Assume the proton and neutron have equal masses . mass of a nucle is `1.6 times 10^(-27)` kg and radius of nucle us `1.5 times 10^(-15)A^(1/3)m`.The approximate radio of the nuclear density and water density is `n times 10^(13)`.the value of n is___________.

Answer

To solve the problem, we need to find the ratio of nuclear density to water density and express it in the form \( n \times 10^{13} \).

### Step-by-Step Solution:

1. **Understanding the Given Data**:
- Mass of a nucleon (proton/neutron) = \( 1.6 \times 10^{-27} \) kg.
- Radius of the nucleus = \( 1.5 \times 10^{-15} A^{1/3} \) m.
- We need to find the ratio of nuclear density to water density.

2. **Calculate the Mass of the Nucleus**:
The mass of the nucleus can be approximated as:
\[ \text{Mass of nucleus} = A \times \text{mass of nucleon} = A \times 1.6 \times 10^{-27} \text{ kg} \]

3. **Calculate the Volume of the Nucleus**:
The volume \( V \) of the nucleus can be calculated using the formula for the volume of a sphere:
\[ V = \frac{4}{3} \pi r^3 \]
Substituting \( r = 1.5 \times 10^{-15} A^{1/3} \):
\[ V = \frac{4}{3} \pi (1.5 \times 10^{-15} A^{1/3})^3 = \frac{4}{3} \pi (1.5^3 \times 10^{-45} A) = \frac{4 \pi \times 3.375 \times 10^{-45} A}{3} = 4.5 \pi \times 10^{-45} A \text{ m}^3 \]

4. **Calculate the Nuclear Density**:
The nuclear density \( \rho_n \) is given by:
\[ \rho_n = \frac{\text{Mass of nucleus}}{\text{Volume of nucleus}} = \frac{A \times 1.6 \times 10^{-27}}{4.5 \pi \times 10^{-45} A} \]
Simplifying this:
\[ \rho_n = \frac{1.6 \times 10^{-27}}{4.5 \pi \times 10^{-45}} = \frac{1.6}{4.5 \pi} \times 10^{18} \text{ kg/m}^3 \]

5. **Calculate the Density of Water**:
The density of water \( \rho_w \) is approximately:
\[ \rho_w = 1000 \text{ kg/m}^3 \]

6. **Calculate the Ratio of Nuclear Density to Water Density**:
The ratio \( R \) is:
\[ R = \frac{\rho_n}{\rho_w} = \frac{\frac{1.6}{4.5 \pi} \times 10^{18}}{1000} = \frac{1.6 \times 10^{15}}{4.5 \pi} \]

7. **Calculate the Value of \( n \)**:
Approximating \( \pi \approx 3.14 \):
\[ R \approx \frac{1.6 \times 10^{15}}{4.5 \times 3.14} \approx \frac{1.6 \times 10^{15}}{14.13} \approx 0.113 \times 10^{15} \approx 11.3 \times 10^{13} \]
Thus, \( n \approx 11.3 \).

### Final Answer:
The value of \( n \) is approximately **11**.