Answer to Question #348228 in Physics for Umme

Question

Find the velocity aquired by an electron in falling through a potential difference of 2000 volts.

Answer 1

$$\frac{mv^2}{2}=eV$$

$$v=\sqrt{\frac{2eV}{m}}=\sqrt{\frac{2*1.6*10^{-19}*2000}{9.1*10^{-31}}}=2.65*10^7\:\rm m/s$$