Question: The mass of an electron is $\mathbf{9}\mathbf{.}\mathbf{10938188}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{31}}\mathbf{}\mathbf{\text{kg}}$. To eight significant figures, find the following for the given electron kinetic energy: (a)localid="1663051516359" $\mathit{\gamma }$and (b)$\mathit{\beta }$localid="1663053404383" $\mathit{\beta }$for$\mathit{K}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0000000}\mathbf{}\mathbf{\text{keV}}$, (c)localid="1663051781874">γand (d)localid="1663051803695" $\mathit{\beta }$for, $\mathit{K}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0000000}\mathbf{}\mathbf{\text{MeV}}$and then (e)localid="1663051835448" $\mathit{\gamma }$and (f)localid="1663051820843" $\mathit{\beta }$for$\mathit{K}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0000000}\mathbf{}\mathbf{\text{GeV}}$.

The mass of the electron,$m=9.10938188\times {10}^{-31}\text{kg}$

The expression to calculate the Lorentz factor is given as follows.

$\mathit{\gamma }\mathbf{=}\frac{\mathbf{K}}{\mathbf{m}{\mathbf{c}}^{\mathbf{2}}}\mathbf{+}\mathbf{1}$ …(i)

Here, is the kinetic energy of electron, is the speed of the light.

The expression to calculate the speed parameter is given as follows.

$\mathit{\beta }\mathbf{=}\sqrt{\mathbf{1}\mathbf{-}\frac{\mathbf{1}}{{\mathbf{\gamma }}^{\mathbf{2}}}}$ …(ii)

Conversion from kiloelectron volt to joule,

$\mathbf{1}\mathbf{}\mathbf{\text{eV}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathbf{}\mathbf{\text{J}}$

(a)

The kinetic energy of the electron,

$$\begin{gathered} K=1\text{keV} \\ K=1\times {10}^3\times 1.6\times {10}^{-19} \text{J} \\ K=1.6\times {10}^{-16}\text{J} \end{gathered}$$

Calculate the Lorentz factor.

Substitute $1.6\times {10}^{-16}\text{JforK,}9.10938188\times {10}^{-31}\text{kgformand}3\times {10}^{8}m/s\text{forc}$into equation (i).

$$\begin{gathered} \gamma =\frac{1.6\times {10}^{-16}\text{J}}{9.10938188\times {10}^{-31}\text{kg}\times {\left(3\times {10}^{8}m/s\right)}^2}+1 \\ \gamma =0.19516\times {10}^{-2}+1 \\ \gamma =1.0019516 \end{gathered}$$

Hence the value of the Lorentz factor is $1.0019516$.

(b)

Calculate the value of the speed parameter.

Substitute for into equation (ii).

$$\begin{gathered} \beta =\sqrt{1-\frac{1}{{1.0019516}^2}} \\ \beta =\sqrt{1-0.99610} \\ \beta =0.06238432 \end{gathered}$$

Hence the value of speed parameter is $0.06238432$.

(c)

The kinetic energy of the electron,

$$\begin{gathered} K=1\text{MeV} \\ K=1\times {10}^6\times 1.6\times {10}^{-19}\text{J} \\ K=1.6\times {10}^{-13}\text{J} \end{gathered}$$

Calculate the Lorentz factor.

Substitute$1.6\times {10}^{-13}\text{JforK,}9.10938188\times {10}^{-31}\text{kgformand}3\times {10}^8\text{forc}$ into equation (i).

$$\begin{gathered} \frac{1.6\times {10}^{-13}\text{J}}{9.10938188\times {10}^{-31}\text{kg}\times {\left(3\times {10}^{8}m/s\right)}^2}+1 \\ \gamma =0.19516955\times 10+1 \\ \gamma =2.9516955 \end{gathered}$$

Hence the value of the Lorentz factor is $2.9516955$.

(d)

Calculate the value of the speed parameter.

Substitute $2.9516955$for $\gamma$into equation (ii).

$$\begin{gathered} \beta =\sqrt{1-\frac{1}{{2.9516955}^2}} \\ \beta =\sqrt{1-0.11477753} \\ \beta =0.94086262 \end{gathered}$$

Hence the value of speed parameter is 0.94086262.

(e)

The kinetic energy of the electron,

$$\begin{gathered} K=1\text{GeV} \\ K=1\times {10}^9\times 1.6\times {10}^{-19}\text{J} \\ K=1.6\times {10}^{-10}\text{J} \end{gathered}$$

Calculate the Lorentz factor.

Substitute$1.6\times {10}^{-10}\text{Jfork,}9.10938188\times {10}^{-31}\text{kgformand}3\times {10}^8\text{forc}$ into equation (i).

$$\begin{gathered} \gamma =\frac{1.6\times {10}^{-10}\text{J}}{9.10938188\times {10}^{-31}\text{kg}\times {\left(3\times {10}^{8}m/s\right)}^2}+1 \\ \gamma =0.19516955\times {10}^4+1 \\ \gamma =1951.6955 \end{gathered}$$

Hence the value of the Lorentz factor is $1951.6955$.

(f)

Calculate the value of the speed parameter.

Substitute for into equation (ii).

$$\begin{gathered} \beta =\sqrt{1-\frac{1}{{1951.6955}^2}} \\ \beta =\sqrt{1-2.62\times {10}^{-7}} \\ \beta =0.99999987 \end{gathered}$$

Hence the value of speed parameter is $0.99999987$.