Calculating the closest possible distance of approach

[rreedde]

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?

 

[diazona]

Try thinking about it in terms of energy.

 

[Winzer]

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.

 

[planauts]

Is the answer to this problem 2.2165 * 10^-13 m?

 

[rwisz]

To calculate the closest possible distance of approach, we can use the Coulomb's Law which states that the force of interaction between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. Mathematically, it can be expressed as:

F = k(q1q2)/r^2

Where F is the force of interaction, k is the Coulomb's constant, q1 and q2 are the charges of the two particles, and r is the distance between them.

In this case, we can calculate the force of interaction between the alpha particle and the gold nucleus using the charges given in the problem and the Coulomb's constant (k = 8.99 x 10^9 Nm^2/C^2). The force of interaction will be equal to the centripetal force acting on the alpha particle, which is given by:

F = mv^2/r

Where m is the mass of the alpha particle and v is its velocity.

Equating these two equations, we can solve for the distance of approach (r):

mv^2/r = k(q1q2)/r^2

r = k(q1q2)/mv^2

Substituting the values given in the problem, we get:

r = (8.99 x 10^9 Nm^2/C^2)(2e)(79e)/(4.00 x 10^-27 kg)(7.0 x 10^6 m/s)^2

r = 4.5 x 10^-14 meters

Therefore, the closest possible distance of approach between the alpha particle and the gold nucleus is 4.5 x 10^-14 meters.