00:01
So in this question, we have an x -ray tube that is similar to a cathode rate tube.
00:06
And this question, this is this question 26 of chapter 18, it is completely based on question 25.
00:14
So if you have not done question 25 yet, please go back and work on question 25 yet, because i will not repeat the procedure and set up of question 25, but that is essential for this question.
00:27
So i would just say from question 25, what we get is we have this expression of the velocity in terms of the potential differences in the machine.
00:44
So this is the velocity for a particle with the charge of magnitude e with the speed of...
00:52
So here, we have the same similar setup here.
00:58
And we want to know through what potential difference would electrons be accelerated, so their speed is 1 % of the speed of light.
01:07
And speed of light, c, is a constant.
01:11
It's 3 times 10 to the 8th meters per second.
01:16
So we first want to rearrange this expression that we already have from question 25, so that it becomes n v square over 2e.
01:25
Then if we want it to be 10 % of the speed of light, so that v is 0 .1c.
01:35
1%, 0 .01c.
01:37
So we have some mass of the electrical times 31 kilogram times 0 .01 times 3 times 10 to 8 meters per second square.
01:58
2 times the electron 1 .6 times 10 to the negative 19 column.
02:05
So this in total gives us 26 volts.
02:11
So that is the potential required in the x -ray machine in order for the electron to achieve the speed.
02:19
And if we look at part b, part b asks us what potential difference should be needed to give the protons the same kinetic energy as the electrons.
02:28
So now the variation we have is this one.
02:34
The mass of the proton is different from the mass of the electron...