What is the speed of an electron whose total energy is equal to the total energy of a proton that is at rest? What is the kinetic energy of this electron?

The given data can be listed below as,

- The mass of electron is, $m_6=9.1\times {10}^{-31}\mathrm{kg}$

- The mass of proton is, $m_{\rho }=1.673\times {10}^{-27}\mathrm{kg}$

- The speed of light is, $c=3\times {10}^8\mathrm{m}/\mathrm{s}$

Kinetic energy is a form of energy that an object or particle retains due to its motion.

The relation of total energy is expressed as,

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

(i)

Here $E$is the total energy of an electron, $m_e$is the mass of electron and $c$is the speed of light..

Here $\gamma$is expressed as,

$\gamma =\frac{1}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}$

Here $v$is the speed of electron.

Substitute the value of $\gamma$in the equation (i).

$E=\frac{m_e{c}^2}{\sqrt{1-{\left(\begin{array}{l}v\\ c\end{array}\right)}^2}}$

The relation of rest energy is expressed as,

$E_R=m_p{c}^2$

(ii)

Here $E_R$is the rest energy, $m_p$is the mass of proton.

The total energy of electron is equal to the rest energy of proton, then it is expressed as,

$\frac{m_e{c}^2}{\sqrt{1-{\left(\begin{array}{l}v\\ c\end{array}\right)}^2}}=m_p{c}^2$

$\frac{m_e}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}=m_p$

$\frac{m_e^2}{1-{\left(\frac{v}{c}\right)}^2}=m_p^2$

$v=\sqrt{\left(1-\frac{m_0^2}{m_p^2}\right)c^2}$

Substitute $9.1\times {10}^{-31}\mathrm{kg}$for $m_e,1.673\times {10}^{27}\mathrm{kg}$for $m_p$, and $3\times {10}^8\mathrm{m}/\mathrm{s}$for $c$in the above equation.

$v=\sqrt{\left(1-\frac{{\left(9.1\times {10}^{-31}\mathrm{kg}\right)}^2}{{\left(1.673\times {10}^{-27}\mathrm{kg}\right)}^2}\right){\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)}^2}$

$=3\times {10}^8\mathrm{m}/\mathrm{s}$

Hence the speed of electron is, $3\times {10}^8\mathrm{m}/\mathrm{s}$.

The kinetic energy of electron is expressed as,

$K=E-E,$

...(iii)

Here $K$is the kinetic energy of an electron, $E$is the total energy and $E_p$is the rest energy of an electron.

Substitute the value of $E$, and $E_p$, in equation (iii)

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

The terms $\frac{m_0{c}^2}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}=m_{\rho }{c}^2$, then the above equation is express as,

$K=m_p{c}^2-m_e{c}^2$

$=\left(m_p-m_e\right)c^2$

Hence the kinetic energy is,$1.50\times {10}^{-15}\mathrm{J}$