Order of magnitude of density of uranium nucleus is , [m = 1.67 xx 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: Understand the formula for density
The density \( d \) is given by the formula:
\[ d = \frac{m}{V} \]
where \( m \) is the mass and \( V \) is the volume.

### Step 2: Determine the volume of the nucleus
Assuming the nucleus is spherical, the volume \( V \) can be expressed as:
\[ V = \frac{4}{3} \pi r^3 \]

### Step 3: Find the radius of the nucleus
The radius \( r \) of the nucleus can be approximated using the formula:
\[ r = r_0 A^{1/3} \]
where \( r_0 \) is a constant (approximately \( 1.25 \) femtometers or \( 1.25 \times 10^{-15} \) meters) and \( A \) is the mass number (approximately \( 238 \) for uranium).

### Step 4: Substitute the radius into the volume formula
Substituting \( r \) 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: Substitute mass and volume into the density formula
The mass \( m \) of the nucleus can be approximated as:
\[ m = A \times m_p \]
where \( m_p \) is the mass of a proton (approximately \( 1.67 \times 10^{-27} \) kg). Thus, the density becomes:
\[ d = \frac{A \cdot m_p}{\frac{4}{3} \pi r_0^3 A} \]
This simplifies to:
\[ d = \frac{m_p}{\frac{4}{3} \pi r_0^3} \]

### Step 6: Substitute the known values
Now substitute the known values:
- \( m_p = 1.67 \times 10^{-27} \) kg
- \( r_0 = 1.25 \times 10^{-15} \) m
- \( \pi \approx 3.14 \)

Calculating \( r_0^3 \):
\[ r_0^3 = (1.25 \times 10^{-15})^3 = 1.953125 \times 10^{-45} \text{ m}^3 \]

Now calculate the volume:
\[ \frac{4}{3} \pi r_0^3 \approx \frac{4}{3} \times 3.14 \times 1.953125 \times 10^{-45} \approx 8.18 \times 10^{-45} \text{ m}^3 \]

### Step 7: Calculate the density
Now substituting back into the density formula:
\[ d = \frac{1.67 \times 10^{-27}}{8.18 \times 10^{-45}} \approx 2.04 \times 10^{17} \text{ kg/m}^3 \]

### Step 8: Determine the order of magnitude
The order of magnitude of the density of the uranium nucleus is approximately:
\[ d \approx 2 \times 10^{17} \text{ kg/m}^3 \]

### Final Answer
Thus, the order of magnitude of the density of the uranium nucleus is \( 2 \times 10^{17} \text{ kg/m}^3 \).

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