A 4.78 -MeV alpha particle from a ^226 Ra decay makes a head-on collision with a uranium nucleus

Name: Problem 22, Chapter 39: University Physics with Modern Physics
Uploaded: 2019-05-12T22:47:29-08:00
Duration: 34 s
Channel: Christopher Provencher

A 4.78 -MeV alpha particle from a ^226 Ra decay makes a...

A 4.78 -MeV alpha particle from a $^{226} \mathrm{Ra}$ decay makes a head-on collision with a uranium nucleus. A uranium nucleus has 92 protons. (a) What is the distance of closest approach of the alpha particle to the center of the nucleus? Assume that the uranium nucleus remains at rest and that the distance of closest approach is much greater than the radius of the uranium nucleus. (b) What is the force on the alpha particle at the instant when it is at the distance of closest approach?

Key Concepts

Energy Conservation

Energy conservation is used to relate the kinetic energy of the incident particle to the potential energy at the point of closest approach. At this point, all the kinetic energy has been converted into electrostatic potential energy, providing the basis for calculations in head-on collisions between charged particles.

Electrostatic (Coulomb) Potential Energy

Electrostatic potential energy arises from the interaction between charged particles and is calculated as the product of the two charges divided by the separation between them, scaled by Coulomb's constant. This concept is fundamental for determining the energy state of two charged particles at a given distance.

Coulomb's Law

Coulomb's Law quantifies the force between two point charges by stating that the force is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. This law is essential in calculating both the potential energy and the force experienced at the closest approach in a collision.

Distance of Closest Approach

The distance of closest approach in a particle collision is the minimum separation between the interacting charged particles, occurring when the entire initial kinetic energy is transformed into electrostatic potential energy. This concept is critical for applying energy conservation principles to determine how close two charged particles will come under repulsive forces.

Transcript

0:00 Okay.

00:02 In this problem, we have an alpha particle with an energy of 4 .78 mega electron volts, moving directly towards a big uranium nucleus, which has a charge of 92 electrons because it has 92 protons.

00:17 And we are asked to find the distance closest that this alpha particle gets to the nucleus before being turned around and deflected.

00:28 And at this particular distance away, we're asked for the strength of the kulom force.

00:34 So the equations we'll use for this problem is first we'll need to know the potential energy of a particle, an electric potential energy, which is simply the charge of the particle times the potential, electric potential it's in.

00:46 And we need to know the electric potential due to a point charge, which is simply the electric constant k, times the charge divided by a distance, r.

00:55 We'll also need the kulum force, f equals k, q1, 2, of our, squared, it's an inverse square law.

01:02 So let's jump into it.

01:04 So to start, we know that when a particle is about to be deflected and sent backwards, its kinetic energy will be zero.

01:14 It'll be completely stopped.

01:18 So its kinetic energy initial will be equal to the potential energy final in this situation.

01:29 Thus, we can replace u with the potential energy, the electric potential energy.

01:36 So k will be equal to qv...

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