Chapter 32, Problem 59GP

To determine

To find: Length of track of tau lepton measured in lab.

Answer to Problem 59GP

  $6.59\times {10}^{-5}\text{m}$

Explanation of Solution

Given data:

Kinetic energy of the tau lepton, $KE=450\text{MeV}$

The rest energy of the tau lepton, $m_r{c}^2=1777\text{MeV}$

The proper time of the tau lepton decay, $t_d=2.91\times {10}^{-13}\text{sec}$

Formula used:

Time dilation formula

  $\Delta t_{\text{lab }}=\frac{\Delta t_0}{\sqrt{1-v^2/c^2}}$

Relativistic kinetic energy

  $\text{KE}=(\gamma -1)m{c}^2$

Calculation:

Relativistic kinetic energy of tau lepton

  $\begin{array}{l}\text{KE}=(\gamma -1)m{c}^2\\ \text{KE}=\left(\frac{1}{\sqrt{1-v^2/c^2}}-1\right)m{c}^2\end{array}$

Rearranging the above equation

  $\begin{array}{l}v=c\sqrt{1-\frac{1}{{\left(\frac{\text{KE}}{m{c}^2}+1\right)}^2}}\\ =c\sqrt{1-\frac{1}{{\left(\frac{450\text{MeV}}{1777\text{MeV}}+1\right)}^2}}\\ =0.603c\end{array}$

Time dilation of decay

  $\begin{array}{c}\Delta t_{\text{lab }}=\frac{\Delta t_0}{\sqrt{1-v^2/c^2}}\\ =\frac{2.91\times {10}^{-13}\text{s}}{\sqrt{1-{(0.603)}^2}}\\ =3.647\times {10}^{-13}\text{s}\end{array}$

Length of track measured in the lab

  $\begin{array}{c}\Delta x_{\text{lab}}=v\Delta t_{\text{lab}}\\ =(0.603)\left(3.00\times {10}^8\text{m}/\text{s}\right)\left(3.647\times {10}^{-13}\text{s}\right)\\ =6.59\times {10}^{-5}\text{m}\end{array}$

Conclusion:

Length of track measured in the lab is $6.59\times {10}^{-5}\text{m}$