Assuming the radius of a hydrogen atom is given by the Bohr radius `r_("Bohr")=5.29xx10^(-1)` m what is the ratio of the nuclear density of a hydrogen atom to its atomic density?

Options

A. `1.2xx10^(-14)`
B. `8.6xx10^(13)`
C. `4.4xx10^(4)`
D. `3.9xx10^(17)`

Answer

To find the ratio of the nuclear density of a hydrogen atom to its atomic density, we will follow these steps:

### Step 1: Understand the Definitions
- **Nuclear Density (ρ_nucleus)**: This is defined as the mass of the nucleus divided by its volume.
- **Atomic Density (ρ_atom)**: This is defined as the mass of the atom divided by its volume.

### Step 2: Identify the Relevant Values
- **Bohr Radius (r_Bohr)**: The radius of a hydrogen atom is given as \( r_{Bohr} = 5.29 \times 10^{-11} \) m.
- **Nuclear Radius (r_nucleus)**: The radius of the hydrogen nucleus (which is essentially a proton) is approximately \( r_{nucleus} = 1.19 \times 10^{-15} \) m.

### Step 3: Calculate the Volumes
- **Volume of the Nucleus (V_nucleus)**:
\[ V_{nucleus} = \frac{4}{3} \pi r_{nucleus}^3 \]

- **Volume of the Atom (V_atom)**:
\[ V_{atom} = \frac{4}{3} \pi r_{Bohr}^3 \]

### Step 4: Calculate the Densities
- **Density of the Nucleus (ρ_nucleus)**:
\[ \rho_{nucleus} = \frac{m_{nucleus}}{V_{nucleus}} = \frac{m_{proton}}{\frac{4}{3} \pi r_{nucleus}^3} \]
Here, we can approximate the mass of the nucleus as the mass of a proton, \( m_{proton} \approx 1.67 \times 10^{-27} \) kg.

- **Density of the Atom (ρ_atom)**:
\[ \rho_{atom} = \frac{m_{atom}}{V_{atom}} = \frac{m_{proton}}{\frac{4}{3} \pi r_{Bohr}^3} \]
Note that the mass of the atom is approximately equal to the mass of the proton since the mass of the electron is negligible.

### Step 5: Calculate the Ratio of Densities
The ratio of nuclear density to atomic density is given by:
\[ \frac{\rho_{nucleus}}{\rho_{atom}} = \frac{\frac{m_{proton}}{V_{nucleus}}}{\frac{m_{proton}}{V_{atom}}} = \frac{V_{atom}}{V_{nucleus}} \]

Substituting the volumes:
\[ \frac{\rho_{nucleus}}{\rho_{atom}} = \frac{\frac{4}{3} \pi r_{Bohr}^3}{\frac{4}{3} \pi r_{nucleus}^3} = \left(\frac{r_{Bohr}}{r_{nucleus}}\right)^3 \]

### Step 6: Substitute the Values
Substituting the values of \( r_{Bohr} \) and \( r_{nucleus} \):
\[ \frac{\rho_{nucleus}}{\rho_{atom}} = \left(\frac{5.29 \times 10^{-11}}{1.19 \times 10^{-15}}\right)^3 \]

### Step 7: Calculate the Final Value
Calculating the ratio:
1. Calculate \( \frac{5.29 \times 10^{-11}}{1.19 \times 10^{-15}} \).
2. Cube the result.

Calculating:
\[ \frac{5.29 \times 10^{-11}}{1.19 \times 10^{-15}} \approx 4.44 \times 10^{4} \]
Cubing this value:
\[ (4.44 \times 10^{4})^3 \approx 8.6 \times 10^{13} \]

### Final Answer
The ratio of the nuclear density of a hydrogen atom to its atomic density is approximately \( 8.6 \times 10^{13} \).

---