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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 N_{o}, the number at time t is given by N = N_{0}e^{-t/\tau}, where \tau is the mean lifetime, equal to 2.2 \mu s. Suppose the muons move at a speed of 0.95c and there are 5.0 X 10^{4} 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
N = N_{0}e^{-t/\tau}
t = d/v
\tau ' = \gamma \tau
\gamma = 1/ \sqrt{1 - v^{2}/c^{2}}
The Attempt at a Solution
t = d/v => 3 km/0.95c = 1.05 X 10^{-5} where c is 3 X 10^{8}
In the Muon frame: \tau = 2.2\mu
In the Earth frame: \tau ' = \gamma \tau
\gamma = 1/ \sqrt{1 - v^{2}/c^{2}} => 1/ \sqrt{1 - (0.95c/c)^{2}}
\gamma = 3.2 \mu s
Plug this back into the equation for \tau '...
\tau ' = 7.04 \mu s
In the back of my book, the solution for part (a) is 7.1 \mu s. This answer (\tau ' = 7.04 \mu s) 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 N = N_{0}e^{-t/\tau}
If I read the problem right, N_{o} = 5.0 X 10^{4}. 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:
N = N_{0}e^{-t/\tau}
N = 5.0 X 10^{4}e^{-1.05 X 10^{-5}/7.04 \mu s}
N = 4.9 X 10^{-4}
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 1.1 X 10^{-4}
Where am I going wrong? Please put me on track, thanks!