00:01
So first let's write down the information that we are given.
00:03
So we know the wavelength lambda of the light source, and it is given to be 235 nanometer or 235 times 10 to the bar minus 9 meter.
00:17
And from table 38 .1, you can find the work function of this metal.
00:30
We'll know in a while why we need to use the swap function.
00:35
So, and this is an electron volt.
00:39
So we need to convert this into joules in order to be, in order to keep our units consistent.
00:46
So one electron volt is 1 .602 times 10.
01:02
So first, let's write down the information that we are given.
01:06
So we know the wavelength, lambda of the light source.
01:10
And that is given to be 235 nanometer or 235 times 10 to the power minus 9 meter and from table 38 point the power minus 19 joules.
01:29
So multiplying this we get our work function phi in jules and that comes out to be equal to 8 .1711 times 10 to the power minus 19.
01:42
Joules and we need the we need to find the velocity of the emitted electrons emitted photo electrons so we need one you can find the work function of this metal you know in a while why we need to use this work function so and this is an electron volt so we need to convert this into joules in order to be in order to be in to keep our units consistent.
02:20
So one electron volt is 1 .602 times 10 to the power minus 19 joules.
02:30
Know that by photoelectric effect, kinetic energy of the emitted electrons is equal to energy of the electron resident or the energy of the source light, source photons, minus work function of the electron.
02:53
So this is the kinetic energy of the emitted electrons.
02:56
This is the total energy of the photons of the light source and we know that kinetic energy is so multiplying this we get our work function phi in joules and that comes out to be equal to 8 .1711 times 10 to the power minus 19 joules and we need the we need to find the velocity of the emitted electrons emitted photo electrons so we know that by photoelectric effect e is h times frequency of the photon then we have minus work function so we don't have the frequency here but we do know the wavelength of the photon.
03:51
So frequency can be formed from the wavelength by dividing the wavelength with the speed of light.
04:01
So let's substitute this expression over here.
04:04
So we have xc over lambda in place of energy now minus our function.
04:11
So now we can solve this equation and find v...