00:01
In this problem, we have been given that a voltage of 120 volt is applied so that in the first case, a proton accelerates and in the second case, an electron accelerates.
00:13
So both are initially at rest.
00:15
So the initial kinetic energy of both these protons and electrons, that will be zero.
00:21
So if we want to figure out the speed that they attain when this potential difference is applied, we use work energy theorem.
00:28
And according to that the network is change in kinetic energy.
00:31
So the change in kinetic energy will be just including the final kinetic energy because initially they were at rest and initial kinetic energy will be zero according to the expression k is equal to half mv squared.
00:44
So for the protons, we see that the work done can be computed for both of them because they carry same charge, work can be computed by using the idea that potential difference is multiplied by the charge.
00:55
So as they contain charge of 1 .6 into 10 raise to minus 19 coulamp.
01:02
So the work in each case is given by 120 times 1 .6 into 10 raise to minus 19 joules.
01:09
So this work should be equated in the first case.
01:14
That's 120 to 1 .6 into 10 raise to minus 19.
01:18
And this is equal to the final kinetic energy.
01:21
That's half times the mass of the proton, which is 1 .6...