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Author: Subject: Quantum Mechanics/Wave Mechanics in Chemistry Beginners Thread
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[*] posted on 23-7-2015 at 10:07


Here she goes!

Energy Quantization of a Particle in a one dimensional box

Higher up I wrote: “Together the right wave function ψ(x) and the right value of E provide a mathematical solution to the Schrodinger equation.”

So far we’ve developed expressions for ψ(x), so now it’s time to look at E:

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/schr.html...

So the Total Energy (eigenvalue) of the particle in a one dimensional box associated with the wave function ψ<sub>n</sub>(x) is:

E<sub>n</sub> = n<sup>2</sup>h<sup>2</sup>/(8mL<sup>2</sup>;)

Slightly rearranged:

E<sub>n</sub> = [h<sup>2</sup>/(8mL<sup>2</sup>;)]n<sup>2</sup>

With n the Quantum Number n = 1, 2, 3,…

The Total Energy of the quantum system is thus quantized: only discrete levels in accordance with the E<sub>n</sub> function are allowed.

Important conclusions:

1. The energies are quantized and can be characterized by a quantum number n.
2. The energy cannot be exactly zero.
3. The smaller the confinement, the larger the energy required.

The lowest state of energy (n = 1), often referred to as the ‘Ground State’ (remember that term!) is thus:

E<sub>1</sub> = h<sup>2</sup>/(8mL<sup>2</sup>;)

The states where n > 1 are often referred to as ‘Excited States’.

[Edited on 23-7-2015 by blogfast25]




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[*] posted on 23-7-2015 at 12:24


So, in this one dimentional box model (got to remember that) :

In the Ground state, the total possible energy in the system is defined by the square of Planck's constant divided by ( 8 times the mass of the particle times the square of the length of the box )

Can i go into an exicited state now ?

Edit :

Just going on the Quantitisation going on, i guess En

will relate to Ψ(n), or Ψ(n)Ψ*(n) in a minute.

[Edited on 23-7-2015 by aga]




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[*] posted on 23-7-2015 at 12:28


Quote: Originally posted by aga  
So, in this one dimentional box model (got to remember that) :

In the Ground state, the total possible energy in the system is defined by the square of Planck's constant divided by ( 8 time the mass of the particle times the square of the length of the box )

Can i go into an exicited state now ?


Excite away!

Q: If the ground state energy is 15.4 eV, what is E for n = 6?

Pocket calculators allowed. No mobile phones, please.

[Edited on 23-7-2015 by blogfast25]




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[*] posted on 23-7-2015 at 12:38


Confooseled for as second there and was going to bleat on about not knowing m or L values, and being evil and everything.

Then it became clear that 15.4 = [h2/(8mL2)] n<sup>2</sup> where n=1

i.e. the whole h2/(8mL2) schmozzle = 15.4

so the answer should be 15.4 n<sup>2</sup> with n = 6

= 15.4 * 6<sup>2</sup>
= 15.4 * 36
= 554.4 eV

Edit :

added the interim step so i can refer back to it when sober

[Edited on 23-7-2015 by aga]




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[*] posted on 23-7-2015 at 12:46


'Evil' I am (Mwhahaha!)

But 554.4 electron volts (eV) it is indeed!

The quantization means that a 'quantum car' would have a bumpy ride. A Classic car can take on any value of kinetic energy (a so-called 'energy continuum'). Your Fiat Quantum would move at 1 mph or 5 or 10 or 15 mph but not at 1.2 or 11.5 or 14.9 mph. Only the SE quantized levels would be allowed.





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[*] posted on 23-7-2015 at 12:56


Woohoo !

Thanks very much for posing a maths question i could actually answer !

This explains everything.

I do indeed own a Fiat Quantum, and n=3 is actually it's top speed.




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[*] posted on 23-7-2015 at 13:23


Quote: Originally posted by aga  


I do indeed own a Fiat Quantum, and n=3 is actually it's top speed.


You need a Time Dependent SE to get it into higher n! I can do you mate's rates, if you want? ;)

[Edited on 23-7-2015 by blogfast25]




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[*] posted on 23-7-2015 at 13:50


A very generous offer.

With my new found maths powers i just removed a lot of 'm' (who really needs back seats ?) and replaced the factory-fitted Planck with a much bigger one and now it's a lot faster.

The wierd thing is that it is harder to find in the car park, as i am Uncertain of where it is.




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[*] posted on 23-7-2015 at 14:01


Quote: Originally posted by aga  
A very generous offer.

