `""_(Z) M^(A)` nucleus is (outer most configuration `3s^(2) 3p^(1)` and A + Z = 40)

Options

A. 4.2 FM
B. `1.4 xx root340 FM`
C. `1.4 xx root2 40 FM`
D. `1.4 xx 40` F M

Explanation

To find the radius of the nucleus given the outermost electron configuration and the relationship \( A + Z = 40 \), we can follow these steps:

### Step 1: Identify the Element
The outermost electron configuration provided is \( 3s^2 3p^1 \). This configuration corresponds to the element Aluminum (Al), which has an atomic number \( Z = 13 \).

### Step 2: Calculate the Atomic Mass (A)
We are given the equation \( A + Z = 40 \). Since we know \( Z = 13 \):

\[ A + 13 = 40 \]

To find \( A \), we can rearrange the equation:

\[ A = 40 - 13 = 27 \]

### Step 3: Use the Formula for Nuclear Radius
The formula for the radius \( r \) of a nucleus is given by:

\[ r = r_0 \cdot A^{1/3} \]

where \( r_0 \) is a constant approximately equal to \( 1.4 \, \text{fm} \) (femtometers).

### Step 4: Substitute Values into the Formula
Now we can substitute \( A = 27 \) into the formula:

\[ r = 1.4 \cdot 27^{1/3} \]

### Step 5: Calculate \( 27^{1/3} \)
Calculating the cube root of 27:

\[ 27^{1/3} = 3 \]

### Step 6: Final Calculation of the Radius
Now substituting back into the radius formula:

\[ r = 1.4 \cdot 3 = 4.2 \, \text{fm} \]

Thus, the radius of the nucleus is \( 4.2 \, \text{fm} \).

### Summary of the Solution
The radius of the nucleus for the element with the outermost configuration \( 3s^2 3p^1 \) and \( A + Z = 40 \) is \( 4.2 \, \text{fm} \).

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