Question

What is the momentum in MeV/c of an electron with a kinetic energy of $\mathbf{2}.\mathbf{00}\mathbf{MeV}$ ?

Short answer

The momentum of the electron is $2.46MeV/c$.

Step-by-step solution

Lorentz factor:

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

$K=m_e{c}^2\left(\gamma -1\right)$

Here $m_e$ is the electron’s rest mass, and $\gamma$ is the Lorentz factor.

Using this relation to determine the Lorentz factor,

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

The Lorentz factor is expressed in terms of speed parameter $\left(\beta =v/c\right)$ as,

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

Hence, the electron is traveling at a speed,

$$\begin{gathered} v=\beta c \\ =0.979c \end{gathered}$$

Relativistic Momentum:

The expression for relativistic momentum is given by

$$\begin{gathered} p=\gamma m_{e}v \\ =\gamma \beta m_{e}c \\ =\frac{\gamma \beta m_e{c}^2}{c} \end{gathered}$$

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

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