Question

How much work must be done to increase the speed of an electron (a) from 0.18c to 0.19c and (b) from 0.98c to 0.99c? Note that the speed increase is 0.01c in both cases.

Short answer

(a) The work for increasing the speed of electron is 1 keV.

(b) The work for increasing the speed of electron is 1.1 MeV.

Step-by-step solution

Identification of given data

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.

Determination of work to increase the speed of electron

(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 $3\times {10}^8\mathrm{m}/\mathrm{s}$, m is the mass of an electron and its value is $9.1\times {10}^{-31}\mathrm{kg}$.

Substitute all the values in the above equation.

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

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

Determination of work to increase the speed of electron

(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.

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

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