Why Are We Bombarded by Muons? Muons are unstable subatomic particles that decay to electrons with a mean lifetime of 2.2 ms. They are produced when cosmic rays bombard the upper atmosphere about 10 km above the earth's surface, and they travel very close to the speed of light. The problem we want to address is why we see any of them at the earth's surface. (a) What is the greatest distance a muon could travel during its 2.2@ms lifetime? (b) According to your answer in part (a), it would seem that muons could never make it to the ground. But the 2.2@ms lifetime is measured in the frame of the muon, and muons are moving very fast. At a speed of 0.999c, what is the mean lifetime of a muon as measured by an observer at rest on the earth? How far would the muon travel in this time? Does this result explain why we find muons in cosmic rays? (c) From the point of view of the muon, it still lives for only 2.2 ms, so how does it make it to the ground? What is the thickness of the 10 km of atmosphere through which the muon must travel, as measured by the muon? Is it now clear how the muon is able to reach the ground?

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Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. So an unstable particle released during a nuclear disintegration has a average lifetime of 2.2 microseconds. A particle release. A particle is released 5.0 kilometers away from a detector. The speed of the particle in the atmosphere approaches the speed of light at V equals 0.995 C I determine the average lifetime of the particle as measured by an observer at rest. And I I, the distance covered by the particle during the time measured by the observer at rest. I I, I use the answers in part I and I I to state what you think about the possibility of the particle hitting the detector. Awesome. So we're given some multiple choice answers. Let's read those off to see what our final set of answers might be. A is I 2.20 multiplied by 10 of the power of negative five S per seconds I I 6.58 kilometers. I I I, the particle hits the detector B I 2. multiplied by 10 of the power of negative five seconds. I I 6.58 kilometers I I I, the particle does not reach the detector C I 2.20, multiplied by 10 to the power of negative seven seconds. I I 65.6 m. I I I, the particle hits the detector and D I 2.20 multiplied by 10 to the power of negative seven seconds. I I 65.6 m. I I I, the particle does not reach the detector. OK. So to solve for, I, we must recall and use the relativistic factor equation and the time dilation equation. So let's write those down. So the relativistic factor equation, it's also known as the Lorentz factor equation states that gamma equals one divided by the square root of one minus velocity squared divided by the speed of light squared. OK. So, and then the time dilation equation states that delta T equals gamma multiplied by delta T zero, delta T subscript zero. OK. So just to note here, delta T zero represents the proper time. OK. OK. So first off, let's start by solving for gamma. And then once we solve for gamma, we can plug it in to sulfur delta T. So let's do that. So this is just officially start solving for I. So let's solve for gamma. So gamma equals plugging in our known variables. So one divided by the square root of one minus and the velocity is given to us in the problem as zero point 995 C divided by C squared. OK. So note that the CS cancel out. So when you plug that into a calculator, you should get 10.125. Now we can plug gamma in for the time dilation equation, the sulfur delta T. So let's do that. OK. So our gamma value was 10. multiplied by. And in the problem, we're given the average lifetime which represents the T zero delta T zero. So the proper time. So it's 2.2 microseconds, but we need to convert microseconds to seconds. So we need to do some quick dimensional analysis to do that. So let's do that really quick because we can't have it in microseconds, we need it in seconds. So to do that using dimensional analysis, we know that in one second there is one million microseconds. So the microseconds cancel out just leaving seconds. So when you plug it into a calculator, the value for delta T that you should get is 2.20, multiplied by 10 to the power of negative five secondss. And this is our answer where I, OK. So now we could start solving for I I. So we need to recall and use the distance traveled equation which states that the distance traveled is equal to the velocity multiplied by delta T. OK. So let's plug in our known variables. So we know that the velocity is 0.995 C. But let's make a quick note here that C is the speed of light. And the numerical value for that is 3.0 multiplied by 10 to the power of eight meters per second. OK. So it was 0.995 C but C equals 3.0, multiplied by 10 to the power of eight m per second, multiplied by delta T which we just discovered delta T to be 2.20, multiplied by 10 to the power of negative five seconds. So the seconds cancel out leaving meters. So when you plug that into a calculator note that the answer is in meters. So it's 657 sorry, 606,575 m. But we need to convert that two kilometers. So we need to use dimensional analysis. So 6575 m multiplied by one kilometer. So in one kilometer, there is 1000 m. So when you plug that into a calculator, you should get 6. kilometers. Awesome. So that is our answer for I I, OK. Finally you could start solving for I I I, so let's do that. OK. So in order to sol for I I, we need to recall the following characteristics of relativistic effects. So relativistic effects reveal that an observer at breast will measure a greater lifespan of the particle. So with that in mind, therefore, the particle will hit the detector. So the particle hits the detector. Awesome. So we solve for everything we need to solve for hooray. We did it. How exciting. OK. So that means our final answer must be A I is 2.20, multiplied by 10 to the power of negative five seconds. I I is 6.58 kilometers and I I I the particle hits the detector. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

Verified Solution

Key Concepts

Muon Lifetime and Decay

Relativistic Time Dilation

Cosmic Rays and Atmospheric Interactions