Question

What is the speed of an electron that has been accelerated from rest through a potential difference of $1000V$?

Short answer

The electron speed is$1.87\times {10}^7\mathrm{m}/\mathrm{s}$.

Step-by-step solution

Given information and concept

Potential difference, $V=1000\mathrm{V}$

Mass of electron, $m_e=9.1\times {10}^{-31}\mathrm{kg}$

Charge of electron, $q=-1.6\times {10}^{-19}\mathrm{C}$

Formula used:

The following is the relationship between potential energy and potential difference:

$U=q\Delta V$

Where,

$U$is the potential energy

q is the charge of electron

$\Delta V$ is the potential difference

Find the speed of electron

Calculate the electron's potential energy.

$U=q\Delta V$

Insert the value in the above equation

$\begin{array}{rcl}U& =& -1.6\times {10}^{-19}\times \left(1000\right)\\ & =& -1.6\times {10}^{-16}\mathrm{J}\end{array}$

All potential energy is turned into kinetic energy according to energy conservation.

$$\begin{gathered} K{E}_{\text{initial}}-K{E}_{\text{final}}=U \\ \frac{1}{2}{m}_e{v}_{\text{initial}}^2-\frac{1}{2}{m}_e{v}_{\text{final}}^2=U \end{gathered}$$

Insert the required value in the above equation

$$\begin{gathered} \frac{1}{2}\times 9.1\times {10}^{-31}\times 0-\frac{1}{2}\times 9.1\times {10}^{-31}\times v^2=-1.6\times {10}^{-16} \\ 0-\frac{1}{2}\times 9.1\times {10}^{-31}\times v^2=-1.6\times {10}^{-16} \\ v=\sqrt{\frac{1.6\times {10}^{-16}}{\frac{1}{2}\times 9.1\times {10}^{-31}}} \\ =1.87\times {10}^7\mathrm{m}/\mathrm{s} \end{gathered}$$