Homework SourseStarting from rest a proton falls through a potential differ
Starting from rest a proton falls through a potential differenceof 1200v. What speed does it aquire in m/s?
Solution
According to work energy theorm ,we have
W = (1/2)mv2
but W =Vq
the speed of the proton is
v = (2Vq/m)
Here m =1.67*10-27kg
q = 1.6*10-19C
V = 1200 V
Substitute the values.
but W =Vq
the speed of the proton is
v = (2Vq/m)
Here m =1.67*10-27kg
q = 1.6*10-19C
V = 1200 V
Substitute the values.
