The ratio of density of nuciel of `O^(16)` and `Ca^(40)` is :

Options

A. `1:1`
B. `1:2`
C. `1:3`
D. `1:4`

Answer

To find the ratio of the densities of the nuclei of \( O^{16} \) (Oxygen-16) and \( Ca^{40} \) (Calcium-40), we can follow these steps:

### Step 1: Understand the formula for density
The density (\( \rho \)) of a nucleus is given by the formula:
\[ \rho = \frac{m}{V} \]
where \( m \) is the mass of the nucleus and \( V \) is its volume.

### Step 2: Calculate the mass of the nuclei
The mass of a nucleus can be expressed in terms of its mass number (\( A \)) and the mass of one nucleon (\( m_n \)):
\[ m = A \times m_n \]
For \( O^{16} \), the mass number \( A = 16 \). For \( Ca^{40} \), the mass number \( A = 40 \).

### Step 3: Calculate the volume of the nuclei
The volume of a nucleus can be approximated using the formula:
\[ V = \frac{4}{3} \pi r^3 \]
where \( r \) is the radius of the nucleus. The radius can be estimated using the formula:
\[ r = r_0 A^{1/3} \]
where \( r_0 \) is a constant (approximately \( 1.2 \, \text{fm} \) or \( 1.2 \times 10^{-15} \, \text{m} \)).

### Step 4: Substitute the radius into the volume formula
Substituting the expression for \( r \) into the volume formula gives:
\[ V = \frac{4}{3} \pi (r_0 A^{1/3})^3 = \frac{4}{3} \pi r_0^3 A \]

### Step 5: Write the density in terms of mass number
Now substituting the mass and volume into the density formula, we get:
\[ \rho = \frac{A \cdot m_n}{\frac{4}{3} \pi r_0^3 A} = \frac{3 m_n}{4 \pi r_0^3} \]
Notice that the density does not depend on \( A \) (the mass number). Thus, the density is a constant for all nuclei.

### Step 6: Calculate the ratio of densities
Since the density is constant for both \( O^{16} \) and \( Ca^{40} \), the ratio of their densities is:
\[ \frac{\rho_{O^{16}}}{\rho_{Ca^{40}}} = 1 \]

### Conclusion
The ratio of the densities of the nuclei of \( O^{16} \) and \( Ca^{40} \) is:
\[ \text{Ratio} = 1 : 1 \]