Solution to 2008 Problem 39

If there are N_0 muons at time t = 0, then at some later time t, there would be
\begin{align*}N(t) = N_0 e^{-t/\tau}\end{align*}
So, at time t, the muons decay at a rate
\begin{align*}-\frac{d N(t)}{dt} = \frac{N_0}{\tau}e^{-t/\tau}\end{align*}
So, in a time interval (t,t+dt),
\begin{align*}\frac{N_0}{\tau}e^{-t/\tau} dt\end{align*}
muons decay. So, the probability that a muon decays in this time interval is
\begin{align*}\frac{e^{-t/\tau}}{\tau} dt\end{align*}
A muon that travels a distance l in the laboratory will decay at time t = l/\gamma v in its rest frame.
\begin{align*}\gamma = \frac{1}{\sqrt{\displaystyle 1 - \frac{\left(\frac{\displaystyle 4c}{\displaystyle 5}\right)^2}{\displ...
So, t = l/\gamma v =  5l/4 \gamma c = 3 l /4c. So, the probability that a muon decays after traveling a distance l is
\begin{align*}\frac{e^{-3l/4c\tau}}{\tau}dt = \frac{e^{-3l/4c\tau}}{\tau}\frac{3dl}{4c}\end{align*}
The mean distance traveled by the muons is therefore,
\begin{align*}\int_0^{\infty}l\frac{ e^{ -3l/4c\tau}}{\tau}\frac{3dl}{4c}\\&= \left[-l e^{ -3l/4c\tau}\right]_0^{\infty} ...
Plugging in \tau = 2.2 \cdot 10^{-6} \mbox{ s} and c = 3\cdot 10^8 \mbox{ m/s} gives
\begin{align*}\frac{4 c \tau}{3} = \frac{4 \cdot 3 \cdot 10^8 \mbox{ m/s} \cdot (11/5) \cdot 10^{-6} \mbox{ s}}{3} = \boxed{8...
Therefore, answer (C) is correct.

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