91Ó°ÊÓ

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.

Here, we will apply the Work-Energy principle to calculate the work done.

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

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

role="math" localid="1663145108939" $$\begin{gathered} KE=Totalenergy-m_o{c}^2 \\ KE=m{c}^2-m_o{c}^2 \\ KE=\mathrm{γ}{m}_o{c}^2-m_o{c}^2=\left(\mathrm{γ}-1\right)m_o{c}^2 \end{gathered}$$

Therefore, work will be done,

$$\begin{gathered} W={\left[\left(\mathrm{γ}-1\right)m_o{c}^2\right]}_f-{\left[\left(\mathrm{γ}-1\right)m_o{c}^2\right]}_i \\ =\left({\mathrm{γ}}_f-{\mathrm{γ}}_i\right)m_o{c}^2 \end{gathered}$$

For part (a) the values of${\mathrm{γ}}_f$and${\mathrm{γ}}_i$are calculated as

$$\begin{gathered} {\mathrm{γ}}_i=\frac{1}{\sqrt{1-\frac{{\left(0.9850c\right)}^2}{c^2}}} \\ =5.797 \\ {\mathrm{γ}}_f=\frac{1}{\sqrt{1-\frac{{\left(0.9860c\right)}^2}{c^2}}} \\ =5.999 \end{gathered}$$

We’ll get,

$W=\left(5.999-5.797\right)×1.67×{10}^{-27}kg×{(3×{10}^{8}m/s)}^2=3.04×{10}^{-11}J$

Therefore, work done in accelerating the electron from$0.9850c$ to$0.9860c$ requires$3.04×{10}^{-11}J$ of energy.