00:01
In this question we have a single slit experiment in which we pass through a photon as well as electrons and we see that they actually end up with the same diffraction pattern where they have the first dark fringe right at the angle which we don't know right we only know that it is at the first duct fringe is the same angle.
00:42
Well what this means is that they must have the same wavelength right in order to have the same diffraction pattern.
00:55
Now we can actually relate the wavelength for the electron to its momentum because it's equals to h over p and h over p where p is related to the kinetic energy of the electron via p square over 2m so p is just 2m times kinetic energy square root this h over square roots of 2m times kinetic energy this is the wavelength of the electron and also the wavem of the photon right so energy of the photon when it's c over lambda substituting in this lambda, which is this expression.
02:03
This goes to c times 2m times square root of k.
02:11
So the only unknown here is really the kinetic energy.
02:15
Right, this c and 2m are constants, right? this speed of light, this is the mass of the electron.
02:31
So we can actually evaluate the factor over here a little bit.
02:38
This is actually just 4 .05 times 10 to the power of minus 7.
02:55
And the units over here is actually joe to the power of half, right? because the kinetic energy will give another joules to the power half because it's in square root...