Question

n electron is travelling at a speed of $\mathbf{0}\mathbf{.}\mathbf{95}\mathbf{c}$in an electron accelerator. An electric force of $\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{13}}\mathbf{}\mathbf{N}$is applied in the direction of motion while the electron travels a distance of $\mathbf{2}\mathbf{}\mathbf{m}$. What is the new speed of the electron?

Short answer

Answer

The new speed of the electron is$0.989c$

Step-by-step solution

- Identification of given data

The applied electric force is$F=1.6\times {10}^{-13}\mathrm{N}$

The distance travelled by an electron is$d=2\mathrm{m}$

- Concept of the speed of an electron

Generally, the travel speed of electron is $2200$kilometers per second. this speed is lower than the travel speed of light. But this speed of electron is fast enough to revolve around the earth in more than $18$seconds.

- Calculation of the new speed of the electron

Here, the new speed of an electron is calculated by using mass-energy principle.

The principle of mass-energy,

$E=\gamma m{c}^2...........\left(1\right)$

where,

mass of an electron $\Rightarrow 9\times {10}^{-31}\mathrm{kg}$

the speed of light in air $\Rightarrow 3\times {10}^8\mathrm{m}/\mathrm{s}$

$\gamma =\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}.......\left(a\right)$

$=\frac{1}{\sqrt{1-(0.95{)}^2}}$

$y=3.20$

Total Energy $E_{\text{total}}=E+W$........... (2) Where,

$E$= Energy

$W$= Work done by applied force

Using Equation (1), we can calculate Energy,

$$\begin{gathered} E=3.20\times 9\times {10}^{-31}\times {\left(3\times {10}^8\right)}^2 \\ =2.59\times {10}^{-13}J \end{gathered}$$

Work done is calculated by the following formula,

$W=F\times d$

Where, $F=$Applied eletric force, role="math" localid="1657971421085" $d=$distance travelled by an electron

$$\begin{gathered} W=1.6\times {10}^{-13}\times 2 \\ =3.2\times {10}^{-13} \end{gathered}$$

From Equation (2),

$$\begin{gathered} E_{total}=2.59\times {10}^{-13}+3.2\times {10}^{-13} \\ =5.79\times {10}^{-13J} \end{gathered}$$

From Equation (1),

$$\begin{gathered} \gamma m{c}^2=5.79\times {10}^{-13}J \\ \gamma =\frac{5.79\times {10}^{-13}}{9\times {10}^{-31}\times {\left(3\times {10}^5\right)}^2} \\ =7.15 \end{gathered}$$

From Equation (a),

$\frac{1}{\sqrt{1-\frac{{\left(v^{\text{'}}\right)}^2}{c^2}}}=7.15$

where,$v^{\text{'}}=$ the new speed of an electron

$$\begin{gathered} \sqrt{1-\frac{(v{)}^2}{c^2}}=\frac{1}{7.15} \\ 1-\frac{(v{)}^2}{c^2}=0.020 \\ \frac{{\left(v^{\text{'}}\right)}^2}{c^2}=1-0.020 \\ \frac{{\left(v^{\text{'}}\right)}^2}{c^2}=0.98 \\ {\left(v^{\text{'}}\right)}^2=0.98{\mathrm{c}}^2 \\ {\mathrm{v}}^{\text{'}}=0.989\mathrm{c} \end{gathered}$$

Hence, the new speed of the electron is$0.989c$