Size of nucleus was obtained by the equation `r=R_(0)A^(1//3)` , Where r is the radius of nucleus of mass no A. and `R_(0)` is a constant whose valie is equal to `1.5xx10^(-15)` metre.
(Given : 1 amu = `1.66xx10^(-24)g`)
What is the density of a nucleus of mass number A ?

Options

A. `(4)/(3)pi(1.5xx10^(-15))^(3)` A
B. `1.17xx10^(17)kg//cm^(3)`
C. `1.17xx10^(-17)kg//m^(3)`
D. none of these

Answer

To find the density of a nucleus with mass number \( A \), we can follow these steps:

### Step 1: Calculate the Radius of the Nucleus
Using the formula for the radius of the nucleus:
\[ r = R_0 A^{1/3} \]
where \( R_0 = 1.5 \times 10^{-15} \) m.

### Step 2: 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 \( r \) from Step 1:
\[ V = \frac{4}{3} \pi (R_0 A^{1/3})^3 = \frac{4}{3} \pi R_0^3 A \]

### Step 3: Calculate the Mass of the Nucleus
The mass \( m \) of the nucleus can be calculated using the mass number \( A \) and the conversion of atomic mass units (amu) to grams:
\[ m = A \times 1 \text{ amu} = A \times 1.66 \times 10^{-24} \text{ g} \]
To convert this to kilograms, we divide by 1000:
\[ m = A \times 1.66 \times 10^{-24} \text{ g} \times \frac{1 \text{ kg}}{1000 \text{ g}} = A \times 1.66 \times 10^{-27} \text{ kg} \]

### Step 4: Calculate the Density of the Nucleus
Density \( \rho \) is defined as mass per unit volume:
\[ \rho = \frac{m}{V} \]
Substituting the expressions for mass and volume from Steps 2 and 3:
\[ \rho = \frac{A \times 1.66 \times 10^{-27}}{\frac{4}{3} \pi R_0^3 A} \]
Notice that \( A \) cancels out:
\[ \rho = \frac{1.66 \times 10^{-27}}{\frac{4}{3} \pi R_0^3} \]

### Step 5: Substitute the Value of \( R_0 \)
Now substituting \( R_0 = 1.5 \times 10^{-15} \) m:
\[ R_0^3 = (1.5 \times 10^{-15})^3 = 3.375 \times 10^{-45} \text{ m}^3 \]
Thus,
\[ \rho = \frac{1.66 \times 10^{-27}}{\frac{4}{3} \pi (3.375 \times 10^{-45})} \]

### Step 6: Calculate the Final Density
Calculating the denominator:
\[ \frac{4}{3} \pi (3.375 \times 10^{-45}) \approx 1.413 \times 10^{-44} \text{ m}^3 \]
Now substituting back:
\[ \rho \approx \frac{1.66 \times 10^{-27}}{1.413 \times 10^{-44}} \approx 1.17 \times 10^{17} \text{ kg/m}^3 \]

### Final Answer
The density of the nucleus is approximately:
\[ \rho \approx 1.17 \times 10^{17} \text{ kg/m}^3 \]