How much work must be done to accelerate an electron

(a) from $\mathbf{0}\mathbf{.}\mathbf{3}\mathit{c}$to $\mathbf{0}\mathbf{.}\mathbf{6}\mathit{c}\mathbf{}$and

(b) from $\mathbf{0}\mathbf{.}\mathbf{6}\mathit{c}$to $\mathbf{0}\mathbf{.}\mathbf{9}\mathit{c}$?

Kinetic energy is associated with the motion of the object It could be determined by subtracting the object’s energy (Internal) at rest from the energy of the moving object.

The Work-Energy principle calculates the work done is,

$\mathit{W}\mathbf{=}\mathit{K}{\mathit{E}}_{\mathbf{f}}\mathbf{-}\mathit{K}{\mathit{E}}_{\mathbf{i}}$

Here, KE_f is the final kinetic energy and KE_iis the initial kinetic energy.

As the electron is traveling at a very high speed we will be considering relativistic Kinetic energy,

$$\begin{gathered} KE=Totalenergy-m_o{c}^2 \\ KE=m{c}^2-m_o{c}^2 \\ KE=\gamma m_o{c}^2-m_o{c}^2=(\gamma -1)m_0{c}^2 \end{gathered}$$

Where, localid="1659082607023" $\gamma$ is lorentz factor, c is the speed of light, and m_o is the rest mass of electron.

Therefore, work will be done,

localid="1659082610813" $\begin{array}{rcl}W& =& {\left[\left(\gamma -1\right)m_e{c}^2\right]}_f-{\left[\left(\gamma -1\right)m_e{c}^2\right]}_i\\ & =& \left({\gamma }_f-{\gamma }_i\right)m_e{c}^2.......\left(1\right)\end{array}$

Here, m_e is the mass of electron which is equal to the rest mass of the electron.

For part (a) the values of ${\gamma }_f$and ${\gamma }_i$are calculated as follows:

localid="1659083018725" $\begin{array}{rcl}{\gamma }_i& =& \frac{1}{\sqrt{1-\frac{{\left(0.3c\right)}^2}{c^2}}}\\ & =& 1.05\\ {\gamma }_f& =& \frac{1}{\sqrt{1-\frac{{\left(0.6c\right)}^2}{c^2}}}\\ & =& 1.25\\ & & \end{array}$

By Substituting the known values into equation (1) we get,

$$\begin{gathered} W=(1.25-1.05)\times 9.1\times {10}^{-31}kg\times {\left(3\times {10}^8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\right)}^2 \\ =1.64\times {10}^{-14}J \\ =86.3keV \end{gathered}$$

Therefore, for part (a) work done in accelerating the electron from$0.3c$to $0.6c$requires $1.64\times {10}^{16}J$of energy.

Similarly, for part (b), values of ${\gamma }_{f}and{\gamma }_i$are

localid="1659083022983" $\begin{array}{rcl}{\gamma }_i& =& \frac{1}{\sqrt{1-\frac{{\left(0.6c\right)}^2}{c^2}}}\\ & =& 1.25\\ {\gamma }_f& =& \frac{1}{\sqrt{1-\frac{{\left(0.9c\right)}^2}{c^2}}}\\ & =& 2.29\\ & & \end{array}$

By Substituting the known values into equation (1) we get,

$\begin{array}{rcl}W& =& \left(2.29-1.25\right)\times 9.1\times {10}^{-31}kg\times {\left(3\times {10}^8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\right)}^2\\ & =& 8.52\times {10}^{-14}J\\ & =& 532.5keV\\ & & \\ & & \end{array}$

Hence, the required energy to accelerate electron from 0.6c to 0.9c is localid="1659083088271" $8.52\times {10}^{-14}J$.