Question

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)?

Short answer

Answer

(a) The speed factor for muon is 0.948.

(b) The kinetic energy of muon is $227\text{MeV}$.

(c) The momentum of muon is $315\text{MeV}/\text{c}$.

Step-by-step solution

Identification of given data

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.

Determination of speed factor of muon

(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.

Determination of kinetic energy of muon

(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.

Determination of momentum of muon

(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 $315\text{MeV}/\text{c}$.