91Ó°ÊÓ

The result of the postulate of special relativity is that the clock runs slower for a moving object when measured from a rest frame. The factor by which the clocks run differently is called the Lorentz factor.

The relativistic kinetic energy relation is given by

$K=m_e{c}^2\left(\mathrm{γ}-1\right)$

Here $m_e$ is the electron’s rest mass, and $\mathrm{γ}$ is the Lorentz factor.

Using this relation to determine the Lorentz factor,

$$\begin{gathered} \mathrm{γ}=\frac{K}{m_e{c}^2}+1 \\ =\frac{2.00×{10}^6×1.6×{10}^{-19}J}{\left(9.1×{10}^{-31}kg\right){\left(3×{10}^{8}m/s\right)}^2}+1 \\ =3.91+1 \\ =4.91 \end{gathered}$$

$$\begin{gathered} \mathrm{γ}=\frac{1}{\sqrt{1-{\mathrm{β}}^2}} \\ \mathrm{β}=\sqrt{1-\frac{1}{{\mathrm{γ}}^2}} \\ =\sqrt{1-\frac{1}{{\left(4.91\right)}^2}} \\ =0.979 \end{gathered}$$

Hence, the electron is traveling at a speed,

$$\begin{gathered} v=\mathrm{β}c \\ =0.979c \end{gathered}$$

The expression for relativistic momentum is given by

$$\begin{gathered} p=\mathrm{γ}{m}_{e}v \\ =\mathrm{γ}\mathrm{β}{m}_{e}c \\ =\frac{\mathrm{γ}\mathrm{β}{m}_e{c}^2}{c} \end{gathered}$$

The rest mass energy $\left(m_e{c}^2\right)$ of the electron is $0.511MeV$ (Table 37-3).

$$\begin{gathered} p=\frac{\mathrm{γ}\mathrm{β}}{c}\left(m_e{c}^2\right) \\ =\frac{\left(4.91\right)\left(0.979\right)}{c}\left(0.511MeV\right) \\ =2.46MeV/c \end{gathered}$$