Order of magnitude of density of uranium nucleus is , [m = 1.67 x 10^(-27) kg]`

Options

A. `10^20 "kg m"^(-3)`
B. `10^17 "kg m"^(-3)`
C. `10^14 "kg m"^(-3)`
D. `10^11 "kg m"^(-3)`

Explanation

To find the order of magnitude of the density of a uranium nucleus, we can follow these steps:

### Step 1: Determine the Mass of the Uranium Nucleus
Uranium-235 has 92 protons and 143 neutrons. The mass of the nucleus can be calculated as follows:

\[ \text{Mass of nucleus} = (92 \times m_p) + (143 \times m_n) \]

Given that the mass of a proton \( m_p \) is approximately \( 1.67 \times 10^{-27} \, \text{kg} \) and the mass of a neutron \( m_n \) is approximately equal to the mass of a proton, we have:

\[ \text{Mass of nucleus} = (92 + 143) \times m_p = 235 \times m_p \]

Substituting the value of \( m_p \):

\[ \text{Mass of nucleus} = 235 \times (1.67 \times 10^{-27}) \, \text{kg} \]

### Step 2: Calculate the Volume of the Uranium Nucleus
The volume \( V \) of the nucleus can be approximated using the formula for the volume of a sphere:

\[ V = \frac{4}{3} \pi r^3 \]

The radius \( r \) can be estimated using the empirical formula:

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

where \( r_0 \) is a constant approximately equal to \( 1.1 \times 10^{-15} \, \text{m} \) and \( A \) is the mass number (235 for uranium-235). Thus:

\[ r = 1.1 \times 10^{-15} \times (235)^{1/3} \]

### Step 3: Substitute the Radius into the Volume Formula
Now substituting the radius into the volume formula:

\[ V = \frac{4}{3} \pi (1.1 \times 10^{-15} \times (235)^{1/3})^3 \]

### Step 4: Calculate the Density
Density \( \rho \) is given by the formula:

\[ \rho = \frac{\text{Mass}}{\text{Volume}} \]

Substituting the mass and volume we calculated:

\[ \rho = \frac{235 \times (1.67 \times 10^{-27})}{\frac{4}{3} \pi (1.1 \times 10^{-15} \times (235)^{1/3})^3} \]

### Step 5: Simplify the Expression
After simplification, the \( 235 \) cancels out from the numerator and denominator, leading to:

\[ \rho \approx \frac{1.67 \times 10^{-27}}{\frac{4}{3} \pi (1.1 \times 10^{-15})^3} \]

### Step 6: Calculate the Final Density Value
After performing the calculations, we find:

\[ \rho \approx 2.29 \times 10^{17} \, \text{kg/m}^3 \]

### Conclusion
Thus, the order of magnitude of the density of a uranium nucleus is approximately \( 10^{17} \, \text{kg/m}^3 \).

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