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blogfast25
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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.phyastr.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 2372015 by blogfast25]


aga
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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 E_{n}
will relate to Ψ(n), or Ψ(n)Ψ*(n) in a minute.
[Edited on 2372015 by aga]


blogfast25
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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 2372015 by blogfast25]


aga
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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 2372015 by aga]


blogfast25
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'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 socalled '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.


aga
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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.


blogfast25
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You need a Time Dependent SE to get it into higher n! I can do you mate's rates, if you want?
[Edited on 2372015 by blogfast25]


aga
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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 factoryfitted 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.


blogfast25
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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 factoryfitted 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...


aga
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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 E_{n} to E_{n+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 E_{n} is inversely proportional to m.


blogfast25
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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...


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


blogfast25
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Here’s another problem.
SCRAPPED as per AAHD pointing out an error. See new version below.
[Edited on 2472015 by blogfast25]


annaandherdad
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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?


blogfast25
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Oooopsie. You are indeed correct. My bad.
[Edited on 2472015 by blogfast25]


blogfast25
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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 2472015 by blogfast25]


aga
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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 so (*) 9 E<sub>3</sub> = E<sub>1</sub>
E<sub>2</sub> = E<sub>1</sub> / 4 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 2472015 by aga]


blogfast25
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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


aga
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Quote: Originally posted by aga  E<sub>3</sub> = E<sub>1</sub> / 9 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 2472015 by aga]
[Edited on 2472015 by aga]


blogfast25
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Only those who make nothing make no mistakes.
Before I move on to the next instalment, just for our ‘math delights’, I give you:
Miniinterlude: 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 rewrite 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 2472015 by blogfast25]


aga
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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 2472015 by aga]


blogfast25
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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 2472015 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?


blogfast25
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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


aga
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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 ?


blogfast25
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Derivatives ate the opposite of integrals, thus VERY closely related.
y === derivation ===> y'
y' === integration ===> y
==========================
Example: (derivatives)
A car travels on the xaxis. 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?


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