Calculating the closest possible distance of approach

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Homework Help Overview

The discussion revolves around calculating the closest possible distance of approach between an alpha particle and a gold nucleus, focusing on the electrostatic repulsion involved in the interaction. The context includes concepts from electrostatics and energy conservation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between the speed of the alpha particle and its energy, questioning how to incorporate this speed into their calculations. There is a suggestion to consider energy transformations at the point of closest approach.

Discussion Status

Some participants have offered guidance on considering energy concepts, indicating a productive direction in the discussion. However, there is no explicit consensus on the final answer, as one participant proposes a specific numerical result without confirmation from others.

Contextual Notes

Participants are navigating the constraints of the problem, particularly regarding the use of speed and energy in their calculations. The original poster expresses uncertainty about how to proceed with the given information.

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Homework Statement


An alpha particle 4 2 He 2+ approaches a stationary gold nucleus 197 79 Au at a speed of 7.0 X 10^6 ms-1.
By considering only electrostatic repulsion, calculate the closest possible distance of apporach between the alpha particle and gold nucleus.

Homework Equations


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The Attempt at a Solution


Charge of Helium particle = 2e
Charge of gold nucleus = 79 e
And I am just stuck here, how do i use the value of the speed of the particle in my working?
 
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Try thinking about it in terms of energy.
 
diazona said:
Try thinking about it in terms of energy.

I agree with this.
The alpha particle stars of with a speed and thus a certain type of energy(hint).
When the particle reaches its closest approach, all the energy is converted into this other type of energy. You should be able to get it from there.
 
Is the answer to this problem 2.2165 * 10^-13 m?
 

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