An αparticle with a kinetic energy of 1.0 MeV is headed straight toward a gold nucleus. (a) Find

An αparticle with a kinetic energy of 1.0 MeV is headed...

An $\alpha$ particle with a kinetic energy of $1.0 \mathrm{MeV}$ is headed straight toward a gold nucleus. (a) Find the distance of closest approach between the centers of the $\alpha$ particle and gold nucleus. (Assume the gold nucleus remains stationary. since its mass is much larger than that of the $\alpha$ particle, this assumption is a fairly good approximation.) (b) Will the two get close enough to "touch"? (c) What is the minimum initial kinetic energy of an $\alpha$ particle that will make contact with the gold nucleus?

An $\alpha$ particle with a kinetic energy of $1.0 \mathrm{MeV}$ is headed straight toward a gold nucleus. (a) Find the distance of closest approach between the centers of the $\alpha$ particle and gold nucleus. (Assume the gold nucleus remains stationary. since its mass is much larger than that of the $\alpha$ particle, this assumption is a fairly good approximation.) (b) Will the two get close enough to "touch"?
(c) What is the minimum initial kinetic energy of an $\alpha$ particle that will make contact with the gold nucleus?

Key Concepts

Coulomb Potential Energy

This is the energy associated with the repulsive force between two charged particles. In this context, when an alpha particle approaches a gold nucleus, its kinetic energy is converted into Coulomb potential energy. The magnitude of this potential is given by the equation V = k(Q1Q2)/r, where Q1 and Q2 are the charges of the interacting particles, k is Coulomb's constant, and r is the distance between their centers.

Conservation of Energy in Particle Scattering

This principle states that the total energy in an isolated system remains constant. When an alpha particle approaches a nucleus, its kinetic energy is gradually converted into potential energy as work is done against the Coulomb force. At the distance of closest approach, all the initial kinetic energy has been converted into Coulomb potential energy, allowing one to set these energies equal to solve for the closest distance.

Distance of Closest Approach

This term refers to the minimum separation between the centers of two interacting particles during a collision process, such as scattering. It is determined by equating the particle's initial kinetic energy to the Coulomb potential energy at that separation, highlighting the conversion of energy as the particle slows down under repulsion.

Nuclear Size and Contact Considerations

Even when two nuclei come close during a scattering event, they must overcome the Coulomb barrier to actually 'touch' or interact via the strong nuclear force. The concept of nuclear radii is crucial here, as contact implies the distance between centers is about the sum of the radii of both nuclei. Calculations involving these radii are necessary to determine whether the particles reach proximity close enough for the strong nuclear force to become significant.

Transcript

00:01 For this question, we're looking at an alpha particle approaching a gold nucleus.

00:10 This is our gold nucleus, which is much heavier than our alpha particle.

00:17 And we're given that it has a certain amount of kinetic energy of 1 m .v.

00:26 And we want to find what is the distance of closest approach.

00:29 Now the distance of closest approach, we are assuming that at that point, all of the kinetic energy has actually been converted into electric potential energy.

00:44 Because they will experience kulom repulsion, and so they gain electric potential energy.

00:50 And it will only keep gaining until the ke actually runs out, you know, and the particle can no longer move closer to each other.

01:00 And so the closest approach will be when k -e is equals to pe, your electric potential energy, or epe.

01:13 Now the epe, we can find, we know, that is actually based on the charges of the two nuclei that we are talking about.

01:25 So the charge of the alpha times the charge of the gold, divide the 5 pi epsilon or not times the distance between them which is the closest approach that we want right that's r so r is the distance between the two different nuclei now we can solve for r since we all have all these values iq alpha is 2 2 times the charge of electron rest for a goal it is actually 79 times the charge of the electron.

02:10 And electron charge is simply 1 .602 times 10 .5 minus 19 globes.

02:18 So we equate this to 1 .0 mv.

02:26 And 1 .0 mv we can convert it into juice.

02:34 6 ,000 times 1 .06 times 10 to power minus 19, juice.

02:44 Working in the si units would be more convenient...

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