00:01
In this exercise, we have a table with data from a photoelectric experiment.
00:06
So we have here the stopping potential v in volts as a function of the wavelength of the incoming light, lambda.
00:17
And in the first exercise, in the first question, a, we have to draw a plot of this data, of the stopping potential as a function of the inverse of the wavelength.
00:33
So the first thing we need to do is you calculate the inverse of the wavelength.
00:37
So for 578 nanometers, the inverse is going to be 1 .73.
00:43
And this is already in 10 to the 6 meters to the minus 1.
00:49
Then we're going to have 1 .83, 2 .29, and 2 .73.
00:58
And now we are in our position to draw the plot, so i'm going to draw it.
01:05
Now in order to save some time, i went ahead and draw the plot and i'm showing it to you.
01:12
So here in the y -axis we have the stopping potential.
01:16
In the x -axis, we have lambda to the minus 1 in units of 10 to the 6 meters to the minus 1.
01:25
And i'm going to go ahead and try to draw a line fitting this.
01:30
Is plot.
01:33
Okay, so it's here.
01:34
It's the best it can do.
01:36
And in question b, we are asked to calculate, or better, not only to calculate, we are asked to look at the graph and extract directly, look at the plots, i'm sorry, and extract directly from it, the values of the work function and the cutoff frequency for this material.
02:01
So you can do it.
02:04
The work function is going to be the minus the intercept of the graph, of the plot.
02:10
And that's because the photoelectric formula is ev equals to hc, lambda to the minus 1 minus phi.
02:26
So that v is going to be hc over e, lambda to the minus 1, minus 5.
02:33
Over e so phi over e is going to be minus the intercept okay so minus the intercept is around here right so around 1 .5 somewhere around 1 .5 electron volts sorry so that phi is somewhere around 1 .5 electron volts.
02:54
Sorry so that phi is somewhere around 1 .5 electron volts.
03:02
Okay that's the approximate work function.
03:06
We can also draw the value of the cutoff wavelength.
03:12
That's when the wavelength for which the the value of the stopping potential is zero.
03:18
So it's here around 1 .3 times 10 to 6 meters to the minus 1.
03:28
So you can calculate the cutoff potential, the cutoff wavelength as 1 over 1 .3 times cent to the minus to the 6th, i'm sorry...