Quantum Mechanics (Chapter 4 of 6)

Once you get past existence coming and going... and virtual particles... and uncertainty... and exclusion, then you are ready to enjoy the REALLY weird stuff in the quantum world.

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  Science & Technology

[raydredX]

5:38 How long do they take to return to a lower energy level? Wouldn't it be almost immediate and cause the absoption spectrum to be filled in with the emission spectrum?

[cassiopeiaproject]

The half-lives of excited states is a well-known time for most atoms. And you are right, it is often pretty fast. But it isn't hard to isolate the incoming beam minus absorbed wavelengths, from the random-direction emissions when the excited states decay.

[JoylessBrotato]

if electrons pop in and out of existence, then how can compounds of elements be so stable, since the bonds between the two elements at some point don't exist.

[cassiopeiaproject]

A chemical bond just changes the shape of the locations that the electron can use in its pop-in-pop-out shell game. The force of the bond itself is just an "average" of the variety of values that it can have as a result of the electrons quantum behavior.

[steeveyneon]

Why does an electron pop in an out of existence over a greater area than a proton just because the electron has less mass?

[cassiopeiaproject]

Yes.

[OBZRV82]

Fascinating info but...

The characters in these video are creepy as hell

N who the hell "Jeeves" ?

[KF81]

Go to Wikipedia and type in Atom. Dont listen to dicks on here as most of them dont know shit, go to science forums and ask questions and also theres nothing wrong in watching these videos if the subject matter interest's you as bit by bit you will pick up information and learn about the world.

All i will say is that the solid world around you is not so solid as you might think when you start looking into smaller distance scales..... ;)

[koldasquare]

Yes I understood. But problem is that in reality could be value of mentioned expression (calculated) 0,56786*(h-bar) and this is SMALLER than h-bar so it has to be (...) >= h-bar over 2. Look at my explanation down there (in fact it is continuation of the above; sorry, YT made problems...).

[KuPaGuitar]

No I know, that is what I had said, I only used the negation of the sign since I was talking about the other side. saying h-bar is the same is less than x*p, in this case, is logically equivalent to saying x*p is greater than or equal to.

In my previous comment I was stating how x*p is greater than or equal to h-bar/2 is probably the equation your familiar with.

And that still holds from the equation given in the video, since h-bar/2 is smaller than h-bar.

Since 5 >= 4 , then 5 >= 4/2

[koldasquare]

and here we go 0,56786*h-bar which is how you can see only little bit more than h-bar divide 2, but less than just h-bar - and that is only one reason why there should be (delta)x*m*(delta)v >= h-bar divide 2. The expression (delta)x*m*(delta)v >= h-bar is not complete wrong but ,,just" not accurate. But this is, of course, bloody hell physx and I do not want to mess with it :-). PS Sorry fot the mess, YT thinks I am robot :-D

[koldasquare]

According statistics and quantum mechanics you can determinated of standard deviation of the position of a particle in this box (for state n = 1) - is equal to 0,180756a (please, you really do not want to know how you can both calculate). In that point you find px, p^2 x and (sigma)px which is in this case h over (2a)...then you calculate uncertainty product...

[koldasquare]

E. g.: x*y >= 0,5 allows numbers 0,5 or bigger, but x*y>= 1 NOT allows number smaler than 1 (which 0,5 definitively is). And here is the problem because you are able to calculate estimated value of (delta)x*(delta)p for particle e.g. in one-dimensional box. Let’s say you have particle in 1D box (where L=a, L is length).

[koldasquare]

Thx for reaction, I am glad :-), but it is wrong. You are probably little bit confused with mentioned expression in this video. The expression (delta)x*(delta)p (or (delta)x*m*(delta)v, that is the same) >= (value) means that the (delta)x*(delta)p is even or BIGGER than some value NOT smaller.