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: Write the formula for the radius of the nucleus
The radius \( r \) of the nucleus is given by the equation:
\[ r = R_0 A^{1/3} \]
where \( R_0 = 1.5 \times 10^{-15} \) meters.

### Step 2: Calculate the volume of the nucleus
The volume \( V \) of a sphere (which we assume the nucleus is) is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
Substituting the expression for \( r \):
\[ V = \frac{4}{3} \pi (R_0 A^{1/3})^3 \]
This simplifies to:
\[ V = \frac{4}{3} \pi R_0^3 A \]

### Step 3: Calculate the mass of the nucleus
The mass \( m \) of the nucleus can be calculated using the mass number \( A \) and the mass of one atomic mass unit (amu):
\[ m = A \times 1 \text{ amu} \]
Given that \( 1 \text{ amu} = 1.66 \times 10^{-27} \) kg, we have:
\[ m = A \times 1.66 \times 10^{-27} \text{ kg} \]

### 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 \( m \) and \( V \):
\[ \rho = \frac{A \times 1.66 \times 10^{-27}}{\frac{4}{3} \pi R_0^3 A} \]
The \( A \) cancels out:
\[ \rho = \frac{1.66 \times 10^{-27}}{\frac{4}{3} \pi R_0^3} \]

### Step 5: Substitute \( R_0 \) and calculate the density
Now substituting \( R_0 = 1.5 \times 10^{-15} \):
\[ \rho = \frac{1.66 \times 10^{-27}}{\frac{4}{3} \pi (1.5 \times 10^{-15})^3} \]
Calculating \( (1.5 \times 10^{-15})^3 \):
\[ (1.5 \times 10^{-15})^3 = 3.375 \times 10^{-45} \]
Now substituting this value back:
\[ \rho = \frac{1.66 \times 10^{-27}}{\frac{4}{3} \pi \times 3.375 \times 10^{-45}} \]
Calculating \( \frac{4}{3} \pi \approx 4.18879 \):
\[ \rho = \frac{1.66 \times 10^{-27}}{4.18879 \times 3.375 \times 10^{-45}} \]
Calculating the denominator:
\[ 4.18879 \times 3.375 \approx 14.1372 \Rightarrow \rho \approx \frac{1.66 \times 10^{-27}}{14.1372 \times 10^{-45}} \approx 1.17 \times 10^{17} \text{ kg/m}^3 \]

### Final Answer
The density of a nucleus with mass number \( A \) is approximately:
\[ \rho \approx 1.17 \times 10^{17} \text{ kg/m}^3 \]

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