Calculating the closest possible distance of approach

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To calculate the closest distance of approach between an alpha particle and a gold nucleus, consider the electrostatic repulsion due to their charges. The alpha particle has a charge of 2e, while the gold nucleus has a charge of 79e. The initial kinetic energy of the alpha particle, derived from its speed of 7.0 x 10^6 m/s, is converted into potential energy at the point of closest approach. The discussion suggests that the final answer for this distance is approximately 2.2165 x 10^-13 m. This approach effectively utilizes energy conservation principles to solve the problem.
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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


-

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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