(a) Use energy methods to calculate the distance of closest...

(a) Use energy methods to calculate the distance of closest approach for a head-on collision between an alpha particle having an initial energy of 0.500 $\mathrm{MeV}$ and a gold nucleus ( 197 $\mathrm{Au} )$ at rest. Assume the gold nucleus remains at rest during the collision. (b) What minimum initial speed must the alpha particle have to get as close as 300 $\mathrm{fm}$ ?

Key Concepts

Energy Conservation

This concept involves equating the total initial kinetic energy of a moving particle to the potential energy it accumulates as it slows down under the influence of a force (in this case, the Coulomb force). In collision problems where the interaction is conservative, the loss in kinetic energy is completely accounted for by a gain in potential energy, allowing for the calculation of turning points such as the distance of closest approach.

Coulomb Potential Energy

Coulomb potential energy describes the energy due to the electric interaction between two point charges. It is given by the formula V = k(q1q2)/r, where k is Coulomb’s constant, q1 and q2 are the charges of the interacting bodies, and r is the separation between them. This concept is essential when analyzing the repulsive or attractive forces in nuclear collision scenarios.

Distance of Closest Approach

The distance of closest approach is defined as the minimum separation between two interacting charged particles during a collision, where the entire kinetic energy of the incoming particle is converted into potential energy. By setting the initial kinetic energy equal to the Coulomb potential energy at this minimum distance, one can solve for r, obtaining an understanding of how close the particles can come under repulsive forces.

Minimum Initial Speed Requirement

When a particle needs to overcome or reach a specific potential energy barrier to achieve a given distance of closest approach, its initial kinetic energy must be sufficiently high. Expressing the kinetic energy in terms of particle mass and initial speed (using the expression 1/2 mv²) and equating it to the Coulomb potential at the desired distance allows for the determination of the minimum initial speed required.

Transcript

00:01 We have over here a stationary gold nuclei and we have an alpha particle that is actually bombarding or moving towards in a head -on collision with this gold particle.

00:18 We are given that the energy that this alpha particle has, and the start is 0 .5 m .ev.

00:29 We want to find out what is the distance of closest approach.

00:36 So this distance of closest approach is when the alpha particle reaches a point where it stops moving, right? so its kinetic energy is zero and that is because the kinetic energy has been converted all into electric potential energy.

00:52 So electrical potential energy between these two charged nuclei can be described as k times q1 q2 over r.

01:06 Where r is the distance between the centers of the two nuclei.

01:14 So we just equate this electric potential energy to the kinetic energy, initial kinetic energy, because its initial kinetic energy has all been converted into this potential energy.

01:26 And we should be able to solve for r.

01:32 So q1 and q2 are the charges of the alpha particle and the gold particle respectively.

01:38 So we charge for the alpha particle.

01:40 This is a charge for the alpha particle.

01:41 2e.

01:42 Fragical particle is 79e.

01:46 It is based on the number of protons that they have.

01:53 Kinetic energy.

01:56 So ke, over here this is a constant 1 over 4 pi epsilon knot, which is also equals to 8 .99 times 10 to power 9.

02:09 Then we have 2 and 79e.

02:23 The kinetic energy will have to convert from m .v into joules...