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Ravenatic20

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## Homework Statement

The muon is an unstable particle that spontaneously decays into an electron and two neutrinos. If the number of muons at t = 0 is [tex]N_{o}[/tex], the number at time t is given by [tex]N = N_{0}e^{-t/\tau}[/tex], where [tex]\tau[/tex] is the mean lifetime, equal to 2.2 [tex]\mu s[/tex]. Suppose the muons move at a speed of 0.95c and there are [tex]5.0 X 10^{4}[/tex] muons at t = 0. (a) What is the observed lifetime of the muons? (b) How many muons remain after traveling a distance of 3.0 km?

## Homework Equations

[tex]N = N_{0}e^{-t/\tau}[/tex]

[tex]t = d/v[/tex]

[tex]\tau ' = \gamma \tau[/tex]

[tex]\gamma = 1/ \sqrt{1 - v^{2}/c^{2}}[/tex]

## The Attempt at a Solution

[tex]t = d/v => 3 km/0.95c = 1.05 X 10^{-5}[/tex] where c is [tex]3 X 10^{8}[/tex]

In the Muon frame: [tex]\tau = 2.2\mu[/tex]

In the Earth frame: [tex]\tau ' = \gamma \tau[/tex]

[tex]\gamma = 1/ \sqrt{1 - v^{2}/c^{2}} => 1/ \sqrt{1 - (0.95c/c)^{2}}[/tex]

[tex]\gamma = 3.2 \mu s[/tex]

Plug this back into the equation for [tex]\tau '[/tex]...

[tex]\tau ' = 7.04 \mu s[/tex]

In the back of my book, the solution for part (a) is [tex]7.1 \mu s[/tex]. This answer ([tex]\tau ' = 7.04 \mu s[/tex]) doesn't have anything to do with the answer to (a), does it?

Next, I need to plug in the numbers I have into N, which is [tex]N = N_{0}e^{-t/\tau}[/tex]

If I read the problem right, [tex]N_{o} = 5.0 X 10^{4}[/tex]. We know t, and we know the new value of tau. So it's plug and chug, right?

If so, this is what I got:

[tex]N = N_{0}e^{-t/\tau}[/tex]

[tex]N = 5.0 X 10^{4}e^{-1.05 X 10^{-5}/7.04 \mu s}[/tex]

[tex]N = 4.9 X 10^{-4}[/tex]

This is not the same answer in the back of the book, nor does it make sense. The answer in the back of the book is [tex]1.1 X 10^{-4}[/tex]

Where am I going wrong? Please put me on track, thanks!