A certain muon detector counts 600 muouns/h at an altitude of 1900m and 380/h at sea level. Given that the muon half-life at rest is 1.5×(10^-6)s, determine the speed of the muons relative to Earth, assuming they all have the same speed.
To determine the speed of the muons relative to Earth, we can use the fact that the number of muons detected is related to the speed and the decay rate of the muons.
First, let's determine the muon decay rate at the given altitudes. We know that the half-life of muons at rest is 1.5×10^-6 s.
At sea level, the decay rate can be calculated by dividing the number of detected muons (380/h) by the half-life:
Decay rate at sea level = 380/h / (1.5×10^-6 s) = 253.33×10^6 s^-1 (approximately)
Similarly, at an altitude of 1900m, the decay rate can be calculated by dividing the number of detected muons (600/h) by the half-life:
Decay rate at 1900m = 600/h / (1.5×10^-6 s) = 400×10^6 s^-1 (approximately)
Now, let's use the concept of time dilation to find the relative speed of the muons. According to time dilation, the time experienced by a moving object is dilated compared to the time experienced in a stationary frame of reference.
The ratio of the decay rates can be equated to the time dilation factor:
Decay rate at sea level / Decay rate at 1900m = Time dilation factor
Let's calculate the time dilation factor:
Time dilation factor = (253.33×10^6 s^-1) / (400×10^6 s^-1) ≈ 0.6333
Now, we need to calculate the relative speed using the time dilation factor. The time dilation factor is related to the Lorentz factor (γ) and the relative speed (v) by the equation:
γ = 1 / √(1 - v^2 / c^2)
Here, c is the speed of light, which we will assume to be 3×10^8 m/s.
Rearranging the equation, we can solve for the relative speed (v):
v = c * √(1 - 1 / γ^2)
Substituting the value of γ from the time dilation factor, we can calculate the relative speed:
v = (3×10^8 m/s) * √(1 - 1 / (0.6333)^2) ≈ 2.33×10^8 m/s
Therefore, the speed of the muons relative to Earth, assuming they all have the same speed, is approximately 2.33×10^8 m/s.
380/600=e^-(t/1.5E-6)
take ln of each side
t = 1.5E-6*.456-6.84E-7 sec
speed=distance/time=1900/time above.