Question

What percentage of the speed of light will an electron be moving after being accelerated through a potential difference of 1,485 Volts?

Answer 1

The voltage generates a force over the electron, this force is the one we use to calculate the acceleration

`\begin{gathered} V=(Fd)/(q) \\ 1485V\cdot1.602\cdot10^(-19)=Fd=2.38\cdot10^(-16) \end{gathered}`
`\begin{gathered} Fd=m(a\cdot d) \\ m(ad)=2.38\cdot10^(-16) \\ ad=(2.38\cdot10^(-16))/(9.109\cdot10^(-31))=2.612\cdot10^(14) \end{gathered}`
`\begin{gathered} vf^2=vi^2+2ad \\ vf=\sqrt{2\cdot2.612\cdot10^(14)} \\ vf=22.85\cdot10^6m/s \end{gathered}`

First, we apply the energy equation to find the Fd result based on voltage. Then we find the acceleration times distance. Finally, we find the speed with a kinematic equation.

Now I'm going to find the percentage of the speed of light.

`(22.85\cdot10^6)/(3\cdot10^8)=0.0762=7.62\%`