Estimate the mean free path of a cosmic ray proton in the atmosphere at sea level.

Name: Problem 151, Chapter 2: Problems and Solutions on Thermodynamics and Statistical Mechanics
Uploaded: 2020-06-16T18:43:41-08:00
Duration: 7 min 39 s
Channel: Donald Albin
Description: Estimate the mean free path of a cosmic ray proton in the atmosphere at sea level.

step 1 In this problem, we're basically told kinetic energy. So I'm going to write that as kinetic ener

Estimate the mean free path of a cosmic ray proton in the...

Key Concepts

Mean Free Path

The mean free path is a measure of the average distance a particle travels before undergoing an interaction or collision. It is typically calculated using the formula ? = 1/(n?), where n is the number density of the target particles and ? is the interaction cross section. This concept helps in understanding how frequently particles interact with the medium they traverse.

Collision Cross Section

The collision cross section quantifies the likelihood of an interaction occurring between a moving particle and a target particle. It represents an effective area that influences the probability of collisions. Larger cross sections indicate a higher chance of interactions, which are critical in determining the mean free path in any medium, including the atmosphere.

Atmospheric Density

Atmospheric density refers to the concentration of particles, such as molecules, in the Earth's atmosphere. At sea level, the density is relatively high, meaning that particles like cosmic ray protons are more likely to collide with air molecules. This density plays a key role in calculating the mean free path, as it directly affects the value of n in the mean free path equation.

Cosmic Ray Particles

Cosmic rays, which include high-energy protons, are particles originating from outer space that interact with the Earth's atmosphere. Their high energies allow them to penetrate deep into the atmospheric layers, where they can collide with atomic nuclei and trigger cascades of secondary particles. Understanding their interactions is crucial for estimating their mean free paths and the subsequent atmospheric phenomena.

Transcript

00:01 In this problem, we're basically told kinetic energy.

00:04 So i'm going to write that as kinetic energy.

00:09 And that is 10 to the 20th power electron volts, which is 10 to the 20th power times 1 .6 times 10 to the negative 9th power.

00:30 I meant 19th power, sorry, kulams times one volt.

00:48 Now, since that is energy, and since we're talking about protons, we can write that this is equal to 1⁄2m2, and the mass is the mass of a proton, which is 1 .67 times 10 to the negative 27 power kilograms.

01:24 Using this, we could figure out a velocity.

01:35 So let's go ahead and do that.

01:37 V is going to be 2 times this, okay? 2 times that divided by the mass.

02:01 We have to take the square root of that.

02:03 So let's go ahead and calculate that.

02:11 2 times 10 to the 20th power times 1 .6 times 10 to the negative 19th power.

02:24 Divided by mass, which is 1 .67 times 10 to the negative 27th power.

02:33 So that gives me a speed of one point, i'm going to use two significant digits here, 38 times 10 to the 14th power meters per second.

02:51 Now, we also know that the magnetic force on a charged particle is qv, lowercase v, b, which we can set to mass times v squared over r.

03:27 So if we solve this for r, we get r equals, and by the way, this and this will cancel out, mv over qb.

03:43 We know v, we know m...

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