Muons at speed 0.9994c are sent round and round a circular storage ring of radius 1500 m.If a muon at rest decays into other particles after an average T = 2.2 × 10−⁶ s,how many trips around the storage ring do we expect the 0.9994c muons to make before they decay? A) 2 B) 4 C) 0.2 D) 6

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Quizplus · Accepted Answer

The time dilation formula is:$$t' = \frac{t}{\gamma}$$where $t'$ is the time measured in the moving frame, $t$ is the time measured in the stationary frame, and $\gamma$ is the Lorentz factor:$$\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$$In this problem, $t = 2.2 	imes 10^{-6} s$, $v = 0.9994c$, and $r = 1500 m$. The circumference of the storage ring is $2\pi r = 2\pi (1500 m) = 9424.8 m$. The speed of the muons is:$$v = \frac{2\pi r}{t} = \frac{2\pi (1500 m)}{2.2 	imes 10^{-6} s} = 0.9994c$$The Lorentz factor is:$$\gamma = \frac{1}{\sqrt{1 - v^2/c^2}} = \frac{1}{\sqrt{1 - (0.9994c)^2/c^2}} = 22.36$$The time dilation factor is:$$t' = \frac{t}{\gamma} = \frac{2.2 	imes 10^{-6} s}{22.36} = 9.84 	imes 10^{-8} s$$The number of trips around the storage ring that the muons make before they decay is:$$n = \frac{t'}{t} = \frac{9.84 	imes 10^{-8} s}{2.2 	imes 10^{-6} s} = 0.0447$$Therefore, the correct answer is A.

Muons at Speed 0