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

Options

A. (a) `10^20kg//m^3`
B. (b) `10^17 kg//m^3`
C. (c) `10^14 kg//m^3`
D. (d) `10^11kg//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 mass of a nucleon
The mass of one nucleon (proton or neutron) is given as:
\[ m = 1.67 \times 10^{-27} \, \text{kg} \]

### Step 2: Define the mass number of uranium
Let the mass number of uranium be represented by \( A \). For uranium, \( A \) is typically around 238 (for Uranium-238).

### Step 3: Calculate the total mass of the uranium nucleus
The total mass \( M \) of the uranium nucleus can be calculated as:
\[ M = A \times m \]
Substituting the values:
\[ M = 238 \times 1.67 \times 10^{-27} \, \text{kg} \]

### Step 4: Calculate the radius of the uranium nucleus
The radius \( R \) of a nucleus can be estimated using the formula:
\[ R = R_0 A^{1/3} \]
where \( R_0 = 1.25 \times 10^{-15} \, \text{m} \).
Substituting the values:
\[ R = 1.25 \times 10^{-15} \times 238^{1/3} \]

### Step 5: Calculate the volume of the uranium nucleus
The volume \( V \) of the nucleus can be calculated using the formula for the volume of a sphere:
\[ V = \frac{4}{3} \pi R^3 \]

### Step 6: Substitute the radius into the volume formula
Substituting \( R \) into the volume formula:
\[ V = \frac{4}{3} \pi \left(1.25 \times 10^{-15} \times 238^{1/3}\right)^3 \]

### Step 7: Calculate the density
The density \( \rho \) of the uranium nucleus is given by:
\[ \rho = \frac{M}{V} \]
Substituting the expressions for \( M \) and \( V \):
\[ \rho = \frac{A \times m}{\frac{4}{3} \pi R^3} \]

### Step 8: Simplify the expression for density
Substituting \( R \) in terms of \( A \):
\[ \rho = \frac{A \times m}{\frac{4}{3} \pi (R_0 A^{1/3})^3} \]
This simplifies to:
\[ \rho = \frac{A \times m}{\frac{4}{3} \pi R_0^3 A} \]
The \( A \) cancels out:
\[ \rho = \frac{3m}{4\pi R_0^3} \]

### Step 9: Substitute the values and calculate
Now substituting \( m = 1.67 \times 10^{-27} \, \text{kg} \) and \( R_0 = 1.25 \times 10^{-15} \, \text{m} \):
\[ \rho = \frac{3 \times 1.67 \times 10^{-27}}{4\pi (1.25 \times 10^{-15})^3} \]

### Step 10: Calculate the numerical value
After performing the calculation, we find:
\[ \rho \approx 2 \times 10^{17} \, \text{kg/m}^3 \]

### Conclusion
The order of magnitude of the density of a uranium nucleus is:
\[ \rho \approx 2 \times 10^{17} \, \text{kg/m}^3 \]