Question

How much work is needed to accelerate a proton from a speed of 0.9850c to a speed of 0.9860c?

Short answer

Work done in accelerating the electron $0.9850c$from to $0.9860c$requires of role="math" localid="1663144858958" $\mathbf{3}\mathbf{.}\mathbf{04}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{11}}\mathbf{J}$energy.

Step-by-step solution

Use the work-energy principle to derive an expression for work relativistically

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

Where, $\gamma$is the Lorentz factor and role="math" localid="1663145054895" $m_o$is the rest mass of a proton.

Therefore, work will be done,

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

Using the above equation, determine work done in accelerating an electron

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

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

We’ll get,

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

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