With my new found maths powers i just removed a lot of 'm' (who really needs back seats ?) and replaced the factory-fitted Planck with a much bigger one and now it's a lot faster.

The wierd thing is that it is harder to find in the car park, as i am Uncertain of where it is.


Nice. :)

Paradoxically perhaps, removing lots of m makes it go faster, yet the Fiat Quantum drives more bumpily. Any ideas why? For 5 points...




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[*] posted on 23-7-2015 at 14:28


First instinct was to shout 'Wavenumber' and hope to get full points and a lollipop.

Had to do some calculationings to actually see.

As mass Increases, the change in En to En+1 Decreases.

Lower mass means a Higher jump in energy levels from one state to the other.

Time to go out on a limb ( 30 year + dormant brain cells activated - Mr Capps would be proud if still living, and if it's right):

The first integral of En is inversely proportional to m.




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[*] posted on 23-7-2015 at 15:33


Quote: Originally posted by aga  

Lower mass means a Higher jump in energy levels from one state to the other.

Time to go out on a limb ( 30 year + dormant brain cells activated - Mr Capps would be proud if still living, and if it's right):



Correct. As m increases the E levels become closer and closer together, eventually resulting in a continuum. A very slow but smooth ride!

Cinqo puntos, Senor. But you still have to buy your own beer... :cool:




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[*] posted on 23-7-2015 at 23:21


Amazing.

It's possible i'll remember some trigonometry as well.




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[*] posted on 24-7-2015 at 05:59


Here’s another problem.

SCRAPPED as per AAHD pointing out an error. See new version below.

[Edited on 24-7-2015 by blogfast25]




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[*] posted on 24-7-2015 at 07:44


Quote: Originally posted by blogfast25  
Here’s another problem.

A scientist studies a QS and from theory finds its energy quantization rule to be:

E<sub>n</sub> = E<sub>1</sub> / ( n + 1 ), with n the quantum number 1, 2, 3, 4, ... and E<sub>1</sub> the ground state energy.



Do you really mean this? If I substitute n=1, I get E1 = E1/2, hence E1=0, hence all En=0.




Any other SF Bay chemists?
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[*] posted on 24-7-2015 at 07:52


Quote: Originally posted by annaandherdad  


Do you really mean this? If I substitute n=1, I get E1 = E1/2, hence E1=0, hence all En=0.


Oooopsie. You are indeed correct. My bad. :)

[Edited on 24-7-2015 by blogfast25]




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[*] posted on 24-7-2015 at 08:06


Ok, problem revised:

A scientist studies a QS and from theory finds its energy quantization rule to be:

E<sub>n</sub> = E<sub>1</sub> / n<sup>2</sup>, with n the quantum number 1, 2, 3, 4, ... ("E<sub>1</sub> divided by the square of n")

Spectroscopically he determines that ΔE = E<sub>3</sub> - E<sub>2</sub> = - 20 eV ("minus 20 eV").

What is the ground state E<sub>1</sub> of the QS?


[Edited on 24-7-2015 by blogfast25]




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[*] posted on 24-7-2015 at 12:18


E<sub>n</sub> = E<sub>1</sub> / n<sup>2</sup>

ΔE = E<sub>3</sub> - E<sub>2</sub> = -20 eV

OMG. It's like a muscle that has not been flexed for so long, it's gone all crystallised.

It's a couple of simultaneous equations !

E<sub>3</sub> = E<sub>1</sub> / 9 &nbsp;&nbsp;so (*) 9 E<sub>3</sub> = E<sub>1</sub>

E<sub>2</sub> = E<sub>1</sub> / 4 &nbsp;&nbsp;so 4 E<sub>2</sub> = E<sub>1</sub>

E<sub>3</sub> - E<sub>2</sub> = -20

E<sub>3</sub> = E<sub>2</sub> - 20

substituting E3 for E2 equivalence in (*)

9 ( E<sub>2</sub> - 20 ) = 4 E<sub>2</sub>

9 E<sub>2</sub> - 180 = 4 E<sub>2</sub>

9 E<sub>2</sub> - 4 E<sub>2</sub> = 180

5 E<sub>2</sub> = 180

E<sub>2</sub> = 36 eV

ΔE = E<sub>3</sub> - E<sub>2</sub> = -20 eV

E<sub>1</sub> = 36 - 20 = 16 eV WRONG cos the delta is MINUS

E<sub>1</sub> Ground State = 56 eV

Edit:

Did the crystals fall off properly ?

