Question

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?

Short answer

The speed of electron is, $3\times {10}^8\mathrm{m}/\mathrm{s}$And the kinetic energy is,

$1.50\times {10}^{-10}\mathrm{J}$

Step-by-step solution

Identification of the given data 

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

Significance of kinetic energy

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

(a) Determination of the work required to change its orbit

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,$1.50\times {10}^{-15}\mathrm{J}$