Question:The mass of a muon is 207 times the electron mass; the average lifetime of muons at rest is $\mathbf{2}\mathbf{.}\mathbf{20}\mathit{\mu }\mathit{s}$. In a certain experiment, muons moving through a laboratory are measured to have an average lifetime of $\mathbf{6}\mathbf{.}\mathbf{90}\mathit{\mu }\mathit{s}$. For the moving muons, what are (a) $\mathbf{\beta }$ , (b) K, and (c) p (in MeV/c)?

The average life time of muons at rest is $t_0=2.20\mu s$

The average life time of moving muons is $t=6,90\mu s$

The mass of the muon is $m=207m_e$

The total energy of a particle has two components. One is the rest energy of particle and other is the kinetic energy of the particle due to the motion of particle.

(a)

The speed factor of muon is given as:

$\beta =\sqrt{1-{\left(\frac{t_0}{t}\right)}^2}$

Substitute all the values in equation.

$$\begin{gathered} \beta =\sqrt{1-{\left(\frac{2.20\mu \text{s}}{6.90\mu \text{s}}\right)}^2} \\ \beta =0.948 \end{gathered}$$

Therefore, the speed factor for muon is 0.948.

(b)

The kinetic energy of muon is given as:

$$\begin{gathered} K=m{c}^2\left(\frac{1}{\sqrt{1-{\beta }^2}}-1\right) \\ K=\left(207m_e\right)c^2\left(\frac{1}{\sqrt{1-{\beta }^2}}-1\right) \\ K=207m_e{c}^2\left(\frac{1}{\sqrt{1-{\beta }^2}}-1\right)\backslash \end{gathered}$$

Here, m_e is the mass of electron and its value is $9.109\times {10}^{-31}\text{kg}$, c is the speed of light and its value is $3\times {10}^{8}m/s$

Substitute these values in equation (3).

$$\begin{gathered} K=207\left(9.109\times {10}^{-31}\text{kg}\right){\left(3\times {10}^8\text{m}/\text{s}\right)}^2\left(\frac{1}{\sqrt{1-{\left(0.948\right)}^2}}-1\right) \\ K=3.635\times {10}^{-11}\text{J} \\ K=\left(3.635\times {10}^{-11}\text{J}\right)\left(\frac{1\text{MeV}}{1.6\times {10}^{-13}\text{J}}\right) \\ K=227\text{MeV} \end{gathered}$$

Therefore, the kinetic energy of muon is 227MeV.

(c)

The momentum of muon is given as:

$$\begin{gathered} p=\left(\frac{\beta }{\sqrt{1-{\beta }^2}}\right)mc \\ p=\left(\frac{\beta }{\sqrt{1-{\beta }^2}}\right)\left(207m_e\right)c \\ p=207\left(\frac{\beta }{\sqrt{1-{\beta }^2}}\right)m_{e}c \end{gathered}$$

Here, m_e is the mass of electron and its energy equivalent is $0.511\text{MeV}/{\text{c}}^2$.

Substitute all the values in above equation.

$$\begin{gathered} p=207\left(\frac{0.948}{\sqrt{1-{\left(0.948\right)}^2}}\right)\left(0.511\text{MeV}/{\text{c}}^2\right)c \\ p=315\text{MeV}/\text{c} \end{gathered}$$

Therefore, the momentum of muon is $315\text{MeV}/\text{c}$.