The ratio of the density of oxygen nucleus(`"^(16)O_8`) and helium nucleus(`"^4He_2`)is

Options

A. 0.16736111111111
B. 0.042361111111111
C. 0.33402777777778
D. 0.084027777777778

Answer

To find the ratio of the density of the oxygen nucleus \(^16O_8\) and the helium nucleus \(^4He_2\), we can follow these steps:

### Step 1: Understand the concept of density in the context of nuclei
The density of a nucleus is defined as its mass divided by its volume. The formula for density (\( \rho \)) can be expressed as:
\[ \rho = \frac{m}{V} \]
where \( m \) is the mass and \( V \) is the volume of the nucleus.

### Step 2: Calculate the volume of the nuclei
The volume \( V \) 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 of a nucleus can be estimated using the empirical formula:
\[ r = r_0 A^{1/3} \]
where \( r_0 \) is a constant (approximately \( 1.2 \times 10^{-15} \) m) and \( A \) is the mass number of the nucleus.

### Step 3: Calculate the radius for both nuclei
- For the oxygen nucleus \(^16O_8\):
- Mass number \( A = 16 \)
- Radius \( r_O = r_0 (16)^{1/3} \)

- For the helium nucleus \(^4He_2\):
- Mass number \( A = 4 \)
- Radius \( r_{He} = r_0 (4)^{1/3} \)

### Step 4: Calculate the volumes
Using the radius, we can find the volumes:
- Volume of oxygen nucleus:
\[ V_O = \frac{4}{3} \pi (r_O)^3 = \frac{4}{3} \pi (r_0 (16)^{1/3})^3 = \frac{4}{3} \pi r_0^3 (16) \]

- Volume of helium nucleus:
\[ V_{He} = \frac{4}{3} \pi (r_{He})^3 = \frac{4}{3} \pi (r_0 (4)^{1/3})^3 = \frac{4}{3} \pi r_0^3 (4) \]

### Step 5: Calculate the densities
Now we can express the densities:
- Density of oxygen nucleus:
\[ \rho_O = \frac{m_O}{V_O} = \frac{m_O}{\frac{4}{3} \pi r_0^3 (16)} \]

- Density of helium nucleus:
\[ \rho_{He} = \frac{m_{He}}{V_{He}} = \frac{m_{He}}{\frac{4}{3} \pi r_0^3 (4)} \]

### Step 6: Find the ratio of densities
The ratio of the densities can be expressed as:
\[ \frac{\rho_O}{\rho_{He}} = \frac{\frac{m_O}{\frac{4}{3} \pi r_0^3 (16)}}{\frac{m_{He}}{\frac{4}{3} \pi r_0^3 (4)}} \]
This simplifies to:
\[ \frac{\rho_O}{\rho_{He}} = \frac{m_O \cdot 4}{m_{He} \cdot 16} \]
Since the mass of the nucleus is approximately proportional to its mass number, we can substitute \( m_O \approx 16 \) and \( m_{He} \approx 4 \):
\[ \frac{\rho_O}{\rho_{He}} = \frac{16 \cdot 4}{4 \cdot 16} = 1 \]

### Final Answer
The ratio of the density of the oxygen nucleus to the helium nucleus is:
\[ \frac{\rho_O}{\rho_{He}} = 1 \]