Question

One of the most impressive experimental confirmations of Time Dilation involves muons. Muons are unstable elementary particles with a lifetime of 2.2 μμs. They are naturally produced in collisions between cosmic radiation and atoms in our upper atmosphere (~3000m above the Earth's surface). A muon travels at 0.98c. In a reference frame that is attached to the moving muon (the muon's reference frame), how far can the muon travel before it decays (recall that v = d/t)?

For an observer at rest on the Earth, what does he measure for the lifetime of the muon (in seconds)?

For an observer at rest on the Earth, how far does the muon travel before it decays?

please help with these 3 questions. It says all three of my answers were incorrect

Answer 1

(a) In a reference frame, how far can the muon travel before it decays?

we know that,   d = v $\tau$

where, v = muon traveling with a speed = 0.98 c

$\tau$ = mean lifetime of the muon = 2.2 x 10^-6 s

then, we get

d = [(0.98 x 3 x 10⁸ m/s) (2.2 x 10^-6 s)]

d = 646.8 m

(b) For an observer at rest on the Earth, what does he measure for the lifetime of the muon?

According to relativity time deliation, we have

$\tau$' = $\tau$ / _$\sqrt{}$1 - (v / c)²

$\tau$' = (2.2 x 10^-6 s) / _$\sqrt{}$1 - [(0.98 c) / (c)]²

$\tau$' = (2.2 x 10^-6 s) / _$\sqrt{}$1 - (0.9604)

$\tau$' = [(2.2 x 10^-6 s) / (0.198)]

$\tau$' = 1.11 x 10^-5 s

(c) For an observer at rest on the Earth, how far does the muon travel before it decays?

we know that,   d' = v $\tau$'

d' = [(0.98 x 3 x 10⁸ m/s) (1.11 x 10^-5 s)]

d' = 3263.4 m