91Ó°ÊÓ

The initial speed of electron is ${\mathrm{u}}_1=0.18\mathrm{c}$

The final speed of electron is ${\mathrm{u}}_2=0.19\mathrm{c}$

The initial speed of electron is $v_1=0.98\mathrm{c}$

The final speed of electron is ${\mathrm{v}}_2=0.99\mathrm{c}$

The work necessary to accelerate any particle from rest is equal to the change in kinetic energy of particle from rest to moving position.

(a)

The work for increasing the speed of rest electron is given as:

${\mathrm{W}}_{\mathrm{a}}={\mathrm{mc}}^2\left(\frac{1}{\sqrt{1-{\left({\displaystyle \frac{{\mathrm{u}}_2}{\mathrm{c}}}\right)}^2}}-\frac{1}{\sqrt{1-{\left({\displaystyle \frac{{\mathrm{u}}_1}{\mathrm{c}}}\right)}^2}}\right)$

Here, c is the speed of light and its value is $3×{10}^8\mathrm{m}/\mathrm{s}$, m is the mass of an electron and its value is $9.1×{10}^{-31}\mathrm{kg}$.

Substitute all the values in the above equation.

$$\begin{gathered} {\mathrm{W}}_{\mathrm{a}}=\left(9.1×{10}^{-31}\mathrm{kg}\right){\left(3×{10}^8\mathrm{m}/\mathrm{s}\right)}^2\left(\frac{1}{\sqrt{1-{\left({\displaystyle \frac{0.19\mathrm{c}}{\mathrm{c}}}\right)}^2}}-\frac{1}{\sqrt{1-{\left({\displaystyle \frac{0.18\mathrm{c}}{\mathrm{c}}}\right)}^2}}\right) \\ {\mathrm{W}}_{\mathrm{a}}=1.612×{10}^{-16}\mathrm{J} \\ {\mathrm{W}}_{\mathrm{a}}=\left(1.612×{10}^{-16}\mathrm{J}\right)\left(\frac{1\mathrm{keV}}{1.6×{10}^{-16}\mathrm{J}}\right) \\ {\mathrm{W}}_{\mathrm{a}}=1\mathrm{keV} \end{gathered}$$

Therefore, the work for increasing the speed of electron is 1 keV.

(b)

The work for increasing the speed of rest electron is given as:

${\mathrm{W}}_{\mathrm{b}}={\mathrm{mc}}^2\left(\frac{1}{\sqrt{1-{\left({\displaystyle \frac{{\mathrm{v}}_2}{\mathrm{c}}}\right)}^2}}-\frac{1}{\sqrt{1-{\left({\displaystyle \frac{{\mathrm{v}}_1}{\mathrm{c}}}\right)}^2}}\right)$

Substitute all the values in the above equation.

$$\begin{gathered} {\mathrm{W}}_{\mathrm{b}}=\left(9.1×{10}^{-31}\mathrm{kg}\right){\left(3×{10}^8\mathrm{m}/\mathrm{s}\right)}^2\left(\frac{1}{\sqrt{1-{\left({\displaystyle \frac{0.99\mathrm{c}}{\mathrm{c}}}\right)}^2}}-\frac{1}{\sqrt{1-{\left({\displaystyle \frac{0.98\mathrm{c}}{\mathrm{c}}}\right)}^2}}\right) \\ {\mathrm{W}}_{\mathrm{b}}=1.687×{10}^{-13}\mathrm{J} \\ {\mathrm{W}}_{\mathrm{b}}=\left(1.687×{10}^{-13}\mathrm{J}\right)\left(\frac{1\mathrm{keV}}{1.6×{10}^{-13}\mathrm{J}}\right) \\ {\mathrm{W}}_{\mathrm{b}}=1.1\mathrm{MeV} \end{gathered}$$

Therefore, the work for increasing the speed of electron is 1.1 MeV.