What are (a) $\mathit{K}$, (b) $\mathit{E}$, and (c) $\mathit{p}$(in $\raisebox{1ex}{$\mathbf{G}\mathbf{e}\mathbf{V}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{c}$}\right.$) for a proton moving at speed $\mathbf{0}\mathbf{.}\mathbf{990}\mathit{c}$? What are (d) $\mathit{K}$, (e) $\mathit{E}$, and (f) $\mathit{p}$(in $\raisebox{1ex}{$\mathbf{M}\mathbf{e}\mathbf{V}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{c}$}\right.$) for an electron moving at speed $\mathbf{0}\mathbf{.}\mathbf{990}\mathit{c}$ ?

The result of the ${\mathbf{2}}^{\mathbf{n}\mathbf{d}}$ 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 Lorentz factor depends only on velocity and not on the particle’s mass. Therefore, the Lorentz factor for proton and electron will be the same as velocity.

$$\begin{gathered} \gamma =\frac{1}{\sqrt{1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}} \\ =\frac{1}{\sqrt{1-{\left({\displaystyle \raisebox{1ex}{$0.990c$}\!\left/ \!\raisebox{-1ex}{$c$}\right.}\right)}^2}} \\ =\frac{1}{\sqrt{1-{\left(0.990\right)}^2}} \end{gathered}$$

$$\begin{gathered} \gamma =\frac{1}{0.1411} \\ =7.1 \end{gathered}$$

The relativistic kinetic energy relation is given by,

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

For proton, rest mass $m_p$ is $1.67\times {10}^{-27}kg$ and its energy equivalent is

$$\begin{gathered} m_p{c}^2=\frac{1.67\times {10}^{-27}kg\times {\left(3\times {10}^{8}m/s\right)}^2}{1.6\times {10}^{-19}J} \\ =9.38\times {10}^{8}eV \\ =938MeV \end{gathered}$$.

Therefore substituting $938MeV$for $m_p{c}^2$ in the above equation, and you get

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

role="math" localid="1663065607531" $$\begin{gathered} =\left(938MeV\right)\left(7.1-1\right) \\ =5721.8MeV \\ =5.72GeV \end{gathered}$$

For Electron, rest mass $m_e$is $\text{9.1}\times {10}^{-31}kg$ and its energy equivalent is

$$\begin{gathered} m_e{c}^2=\frac{9.1\times {10}^{-31}kg\times {\left(3\times {10}^{8}m/s\right)}^2}{1.6\times {10}^{-19}J} \\ =51.1875\times {10}^{4}eV \\ \approx 0.511MeV \end{gathered}$$

Therefore substituting $\text{0.511}MeV$for $m_e{c}^2$ in the above equation, and you get

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

$$\begin{gathered} =\left(0.511MeV\right)\left(7.1-1\right) \\ =3.12MeV \\ =5\times {10}^{-13}J \end{gathered}$$

The relativistic total energy relation is given by

$E=\gamma m{c}^2$

For proton, rest mass is $1.67\times {10}^{-27}kg$and its energy equivalent is $938MeV$. Therefore substituting $938MeV$ for $m_p{c}^2$ in the above equation, you obtain,

$E=\gamma m_p{c}^2$

role="math" localid="1663066308149" $$\begin{gathered} =\left(7.1\right)\left(938MeV\right) \\ =6.66GeV \end{gathered}$$

For Electron, rest mass is $\text{9.1}\times {10}^{-31}kg$ and its energy equivalent is $\text{0.511}MeV$. Therefore substituting $\text{0.511}MeV$ for $m_e{c}^2$in the above equation, you have

$E=\gamma m_e{c}^2$

$$\begin{gathered} =\left(7.1\right)\left(0.511MeV\right) \\ =3.63MeV \end{gathered}$$

The relativistic momentum relation is given by

$p=\gamma mu$

$$\begin{gathered} =\gamma m\left(\beta c\right) \\ =\gamma \beta \left(\frac{m{c}^2}{c}\right) \end{gathered}$$

For proton, rest mass is $1.67\times {10}^{-27}kg$and its energy equivalent is $938MeV$. Therefore substituting $938MeV$ for $m_p{c}^2$in the above equation, you get

$p=\left(7.1\right)\left(0.990\right)\left(\frac{938MeV}{c}\right)$

$=6.6\frac{GeV}{c}$

For Electron, rest mass is $\text{9.1}\times {10}^{-31}kg$ and its energy equivalent is $\text{0.511}MeV$. Therefore substituting $\text{0.511}MeV$ for $m_e{c}^2$in the above equation, and you have

$$\begin{gathered} p=\left(7.1\right)\left(0.990\right)\left(\frac{0.511MeV}{c}\right) \\ =3.6\frac{MeV}{c} \end{gathered}$$