Question

An electron is accelerated by a potential difference of 463.55 kV. How fast is the electron moving if it started from rest?

Answer 1

`\begin{gathered} V=((ma)\cdot d)/(q) \\ 463.55kV=9.11\cdot(10^(-31)kg)/(1.602\cdot10^(-19)C)\cdot a\cdot d \\ 8.15\cdot(10^(16)m^2)/(s^2)=a\cdot d \\ \\ \end{gathered}`
`\begin{gathered} vf^2=2\cdot ad \\ vf=\sqrt[2]{2\cdot8.15\cdot10^(16)}=4.03\cdot10^8m/s \end{gathered}`

First you use the relation between energy and voltage to get the value of axd, then you use kinematics to solve the final speed