The density of a nucleus of mass number A is proportional to

Options

A. `A^(3)`
B. `A^(1//3)`
C. `A`
D. `A^(0)`

Explanation

To determine how the density of a nucleus is related to its mass number \( A \), we can follow these steps:

### Step 1: Understand the definition of density
The density \( D \) of an object is defined as its mass \( m \) divided by its volume \( V \):
\[ D = \frac{m}{V} \]

### Step 2: Relate mass number \( A \) to mass
In nuclear physics, the mass number \( A \) is approximately equal to the mass of the nucleus in atomic mass units (u). Thus, we can express the mass of the nucleus as:
\[ m = A \]

### Step 3: Determine the volume of the nucleus
The volume \( V \) of a nucleus can be modeled as a sphere. The volume of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
where \( r \) is the radius of the nucleus.

### Step 4: Relate the radius \( r \) to the mass number \( A \)
The radius \( r \) of a nucleus is often modeled as being proportional to the cube root of the mass number \( A \):
\[ r = r_0 A^{1/3} \]
where \( r_0 \) is a constant.

### Step 5: Substitute the radius into the volume formula
Substituting \( r \) into the volume formula, we get:
\[ V = \frac{4}{3} \pi (r_0 A^{1/3})^3 = \frac{4}{3} \pi r_0^3 A \]

### Step 6: Substitute mass and volume into the density formula
Now, substituting \( m = A \) and \( V = \frac{4}{3} \pi r_0^3 A \) into the density formula, we have:
\[ D = \frac{A}{\frac{4}{3} \pi r_0^3 A} \]

### Step 7: Simplify the expression
When we simplify this expression, the \( A \) in the numerator and denominator cancels out:
\[ D = \frac{3}{4 \pi r_0^3} \]

### Step 8: Conclusion about density
The resulting expression shows that the density \( D \) is independent of the mass number \( A \) and is instead a constant:
\[ D \propto A^0 \]

### Final Answer
Thus, the density of a nucleus is proportional to \( A^0 \), indicating that it is independent of the mass number.

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