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 find the distance of closest approach \( D \) of an alpha particle bombarding a heavy nucleus, we can use the principles of conservation of energy. Here’s a step-by-step solution:

### Step 1: Understand the System
The alpha particle has a mass \( m \) and a charge of \( +2e \). The heavy nucleus has a charge of \( Ze \). When the alpha particle approaches the nucleus, it experiences a repulsive electrostatic force due to the positive charges.

### Step 2: Initial Energy Calculation
Initially, the alpha particle is moving with a velocity \( v \). Therefore, its initial kinetic energy \( KE_i \) is given by:
\[ KE_i = \frac{1}{2} mv^2 \]
The potential energy \( PE_i \) at an infinite distance is considered to be zero:
\[ PE_i = 0 \]
Thus, the total initial energy \( E_i \) is:
\[ E_i = KE_i + PE_i = \frac{1}{2} mv^2 + 0 = \frac{1}{2} mv^2 \]

### Step 3: Energy at the Closest Approach
At the distance of closest approach \( D \), the alpha particle momentarily comes to rest, so its kinetic energy \( KE_f \) is zero:
\[ KE_f = 0 \]
The potential energy \( PE_f \) at this distance is given by the electrostatic potential energy formula:
\[ PE_f = \frac{k \cdot (2e)(Ze)}{D} = \frac{2kZe^2}{D} \]
where \( k \) is Coulomb's constant.

### Step 4: Apply Conservation of Energy
According to the conservation of energy, the total initial energy equals the total final energy:
\[ E_i = E_f \]
Substituting the values we have:
\[ \frac{1}{2} mv^2 = 0 + \frac{2kZe^2}{D} \]

### Step 5: Solve for Distance \( D \)
Rearranging the equation to solve for \( D \):
\[ \frac{2kZe^2}{D} = \frac{1}{2} mv^2 \]
\[ D = \frac{4kZe^2}{mv^2} \]

### Step 6: Analyze Dependence on Mass \( m \)
From the equation \( D = \frac{4kZe^2}{mv^2} \), we can see that \( D \) is inversely proportional to the mass \( m \):
\[ D \propto \frac{1}{m} \]

### Conclusion
Thus, the distance of closest approach \( D \) depends inversely on the mass \( m \) of the alpha particle.

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

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