What is the speed of a proton after being accelerated from r

What is the speed of a proton after being accelerated from rest through a 90 x 10^6 V potential difference?

Solution

when proton is accelerated through potential difference V then it KE increases as follow

change in KE = q (delteV)

hence

mv^2 /2 = q (deltaV)

1.67 x 10^-27 x v^2 /2 = (1.6 x 10^-19 ) (90 x 10^6)

v= 1.31 x 10^8 m/s