Question

If a muon is traveling at 70% of the speed of light, how long does it take to decay in the observer's rest frame (i.e., what is the observed lifetime τμ of the muon)? express your answer in microseconds to two significant figures.

Answer 1

Answer:

Observed lifetime $= 3.0\mu s$

Explanation:

The lifetime of muon in the muon’s reference frame.

$t^{_{0}}=2.2\mu s$

The lifetime of muon in observer’s rest frame.

$t=\gamma t_{0}$

Here the  

$\gamma =\dfrac{1}{\sqrt{1-\frac{V^2}{C^2}}}$

$\gamma$ is the relativistic factor.  

V = Speed of muon

C = Speed of light

The muon’s speed of 70% of light speed.  

Hence,

$V=\dfrac{70}{100}C$

V = 0.7C

$=\dfrac{1}{\sqrt{1-\frac{(0.7C)^2}{C^2}}}$

$=\dfrac{1}{\sqrt{1-0.49}}$

$=\dfrac{1}{\sqrt{0.51}}$

$=\dfrac{1}{0.71}$

= 1.4

The lifetime of muon in observer’s rest frame.

$t=\gamma t_{0}$

$t=1.4\times 2.2\mu s$

$t=3.0\mu s$

Further explanation:

The muon is a lepton which decays to form an electron or positron. The lifetime of the muon is 2.20 microseconds. The muon lifetime in the muon reference frame is .  

But the observed lifetime $= \gamma =\frac{1}{\sqrt{1-\frac{V^2}{C^2}}}$

Learn more:

1. Speed of muon brainly.com/question/10048817 (answer by skyluke89)

2. Muon brainly.com/question/13198853 answer by skyluke89 )

Keywords: Muon, Speed of light, life time.

Answer 2

The muon lifetime in the muon reference frame is
$\tau_0 = 2.2 \mu s$

In the observer's rest frame, the muon lifetime is instead given by
$\tau = \gamma \tau_0$
where
$\gamma = \frac{1}{ \sqrt{1- \frac{v^2}{c^2} } }$ is the relativistic factor, with v being the muon speed and c the speed of light.

Since the muon is traveling at 70 % of the speed of light, 
$v=0.70 c$
and the relativistic factor is
$\gamma = \frac{1}{ \sqrt{1- \frac{(0.70 c)^2}{c^2} } }=1.4$

Therefore, the muon lifetime in the observer's rest frame is
$\tau = \gamma \tau_0 = (1.4)(2.2 \mu s)=3.1 \mu s$