The Muon is an elementary radioactive particle created when cosmic rays crash into the upper atmosphere of earth. It had half-life 1.5μs = 1.5E^-6s in its rest frame. Muons decay randomly into positron and anti-neutrino
Only muons with speed of 0.995c are stopped in the detector and allowed to sit around until they decay.
Here we are at rest along side the mountain and we observe the muons speeding towards us at speed of 0.995c.
1. Calculate the gamma factor for muons, y=1/√1-v^2/c^2
2.Calculate the time (in μ s) for half of the 568 muons to decay
3. Calculate the distance a typical muon travels in the time of 15 μs
Here the muon is at rest and sees the earth coming up towards it at speed of 0.995c.
1. Use the gamma factor, y, and calculate the height of the mountain as it seen to pass by.
2 Compare answer (1) with the height we normally ascribe to mount washington
3.How far does the earth move towards the muons in a time of 1.5μs. Comment on how many of the muons should be hit by the earth.
4. Given the experimental fact that 412 muons reach the foot of the mountain, calculate the fraction of one-half life that elapsed, and so calculate the elapsed time in μs from the muon's reference frame. For us (at rest on the earth) the time is clearly 6.4μs.
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1. The gamma factor for muons is given by the equation: `gamma=1/(sqrt(1-v^2/c^2))` Here, speed of the muons, v=0.995c. Substituting it in the above equation we get:`gamma=1/sqrt( 1-(0.995)^2)=10.01252~~10`
2. The equation for the exponential decay process is `N(t)=N_0/2^gamma` where `gamma=t/t_g` , N_0 is the initial no. of particles, N_t is the count after time t and t_g is the half life. To calculate the time for half of the 568 muons to decay we will use the equation `gamma=(t/t_g)=10` (from answer 1.)and `t_g=1.5 mus` . Plugging in these values we get: t=10*1.5=15 `mus` .
3. Speed of light(c) = 299792458 m/s. So, Speed of the muon=0.995c = 0.995*299792458 m/s = 298293496 m/s. Hence, the distance a typical muon travels in the time of 15 mus or 1.5*10^-5 s =speed*time=298293496*1.5*10^-5 =4474.4024 m. =4.4744 km.
1. From the reference frame of the muons, which are not moving at all, length contraction of the mountain occurs: `d_v = 1/gamma d_0 =1/10 d_0 =0.1* d_0=0.1*1920 m=192` m.
2. The answer, as obtained here, is clearly one-tenth of the height we normally ascribe to Mount Washington.
3. Now it is Mount Washington alongwith the Earth, that moves, going upward at a speed of 0.995c relative to the muons. The distance that the earth moves towards the muons in a time of `1.5mus` is actually 0.995*299792458*1.5*10^(-6) m =447.44 m. The time for the muons to go 192 m(relative height of the mountain) is 192/(0.995c)=192/298293496 s=0.64 `mus` . Number of muons hitting earth can be obtained as: `N_t=N_0/(2^(0.64/1.5)) =422` .
4. Given the experimental fact that 412 muons reach the foot of the mountain, the fraction of one-half life that elapsed is obtained as:
`rArr 2^f*412=568 `
`rArr f=(ln568-ln412)/ln2 `
So, about 0.463 fraction of one half-life has elapsed. Elapsed time in `mus` from the muon's reference frame is 0.463*1.5=0.69.
For us (at rest on the earth) the time is clearly `6.4mus` . This is an example of time dilation.
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