00:01
So in the first part, we need to find the threshold wavelength.
00:06
And in the second part, we need to find the work function for the surface.
00:11
And in order to do that, so in order to find the threshold wavelength, we can use this equation where wavelength is equal to speed of light over threshold frequency.
00:30
Now let's call this equation one.
00:33
So from here we know that if we find the threshold frequency, we can find the threshold wavelength.
00:39
So in order to find the frequency, the threshold frequency, we will use the following equation, the equation for photoelectric effects.
00:47
So maximum kinetic energy is given by total energy.
00:52
And here the energy is at the threshold frequency minus what function.
00:58
So we can apply another condition here, since we are using threshold frequency, we know that at threshold frequency, maximum kinetic energy is zero because there are, there may be emission of photo electrons, but there is no kinetic energy associated with the emitted photo electrons.
01:17
So thus, this means that phi is equal to the threshold energy.
01:26
So we need to find phi in order to find the threshold frequency.
01:30
Now let's call this equation two.
01:32
We will need to use this later.
01:35
Now to find phi the work function we will use the following equation.
01:39
This is also equation for photoelectric effect but here we have the kinetic energy expressed in terms of the stopping potential and this is equal to the energy of the photo of energy of light or energy of photons minus work function.
01:58
So the first equation for photoelectric effect we solved it for the threshold case and this the second equation is for any general case so this means that phi is equal to hf times e v0 where v0 is the stopping potential and we also know that from this equation we know that frequency is c over lambda and this is the same equation and this is the same equation that we used here and applied it for the threshold case so now we have all the information that we now we have all the information we need to solve this equation so we know plan constant 6 .6 to 6 .6 times 10 to the power minus 34 jule second speed of light is 3 times 10 to the power 8 meter per second lambda is given to be 254 nanometers so 254 times 10 to the power minus 8 sorry minus 9 meter and v0 is a stopping potential and that's given to be 0 .181 volts and one last thing charge of the electron and that is 9 point that is yes, nine.
04:01
Okay, so we don't need that because e times we will have this.
04:07
After multiplying it with the charge of electron, we will have it in units of volts.
04:13
But we can just keep e as it is, and then we will have this ev not expressed in terms of electron volt.
04:23
And we know that one electron volt is 1 .602 times 10 to the par minus 19.
04:29
In jules.
04:30
So we need to convert this factor with this in order to have this factor ev not expressed in terms of joules because we are using h, the plan constant in units of juice.
04:42
So in order to, in order for our units to be consistent, we need to solve this.
04:50
We need to have the ev not the stopping potential energy in terms of jules...