If the distance between nuclei is `2 xx 10^(-13)` cm , the density of nuclear material is (Assume that nuclear mass is order of `(10^(-27))`

Options

A. `3.21 xx 10^(-12) kg// m^(3)`
B. `1.6 xx 10^(-3) kg//m^(3)`
C. `2 xx 10^(9) kg//m^(3)`
D. `1 xx 10^(17) kg //m^(3)`

Explanation

To find the density of nuclear material given the distance between nuclei and the order of nuclear mass, we can follow these steps:

### Step 1: Understand the given values
- Distance between nuclei, \( d = 2 \times 10^{-13} \) cm
- Nuclear mass, \( m \approx 10^{-27} \) kg

### Step 2: Convert the distance from centimeters to meters
Since density is usually expressed in kg/m³, we need to convert the distance from centimeters to meters:
\[ d = 2 \times 10^{-13} \text{ cm} = 2 \times 10^{-13} \times 10^{-2} \text{ m} = 2 \times 10^{-15} \text{ m} \]

### Step 3: Calculate the volume occupied by one nucleus
Assuming that the nuclei are spherical, we can calculate the volume \( V \) of a sphere using the formula:
\[ V = \frac{4}{3} \pi r^3 \]
where \( r \) is the radius of the sphere. Here, the distance between the nuclei can be considered as the diameter, so the radius \( r \) is:
\[ r = \frac{d}{2} = \frac{2 \times 10^{-15}}{2} = 1 \times 10^{-15} \text{ m} \]
Now we can calculate the volume:
\[ V = \frac{4}{3} \pi (1 \times 10^{-15})^3 \]
Calculating \( (1 \times 10^{-15})^3 \):
\[ (1 \times 10^{-15})^3 = 1 \times 10^{-45} \text{ m}^3 \]
Thus,
\[ V = \frac{4}{3} \pi (1 \times 10^{-45}) \approx \frac{4 \times 3.14}{3} \times 10^{-45} \approx 4.19 \times 10^{-45} \text{ m}^3 \]

### Step 4: Calculate the density
Density \( \rho \) is defined as mass per unit volume:
\[ \rho = \frac{m}{V} \]
Substituting the values:
\[ \rho = \frac{10^{-27} \text{ kg}}{4.19 \times 10^{-45} \text{ m}^3} \]
Calculating this gives:
\[ \rho \approx \frac{10^{-27}}{4.19 \times 10^{-45}} \approx 2.39 \times 10^{17} \text{ kg/m}^3 \]

### Step 5: Round off the answer
For simplicity, we can round this to:
\[ \rho \approx 10^{17} \text{ kg/m}^3 \]

### Conclusion
The density of the nuclear material is approximately \( 10^{17} \text{ kg/m}^3 \).

---