00:01
So here we have two sources of light being fallen on a metal surface.
00:08
So let's call those two sources to have wavelength lambda 1 and lambda 2.
00:15
So lambda 1 is given to be 400 nanometer and lambda 2 is 300 nanometer.
00:21
Now after they fall on the middle surface, the maximum kinetic energy is measured to be 1 .10 electron volt in case of the wavelength to be equal to lambda 1.
00:37
So let's say the corresponding kinetic energy is k -e -1 and that is equal to 1 .10 electron volt.
00:46
And we need to find the kinetic energy, the maximum kinetic energy, corresponding to the second wavelength.
00:55
Now, in order to do that, we will use a photoelectric effect.
01:02
So we know that maximum kinetic energy is equal to the energy of the source, of the fallen source, so that is h times f minus work function of the metal.
01:17
And f is c over lambda, so we can substitute that because we have lambda given in the problem and not frequency, so that will save some time to first find frequency and then substitute here.
01:31
So we can directly substitute lambda over here minus phi.
01:38
Now, there are two methods to solve this problem.
01:43
So method one could be to find the work function using information from the first case where you have both the kinetic energy and wavelength given.
02:00
So you can find work function.
02:03
So phi.
02:04
And this is going to be the same.
02:07
Work function for the second case as well because you are using the same metal.
02:12
So the work function for both the cases has to be same.
02:15
So once you find the work function, you can apply it to the second case and use this same equation to solve for the maximum kinetic energy for the second case.
02:28
So find phi first and then find k .e .2.
02:33
So that is one method.
02:34
And another method would be that you don't need to solve for the work function.
02:43
So i'm going to follow the second method because i think this is relatively simpler than with respect to calculations.
02:55
This is going to be relatively simple than the first one.
02:58
So i'll be using this second method over here.
03:02
So you get an idea of applying this method in other cases.
03:08
So let me just write this same equation.
03:14
This one, let me call this stars.
03:17
So let me apply this equation to both of these two cases.
03:22
So we have k -e -1 to be equal to h -c.
03:28
So these two are constant.
03:29
So they don't change with wavelength or, frequency.
03:36
So we, sorry, they don't change with wavelength or frequency.
03:40
So we have hc, which is a constant, and then lambda for the case one is lambda 1 minus 5.
03:48
So, phi is the work function of the metal.
03:51
So we are using the same notation.
03:54
We will be using the same notation for both the cases because the metal that we are using is same.
04:00
And then for the second case, ke2, we have hc over lambda 2 minus 5.
04:10
So as you can see, these two are the different equations of photoelectric effect for these two cases, lambda 1 and lambda 2.
04:21
So now let's subtract one equation from the other taking care of the corresponding sides.
04:32
So we are going to subtract the corresponding sides of the equations.
04:38
So k -e -1 minus k -e -2 is going to be equal to h -c over lambda 1 minus h -c over lambda 2.
04:47
And since phi is the same in both equations, so that just cancels out when we subtract these two equations.
04:55
So now from this equation, let me just write that again.
05:01
So k -e -1 minus k -e -2 is hc1 over lambda 1 minus 1 over lambda 2.
05:11
So from here we can directly solve for the maximum kinetic energy corresponding to lambda 2 without having to solve for 5, which is the warp function...