00:01
So here we have the width of the slit equal to w, lambda being the wavelength of the incoming photons, and delta x is the uncertainty in the position of the photons, assuming, and we assume that it's one half the slit width.
00:14
Use the uncertainty principle here to find the minimum range of angles.
00:18
Well, the minimum range of angles, which is the angle theta, can be found using the fact that tangent theta is equal to the uncertainty in the momentum, divided by the momentum.
00:30
Okay, so first thing, first let's find the uncertainty in the momentum.
00:33
Well, this is where heisenberg's uncertainty principle comes in.
00:36
Delta x, delta p is equal to h over 4 pi.
00:43
Therefore, we can find delta p from this, since we know delta x, so we have delta p is equal to h over 4 pi delta x, where delta x is 1 half w.
00:55
So this is is h over 2 pi times w.
01:01
Or in other words, this is equal to 5 .27 times 10 to the minus 32 kilograms times meter per second.
01:20
Now to find the momentum, we can use the fact that lambda is equal to h over p for a photon.
01:29
If it's debrilj wavelength, since it's age over mv, where mv is p.
01:33
So therefore rearranging this, we find p is equal to h over lambda, which is therefore equal to 1 .04 times 10 to the minus 27...