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

Answer

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 scenario
An alpha particle (which has a charge of \( +2e \)) is approaching a heavy nucleus with charge \( Ze \). As the alpha particle gets closer to the nucleus, it experiences a Coulombic repulsion due to the positive charges.

### Step 2: Set up the energy conservation equation
Initially, the alpha particle has kinetic energy given by:
\[ KE = \frac{1}{2} mv^2 \]
At the closest approach, all of this kinetic energy will be converted into potential energy due to the electrostatic force between the alpha particle and the heavy nucleus. The potential energy \( PE \) at the closest approach can be expressed using Coulomb's law:
\[ PE = \frac{k \cdot (2e) \cdot (Ze)}{d} \]
where \( k \) is Coulomb's constant, \( d \) is the distance of closest approach, and \( 2e \) is the charge of the alpha particle.

### Step 3: Set kinetic energy equal to potential energy
At the point of closest approach, we can equate the kinetic energy to the potential energy:
\[ \frac{1}{2} mv^2 = \frac{k \cdot (2e) \cdot (Ze)}{d} \]

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

### Step 5: Analyze the proportionality
From the equation \( d = \frac{2k \cdot Ze^2}{mv^2} \), we can see that \( d \) is directly proportional to \( Ze^2 \) and \( k \) (which are constants in this scenario) and inversely proportional to \( v^2 \).

### Conclusion
Thus, the distance of closest approach \( d \) is inversely proportional to the square of the velocity of the alpha particle \( v \):
\[ d \propto \frac{1}{v^2} \]