An alpha nucleus of energy `(1)/(2)m nu^(2)` bombards a heavy nucleus of charge `Ze` . Then the distance of closed approach for the alpha nucleus will be proportional to

Options

A. `v^(2)`
B. `1/m`
C. `1//v^(4)`
D. `1/(Ze)`

Explanation

To solve the problem of finding the distance of closest approach for an alpha nucleus bombarding a heavy nucleus of charge \( Ze \), we can follow these steps:

### Step 1: Understand the Initial Conditions
The alpha particle has a kinetic energy given by:
\[ E_i = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the alpha particle and \( v \) is its velocity.

### Step 2: Energy Conservation Principle
As the alpha particle approaches the heavy nucleus, it experiences a repulsive force due to the electrostatic interaction between the charges. At the closest approach, the kinetic energy of the alpha particle is converted entirely into potential energy. Thus, we can write:
\[ E_i = E_f \]
where \( E_f \) is the potential energy at the closest distance \( d \).

### Step 3: Calculate the Potential Energy
The potential energy \( U \) between the alpha particle (which has a charge of \( 2e \)) and the heavy nucleus (which has a charge of \( Ze \)) at a distance \( d \) is given by:
\[ U = \frac{k \cdot (2e) \cdot (Ze)}{d} \]
where \( k \) is Coulomb's constant.

### Step 4: Set Up the Equation
Equating the initial kinetic energy to the potential energy at the closest approach:
\[ \frac{1}{2} mv^2 = \frac{k \cdot (2e) \cdot (Ze)}{d} \]

### Step 5: Solve for Distance \( d \)
Rearranging the equation to solve for \( d \):
\[ d = \frac{2k \cdot (2e) \cdot (Ze)}{mv^2} \]
This simplifies to:
\[ d = \frac{4k \cdot e^2 \cdot Z}{mv^2} \]

### Step 6: Determine Proportionality
From the expression for \( d \), we can see that:
- \( d \) is directly proportional to \( Z \) (the charge of the heavy nucleus).
- \( d \) is inversely proportional to \( m \) (the mass of the alpha particle).
- \( d \) is inversely proportional to \( v^2 \) (the square of the velocity of the alpha particle).

### Conclusion
Thus, the distance of closest approach \( d \) is proportional to \( Z \) and inversely proportional to \( m \) and \( v^2 \). Therefore, the correct answer is that \( d \) is proportional to \( \frac{Z}{mv^2} \).