00:01
In this problem, you have a particle confined to be in the region x squared and equal to zero.
00:06
It means the wave function will be zero for negative x, and they give us a normalized wave function, where x though is in nanometers.
00:16
And that's fine.
00:17
X in nanometers divided by this 1 .0 nanometers, where unit lists inside the exponential, which we must have.
00:23
Cannot have units in here.
00:25
You might as wonder where this 1 .414 comes from? it's a square root of two.
00:30
Skirt or two when you normalize.
00:32
Now to cut down them a little bit on my writing, it multiplies the exponential, i'm going to call capital a, and the 1 .0 nanometers in the exponential, i'll call a little b.
00:44
So my wave function will look like this, ae to minus x over b.
00:53
That's what we have.
00:55
Okay.
00:56
First thing it wants is the probability.
00:59
You're at x equals 1 .80 nanometers but you have a region of 0 .010 nanometers at that point and they want to the probability of being in that region now you could do integration but it's such a small region relatively speaking it's such a small region that you can you don't need to do the integration now let's talk about this in detail this is the probability density times dx.
01:43
This is effectively the probability of finding it between x and x plus dx.
01:48
So let me write that out.
01:56
Probability at the particle is between x and x plus dx.
02:13
So if you think that if your region at that x is very small, relatively speaking, then you can use this.
02:26
You can almost think, if you want to think about the integration idea, you think, well, such a small region, will this probability density change much? i almost think of that as a constant times times delta x, and that is a reasonable mindset.
02:44
In our problem with x equals 1 .8 and then 0 .01, if you were to put that in the exponential, it changes in the third decimal position.
02:54
5 .0.
02:56
The difference between 1 and the other 0 .001.
03:02
So you really are, you are really are seeing this, for all this purpose as being constant.
03:11
So it's a constant times delta x.
03:15
So our x is 1 .80.
03:18
And it means, but this again, only true for a small region, small region where you can think it like this, where you can think that you really still are just at x.
03:28
Even though this is a, we have a, it's a finite region, but a small, small finite region and our dx or our delta x if you like 0 .010 nanometers so we can put this all together now remember this is the modulus squared which is size to our side but our wave function is real so we don't have to worry about any complex conjugation so we just are really squaring the wave function a squared e minus 2x or b, dx, 1 .414 nanometers minus one -half.
04:11
Only the square root, the nanometers is the only thing that has the power of one -half.
04:17
The 1 .414 is not square rooted.
04:20
It actually was, as i mentioned, it is actually a square root of two.
04:27
So it's really the units r1 over a square root of nanometers.
04:32
So, e to minus 2, 1 .80 nanometers or 1 .0 nanometers times the dx, 0 .010 nanometers.
04:46
Now, unit -wise, refine inside the exponential, i already spoke of that.
04:51
Now, we have one over a square root of nanometers here.
04:54
Oh, this is squared.
04:56
One over a square root squared...