The Question is from neet-physics nuclei alpha-ray-scattering, and it was created by the author Mr. ANIL KUMAR.G

Question

An elementary particle of mass $m$ and charge $ + e$ is projected with velocity $v$ at a much more massive particle of charge $Ze,$ where $Z > 0.$ What is the closest possible approach of the incident particle

Mr. ANIL KUMAR.G · Accepted Answer

## CORE INFORMATION
✓ **Answer**: A
## CONCEPT FRAMEWORK
	**Primary Concept**: Conservation of Energy
	**Related Topics**: Electrostatic Potential Energy, Kinetic Energy
	**Prerequisites**: Understanding of kinetic energy, electrostatic potential energy, and conservation of energy.
## SOLUTION PATHWAY
**Given Information**:
	* Mass of the incident particle: $m$
	* Charge of the incident particle: $+e$
	* Initial velocity of the incident particle: $v$
	* Charge of the massive particle: $Ze$
	* The massive particle is much more massive, so it can be considered stationary.
	* We are looking for the closest distance of approach.
**Method & Steps**:
	1.  Initial energy of the incident particle is purely kinetic energy. At the closest approach, the particle momentarily stops, so all of its initial kinetic energy is converted to electrostatic potential energy.
		Reasoning: Energy conservation principle.
		Formula/Rule: $KE = \frac{1}{2}mv^2$,  $U = \frac{k q_1 q_2}{r}$, where $k = \frac{1}{4\pi \epsilon_0}$
	2.  Equate initial kinetic energy to final potential energy.
		Working:  $\frac{1}{2}mv^2 = \frac{1}{4\pi\epsilon_0} \frac{(e)(Ze)}{r_{min}}$
		Key Point:  At the closest approach, the kinetic energy is zero and all the initial kinetic energy is converted into potential energy.
	3.  Solve for the closest distance of approach, $r_{min}$.
		Result:  $r_{min} = \frac{2Ze^2}{4\pi\epsilon_0 mv^2} = \frac{Ze^2}{2\pi\epsilon_0 mv^2}$
		Verification: Dimensional analysis confirms the answer is in units of distance.
 ## OPTION ANALYSIS
[A] ✓: This is the correct result obtained by equating initial kinetic energy to final potential energy at the closest approach.
[B] :x: This option has the correct variables but is missing a factor of 2 in the denominator.
[C] :x: This option has an extra factor of 2 in the denominator. It also has $Z e^2$ instead of $Ze$.
[D] :x: This option is missing a factor of $e$ and has an extra factor of 2 in the denominator.
 ## LEARNING AIDS
	**Common Mistakes**:
		1. [Error 1]: Forgetting to use the correct electrostatic potential energy formula ($U = \frac{k q_1 q_2}{r}$). [How to avoid]: Always double-check the formula and pay attention to units.
		2. [Error 2]: Incorrectly equating the initial and final energies. [How to avoid]: Clearly define the initial and final states and understand energy conservation.
	**Quick Tips**:
		1. [Helpful shortcut]: In problems involving closest approach, remember that kinetic energy is converted to potential energy.
		2. [Memory aid]: Use the conservation of energy principle as your starting point for these types of problems.

Mr. ANIL KUMAR.G · Answer

The correct Option is "$\frac{{Z{e^2}}}{{2\pi {\varepsilon _0}m{v^2}}}$"

Mr. ANIL KUMAR.G · Answer

$\frac{{Ze}}{{4\pi {\varepsilon _0}m{v^2}}}$

Mr. ANIL KUMAR.G · Answer

$\frac{{Z{e^2}}}{{8\pi {\varepsilon _0}m{v^2}}}$

Mr. ANIL KUMAR.G · Answer

$\frac{{Ze}}{{8\pi {\varepsilon _0}m{v^2}}}$

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