The Question is from neet-physics , and it was created by the author P.NARASIMHA

Question

When an  ${\rm{\alpha }}$ - particle of mass ‘m’ moving with the velocity ‘v’ bombards on a heavy nucleus of charge ‘Ze’, its distance of closest approach from the nucleus depends on m as:

P.NARASIMHA · Accepted Answer

## CORE INFORMATION
✓ **Answer**: A
## CONCEPT FRAMEWORK
	**Primary Concept**: Conservation of energy during alpha particle scattering.
	**Related Topics**: Electrostatic potential energy, Kinetic energy, Distance of closest approach.
	**Prerequisites**: Understanding of energy conservation, Electrostatic potential energy formula, Kinetic energy formula.
## SOLUTION PATHWAY
**Given Information**:
	* Alpha particle mass = m
	* Alpha particle velocity = v
	* Nucleus charge = Ze
	* We need to find the dependence of distance of closest approach on mass (m)
**Method & Steps**:
	1. **Initial Kinetic Energy**: The initial kinetic energy of the alpha particle is given by  $KE = \frac{1}{2}mv^2$.
		Reasoning: This is the standard formula for kinetic energy.
		Formula/Rule: $KE = \frac{1}{2}mv^2$
	2. **Electrostatic Potential Energy at Closest Approach**: At the distance of closest approach (r), all the initial kinetic energy of the alpha particle is converted into electrostatic potential energy. The electrostatic potential energy is given by $PE = \frac{k(2e)(Ze)}{r}$, where k is the Coulomb's constant, 2e is the charge of the alpha particle, Ze is the charge of the nucleus, and r is the distance of closest approach.
		Reasoning: Energy conservation, initial kinetic energy is equal to final potential energy.
		Formula/Rule: $PE = \frac{k(2e)(Ze)}{r}$
	3. **Equating Kinetic and Potential Energy**: By the conservation of energy, we have $\frac{1}{2}mv^2 = \frac{k(2e)(Ze)}{r}$.
		Reasoning: Conservation of energy
	4. **Solving for r**: Rearranging the equation to find r, we get  $r = \frac{2kZe^2}{\frac{1}{2}mv^2} = \frac{4kZe^2}{mv^2}$.
		Working: Algebraic manipulation.
	5. **Dependence on m**: From the equation, we can see that the distance of closest approach 'r' is inversely proportional to the mass 'm' i.e. $r \propto \frac{1}{m}$.
		Result: Distance of closest approach is inversely proportional to mass (m)
 ## OPTION ANALYSIS
[A] ✓: The distance of closest approach, 'r' is inversely proportional to the mass 'm', hence $r \propto \frac{1}{m}$.
[B] :x:  The distance of closest approach is inversely proportional to m, not $\sqrt{m}$
[C] :x: The distance of closest approach is inversely proportional to m, not $m^2$
[D] :x: The distance of closest approach is inversely proportional to m, not directly proportional to m
 ## LEARNING AIDS
	**Common Mistakes**:
		1. **Incorrect Potential Energy Formula**: Forgetting to use the correct charges in the potential energy formula. (Remember the alpha particle has a charge of 2e).
		2. **Incorrectly Solving for r**: Making algebraic errors while solving for r.
	**Quick Tips**:
		1. Remember the concept of energy conservation.
		2. The distance of closest approach is when all KE is converted to PE.

P.NARASIMHA · Answer

The correct Option is "$\frac{1}{{\rm{m}}}$"

P.NARASIMHA · Answer

$\frac{1}{{\sqrt {\rm{m}} }}$

P.NARASIMHA · Answer

$\frac{1}{{{{\rm{m}}^2}}}$

P.NARASIMHA · Answer

$m$

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