Predict/Calculate An αparticle with a kinetic energy of 0.85...

Predict/Calculate An $\alpha$ particle with a kinetic energy of 0.85 MeV approaches a stationary gold nucleus. (a) What is the speed of the $\alpha$ particle? (b) What is the distance of closest approach between the $\alpha$ particle and the gold nucleus? (c) If this same $\alpha$ particle were fired at a copper nucleus instead, would its distance of closest approach be greater than, less than, or the same as that found in part (b)? Explain. (To obtain the mass of an alpha particle, refer to Appendix $F$ and subtract the mass of two electrons from the mass of $_{2}^{4}$ He.)

Predict/Calculate An $\alpha$ particle with a kinetic energy of 0.85 MeV approaches a stationary gold nucleus. (a) What is the speed of the $\alpha$ particle? (b) What is the distance of closest approach between the $\alpha$ particle and the gold nucleus? (c) If this same $\alpha$ particle were fired at a copper nucleus instead, would its distance
of closest approach be greater than, less than, or the same as that
found in part (b)? Explain. (To obtain the mass of an alpha particle, refer to Appendix $F$ and subtract the mass of two electrons from the mass of $_{2}^{4}$ He.)

Key Concepts

Kinetic Energy and Velocity Relation

This concept involves using the classical mechanics formula KE = ½mv² to determine the speed of a particle given its kinetic energy and mass. It is fundamental in connecting the energy content of a moving particle to its velocity by rearranging the formula to solve for v. This conversion is essential in many areas of physics where energy and movement are analyzed.

Coulomb Potential Energy

Coulomb potential energy describes the energy between two charged particles and is given by U = k(q?q?)/r. In scattering scenarios, this concept is vital because the distance of closest approach is determined by equating the particle's kinetic energy with its potential energy when it is at the turning point in its trajectory under repulsive forces.

Conservation of Energy in Scattering

In nuclear scattering experiments, the total energy of the system is conserved. Initially, the particle has kinetic energy, and at the point of closest approach, all this kinetic energy is converted into electrostatic potential energy from the nuclear Coulomb repulsion. Understanding this energy conservation is key to analyzing and predicting the behavior of particles during scattering.

Charge Dependence in Coulomb Repulsion

The strength of the Coulomb repulsion between two charged particles depends directly on the product of their charges. In the context of scattering, a nucleus with a higher positive charge will exert a stronger repulsive force on a positively charged projectile, thereby reducing the distance of closest approach compared to a nucleus with a lower charge. This principle explains differences in behavior when the projectile interacts with nuclei of different elements.

Transcript

0:00 Question 7.

00:01 Chapter 32 states an alpha particle with kinetic energy of 0 .85 m .e .v.

00:07 Approaches a stationary gold nucleus.

00:10 Part a asks, what is the speed of the alpha particle? we'll start there.

00:16 Moving with speed.

00:18 So we're given the particle's kinetic energy.

00:21 I'll call k.

00:22 And remember from our mechanics, that's energy and, sorry, kinetic energy and velocity relationship is just given as 1 .5 mv squared.

00:35 So we can rearrange this equation to solve for velocity.

00:41 I'll do the right hand side here, so that's just 2 times the kinetic energy over mass, all square rooted.

00:54 So we can use this.

00:56 The mass, we can deal with it in terms of kilograms, but it's easier to deal with it in terms of its atomic mass units, which will become evident because we are dealing with kinetic energy in terms of m .e .v.

01:10 So if we just write our denominator out here, i'll do that before we draw anything else.

01:15 So our atomic mass of helium -4.

01:18 We look up in appendix f is 4 .00 -2603 u.

01:27 And we just look ahead a little bit.

01:29 Again, our numerator is going to have m -e -v here.

01:33 So we want to get rid of that.

01:34 So we can use our mass energy conversion, 931 .494 m -ev per.

01:44 C squared.

01:47 So usually when our energy equations the c squared will cancel out, but we don't have that.

01:50 So we need to include that c squared in our calculation, 2 times 0 .85 m .ev in our numerator.

02:02 And again, this whole term is square rooted.

02:07 So our units for energy mep cancel out.

02:10 But again, we do have this c squared term.

02:12 We need to take to account.

02:14 So we know the speed of light is 3 .00 times the 8 meters per second.

02:19 So if we include that with the rest of our calculation, we find that our resulting velocity here to two significant figures is simply 6 .4 times 10 to the 6 meters per second.

02:39 Again, so this having this c squared term effectively on the numerator, c squared having here, multiplying that term out if we remove it from a denominator, just to make it a little more explicit.

02:56 Part b, this question asks, what is the distance of closest approach between the alpha particle and the gold nucleus? so the scenario, we have our alpha particle has a kinetic energy, again, which calls k.

03:10 It will approach the gold nucleus and then remain at a certain distance away.

03:15 So that distance between them is described as the electric potential energy.

03:19 So this kinetic energy is transferred to potential energy.

03:25 And whilst it's transferred, it's simply the kinetic, the energy that the particle has goes from kinetic to potential.

03:35 So we can just state that these two stacks are equal.

03:40 And if remember what our kinetic energy, sorry, our electric potential energy is for two particles.

03:47 We have our cool -oom constant little k -e...

Please give Ace some feedback