When an `alpha` particle of mass m moving with velocity v bombards on a heavy nucleus of charge Ze, its distance of closest approach form the nucleus depends on m as

Options

A. `(1)/(sqrt(m))`
B. `(1)/(m^(2))`
C. m
D. `(1)/(m)`

Explanation

To solve the problem of determining how the distance of closest approach \( r_0 \) of an alpha particle to a heavy nucleus depends on its mass \( m \), we can follow these steps:

### Step 1: Understand the Concept of Closest Approach
When an alpha particle, which has a positive charge, approaches a heavy nucleus (also positively charged), it experiences a repulsive electrostatic force. The distance of closest approach is the point where the kinetic energy of the alpha particle is completely converted into electrostatic potential energy.

### Step 2: Write the Energy Conservation Equation
At the closest approach, the kinetic energy (KE) of the alpha particle is equal to the potential energy (PE) due to the electrostatic force between the alpha particle and the nucleus.

- Kinetic Energy (KE) of the alpha particle:
\[ KE = \frac{1}{2} mv^2 \]

- Potential Energy (PE) at the closest approach:
\[ PE = \frac{k \cdot Ze \cdot 2e}{r_0} \]
where \( k \) is Coulomb's constant, \( Ze \) is the charge of the nucleus, and \( 2e \) is the charge of the alpha particle (since it consists of 2 protons).

### Step 3: Set KE Equal to PE
At the closest approach, we can set the kinetic energy equal to the potential energy:
\[ \frac{1}{2} mv^2 = \frac{k \cdot Ze \cdot 2e}{r_0} \]

### Step 4: Rearrange to Find \( r_0 \)
Rearranging the equation to solve for \( r_0 \):
\[ r_0 = \frac{2k \cdot Ze \cdot 2e}{\frac{1}{2} mv^2} \]
\[ r_0 = \frac{4k \cdot Ze^2}{mv^2} \]

### Step 5: Analyze the Relationship
From the equation \( r_0 = \frac{4k \cdot Ze^2}{mv^2} \), we can see that \( r_0 \) is inversely proportional to the mass \( m \):
\[ r_0 \propto \frac{1}{m} \]

### Conclusion
Thus, the distance of closest approach \( r_0 \) depends inversely on the mass \( m \) of the alpha particle. Therefore, we can conclude that:
\[ r_0 \propto m^{-1} \]

### Final Answer
The distance of closest approach \( r_0 \) is inversely proportional to the mass \( m \) of the alpha particle.

---