The Question is from neet-physics atoms-nuclei alpha-ray-scattering, and it was created by the author SUBHANI

Question

An alpha particle is projected towards a gold nucleus with impact parameter 'b'.  If the distance of closest approach is found to be equal to 'b', which of the following is true about the initial kinetic energy ($K$) of the alpha particle?

SUBHANI · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
	**Primary Concept**: Conservation of Energy in Electrostatic Interactions
	**Related Topics**: Electrostatic Potential Energy, Kinetic Energy, Impact Parameter, Distance of Closest Approach
	**Prerequisites**: Understanding of kinetic energy, electrostatic potential energy, and the concept of distance of closest approach in Rutherford scattering.
## SOLUTION PATHWAY
**Given Information**:
	* Impact parameter = b
	* Distance of closest approach = b
	* Alpha particle charge = 2e
	* Gold nucleus charge = 79e
**Method & Steps**:
	1.	At the distance of closest approach, the initial kinetic energy of the alpha particle is completely converted into electrostatic potential energy. The distance of closest approach 'r' is given to be equal to the impact parameter 'b'.
		Reasoning: Energy conservation dictates that initial kinetic energy equals potential energy at closest approach.
		Formula/Rule:  $K = U$, where K is kinetic energy and U is potential energy.
	2.	The electrostatic potential energy between two point charges q1 and q2 separated by a distance r is given by  $U = \frac{1}{4πε₀} \frac{q_1q_2}{r}$. In this case, q1 = 2e (alpha particle) and q2 = 79e (gold nucleus), and r = b.
		Working: Substituting values, $U = \frac{1}{4πε₀} \frac{(2e)(79e)}{b} = \frac{1}{4πε₀} \frac{158e^2}{b}$
		Key Point:  At closest approach, all kinetic energy is converted to potential energy
	3.	Since  $K=U$, the initial kinetic energy K of the alpha particle is $K = \frac{1}{4πε₀} \frac{158e^2}{b}$
		Result: The initial kinetic energy of the alpha particle is $K = \frac{1}{4πε₀} \frac{158e^2}{b}$
		Verification: The calculated kinetic energy matches option B.
 ## OPTION ANALYSIS
[A] :x: Incorrect. This would be the kinetic energy if the gold nucleus had a charge of 79e/2 or if there was a mistake in calculations.
[B] ✓: Correct.  The kinetic energy is equal to the electrostatic potential energy at the distance of closest approach which is equal to the impact parameter b. The potential energy is calculated using the given charges and distance.
[C] :x: Incorrect. The calculation of the electrostatic potential energy incorrectly uses 316e^2 in the numerator. It should be 2e * 79e = 158e^2.
[D] :x: Incorrect. The electrostatic potential energy calculation uses an incorrect charge value resulting in 2e^2 in the numerator. It should be 2e * 79e = 158e^2.
 ## LEARNING AIDS
	**Common Mistakes**:
		1.[Error 1]: Incorrectly using the impact parameter in the potential energy formula. The distance in the potential energy is the distance of closest approach, which in this case is given to be equal to the impact parameter.
			[How to avoid]: Carefully consider the distance at which potential energy is being calculated - at the closest approach.
		2.[Error 2]: Errors in calculating the product of charges. The alpha particle has a charge of +2e, and the gold nucleus has a charge of +79e. The product is (2e)(79e) = 158e^2.
			[How to avoid]: Double-check the charge values and their product.
	**Quick Tips**:
		1.[Helpful shortcut]:  Remember that at the closest approach, all kinetic energy becomes electrostatic potential energy.  This simplifies the problem. 
		2.[Memory aid]:  The electrostatic potential energy between two charges is always proportional to the product of the charges and inversely proportional to the distance between them.

SUBHANI · Answer

K = (1/4πε₀) * (79e²) / b

SUBHANI · Answer

The correct Option is "K = (1/4πε₀) * (158e²) / b"

SUBHANI · Answer

K = (1/4πε₀) * (316e²) / b

SUBHANI · Answer

K = (1/4πε₀) * (2e²) / b

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