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Chapter 26: Problem 75

Refer to Example \(26.2 .\) One million muons are moving toward the ground at speed \(0.9950 c\) from an altitude of \(4500 \mathrm{m} .\) In the frame of reference of an observer on the ground, what are (a) the distance traveled by the muons; (b) the time of flight of the muons; (c) the time interval during which half of the muons decay; and (d) the number of muons that survive to reach sea level. [Hint: The answers to (a) to (c) are not the same as the corresponding quantities in the muons' reference frame. Is the answer to (d) the same?]

Short Answer

Expert verified
Answer: The time interval during which half of the muons decay in the reference frame of an observer on the ground can be found in step 4, where we calculate the time dilation of the decay time using the Lorentz factor and the decay time in the muons' reference frame. The time interval is represented as ∆t = γ × τ, where τ represents the known decay time in the muons' reference frame (2.2 × 10^(-6) s), and γ represents the Lorentz factor found in step 1.

Step by step solution

01

Calculate the Lorentz factor

Given the speed of the muons as \(0.9950c\), first we need to calculate the Lorentz factor, which quantifies the time dilation and length contraction. The Lorentz factor \(\gamma\) can be calculated by the following formula: \(\gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\) With \(v = 0.9950c\), we can find \(\gamma\).
02

Calculate the distance traveled in the reference frame of an observer on the ground

The distance traveled by the muons in the reference frame of an observer on the ground should be the same as the altitude they drop from, which is \(4500m\).
03

Calculate the time of flight in the reference frame of an observer on the ground

Using the equation \(d=v\cdot t\), we can find the time of flight in the reference frame of an observer on the ground: \(t=\frac{d}{v}\) With \(d=4500\mathrm{m}\) and \(v=0.9950c\), we can find the time of flight \(t\).
04

Calculate the proper time interval for half of the muons to decay

In the muons' reference frame, we know that their decay time is \(\tau=2.2\cdot10^{-6}\textrm{s}\). From the Lorentz factor, we can find the time dilation of the decay time in the reference frame from an observer on the ground: \(\Delta t=\gamma\cdot\tau\) Using the calculated \(\gamma\), we can find the time interval \(\Delta t\) during which half of the muons decay in the reference frame of an observer on the ground.
05

Calculate the number of muons that survive to reach sea level

With the information from step 4, we can now calculate the number of muons that survive to reach sea level. Since half of the muons decay in time interval \(\Delta t\), we can find the number of muons decayed during the time of flight \(t\) from step 3: \(N_\textrm{decayed}=\frac{10^6}{2}\left(1-e^{-\frac{t}{\Delta t}}\right)\) And the surviving muons can be found by subtracting the decayed muons from the initial quantity: \(N_\textrm{survived}= 10^6 - N_\textrm{decayed}\)

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Most popular questions from this chapter

Suppose your handheld calculator will show six places beyond the decimal point. How fast (in meters per second) would an object have to be traveling so that the decimal places in the value of \(\gamma\) can actually be seen on your calculator display? That is, how fast should an object travel so that \(\gamma=1.000001 ?\) [Hint: Use the binomial approximation.]
Radon decays as follows: $^{222} \mathrm{Rn} \rightarrow^{218} \mathrm{Po}+\alpha .$ The mass of the radon-222 nucleus is 221.97039 u, the mass of the polonium- 218 nucleus is \(217.96289 \mathrm{u},\) and the mass of the alpha particle is 4.00151 u. How much energy is released in the decay? \(\left(1 \mathrm{u}=931.494 \mathrm{MeV} / \mathrm{c}^{2} .\right)\)
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