[Edited on 24-7-2015 by aga]




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[*] posted on 24-7-2015 at 12:31


Good effort, aga, but you’re ‘overthinking it’:

ΔE = E<sub>3</sub> - E<sub>2</sub> = E<sub>1</sub>/9 - E<sub>1</sub>/4 = (4E<sub>1</sub> – 9E<sub>1</sub>;) / 36 = - 5E<sub>1</sub>/36 = - 20 eV

E<sub>1</sub> = 144 eV

E<sub>2</sub> = 144/4 = 36 eV

E<sub>3</sub> = 144/9 = 16 eV

ΔE = - 20 eV




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[*] posted on 24-7-2015 at 12:42


Quote: Originally posted by aga  
E<sub>3</sub> = E<sub>1</sub> / 9 &nbsp;&nbsp;so (*) 9 E<sub>3</sub> = E<sub>1</sub>

...

substituting E3 for E2 equivalence in (*)

9 ( E<sub>2</sub> - 20 ) = 4 E<sub>2</sub>


It would help a lot if i read what i typed.

DOH !

Edit:

Double DOH. My initial assumption is obviously wrong as well.

The delta E is not linear between n values AND I have already seen and understood that just yesterday.

Feck.

[Edited on 24-7-2015 by aga]

[Edited on 24-7-2015 by aga]




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[*] posted on 24-7-2015 at 12:52


Only those who make nothing make no mistakes.

Before I move on to the next instalment, just for our ‘math delights’, I give you:

Mini-interlude: an alternative way of determining the E levels for a particle in a 1D box:

For such a particle, U(x) = 0 and the SE is:

- (ћ<sup>2</sup>/2m) δ<sup>2</sup>ψ(x)/δx<sup>2</sup> = Eψ(x)

Which I’ll re-write slightly as:

- aψ” = Eψ (with a = ћ<sup>2</sup>/2m)

From higher up we know that:

Ψ<sub>n</sub>(x) = √(2/L) sin(nπx/L)

It’s easy to show that Ψ’<sub>n</sub>(x) = √(2/L) (nπ/L) cos(nπx/L) [first derivative]

And thus Ψ”<sub>n</sub>(x) = - √(2/L) (nπ/L)<sup>2</sup> sin(nπx/L) = - (nπ/L)<sup>2</sup> Ψ<sub>n</sub>(x) [second derivative]

Insert into the SE and Ψ<sub>n</sub>(x) drops out:

a (nπ/L)<sup>2</sup> = E<sub>n</sub> and with a = ћ<sup>2</sup>/2m:

E<sub>n</sub> = n<sup>2</sup>h<sup>2</sup>/(8mL<sup>2</sup>;)

The part - aψ” is often called a 'Quantum Operator'.


[Edited on 24-7-2015 by blogfast25]




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[*] posted on 24-7-2015 at 13:03


Incredible as it may sound, i follow all of that apart from the 'derivative' bit.

What exactly is a First and Second (presumably Third and so on) derivative ?

[Edited on 24-7-2015 by aga]




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[*] posted on 24-7-2015 at 13:19


Quote: Originally posted by aga  
Incredible as it may sound, i follow all of that apart from the 'derivative' bit.

What exactly is a First and Second (presumably Third and so on) derivative ?

[Edited on 24-7-2015 by aga]


Have a look at the earlier part of the course:

http://www.sciencemadness.org/talk/viewthread.php?tid=62973&...

Does this answer your question, at least partly?




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[*] posted on 24-7-2015 at 13:26


Say y = α sin(βx)

Then y’ = α β cos(βx)

And y” = - α β<sup>2</sup> sin(βx) = - β<sup>2</sup> y




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[*] posted on 24-7-2015 at 13:38


Thanks for the link (blush)

To be honest, my current environment is less than optimally conducive towards sustained concentration.

Would it be fair to compare Derivatives to what i imagine are Integrals ?

E.g. speed, accelleration, rate of change of accelleration and so on ?




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[*] posted on 24-7-2015 at 14:13


Derivatives ate the opposite of integrals, thus VERY closely related.


y === derivation ===> y'

y' === integration ===> y

==========================

Example: (derivatives)

A car travels on the x-axis. x represents position. Speed is v:

v =dx/dt (first derivative of x in time t)

Acceleration is a:

a= dv/dt = d<sup>2</sup>x/dt<sup>2</sup>

Derivatives show up when there is change. E.g. in QP because ψ(x) changes with x.

Wife having another little aga? :o




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