Question

The mass of an electron is $\mathbf{9}\mathbf{.}\mathbf{10938188}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{31}}\mathbf{}\mathbf{kg}$. To six significant figures, find (a) data-custom-editor="chemistry" $\mathbf{\gamma }$and (b) data-custom-editor="chemistry" $\mathbf{\beta }$ for an electron with kinetic energy K = 100.000 MeV.

Short answer

(a) The Lorentz factor for electron is 196.159.

(b) The speed factor for electron is 0.999987.

Step-by-step solution

Identification of given data

The mass of electron is $\mathrm{m}=9.10938188\times {10}^{-31}\mathrm{kg}$

The kinetic energy of electron is data-custom-editor="chemistry" $\mathrm{K}=100.000\mathrm{MeV}$

The Lorentz factor and speed factor is the comparison of speed of particle with the speed of light. These factors are used to modify the formulas of classical mechanics.

Determination of Lorentz factor for electron

(a)

The Lorentz factor is given as:

$\mathrm{K}={\mathrm{mc}}^2\left(\mathrm{\gamma }-1\right)$

Here, c is the speed of light and its value is $3\times {10}^8\mathrm{m}/\mathrm{s}$.

Substitute all the values in the above equation.

$\begin{array}{rcl}\left(100.000\mathrm{MeV}\right)\left(\frac{1.6\times {10}^{-13}\mathrm{J}}{1\mathrm{MeV}}\right)& =& \left(9.10938188\times {10}^{-31}\mathrm{kg}\right)\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)\left(\mathrm{\gamma }-1\right)\\ \mathrm{\gamma }-1& =& 195.159\\ \mathrm{\gamma }& =& 196.159\end{array}$

Therefore, the Lorentz factor for electron is 196.159.

Determination of speed factor of electron

(b)

The speed factor for electron is given as:

$\mathrm{\beta }=\sqrt{1-\frac{1}{{\mathrm{\gamma }}^2}}$

Substitute all the values in the above equation.

$$\begin{gathered} \mathrm{\beta }=\sqrt{1-\frac{1}{{\left(196.159\right)}^2}} \\ \mathrm{\beta }=0.999987 \end{gathered}$$

Therefore, the speed factor for electron is 0.999987.