Question

What potential difference is needed to accelerate an electron from rest to a speed of $2.0\times {10}^{6}m/s$?

Short answer

The potential difference needed is$V_{1\to 2}=\boxed{\mathbf{11}\mathbf{.}\mathbf{4}\mathbf{}\mathit{V}}$

Step-by-step solution

Given information and theory used  

Given : To accelerate an electron from rest to a speed of : $2.0\times {10}^{6}m/s$

Theory used :

According to the law of conservation of energy, $U_1=U_2+\frac{1}{2}m{v}^2$. ...(1)

Calculating potential difference needed 

Allow the electron to accelerate from point 1 to point 2, where it will reach $v=2.0\times {10}^{6}m/s$.

Rewriting the law of conservation of energy, in equation (1) :

$\frac{1}{2}m{v}^2=-(U_2-U_1)$

Through $(U_2-U_1)=q(V_2-V_1)=q{V}_{1\to 2}$, the potential difference is connected to the potential energy difference.

Because electron $q=-e$, produces

$$\begin{gathered} \frac{1}{2}m{v}^2=e{V}_{1\to 2} \\ \Rightarrow \boxed{V_{1\to 2}=\frac{m{v}^2}{2e}} \end{gathered}$$

We get $V_{1\to 2}=\boxed{\mathbf{11}\mathbf{.}\mathbf{4}\mathbf{}\mathit{V}}$by substituting$e=1.60\times {10}^{-19}C\mathrm{and}m=9.11\times {10}^{-31}kg$