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Posts posted by ChemSiddiqui


For a start: I think U is not an operator but a terrible mess . Do you know what U means? I don't. Is it really the operator as it is given in the question/book or did you rearrange some letters?
E.g. if U was [math] \frac{\sin \theta}{\cos \theta} \frac{d}{d\theta}[/math] then f would obviously be an eigenfuntion.
You know what atheist, you got it right. The way the question was written down I didnt understand it . U is exactly as you suggested;
[math] \frac{\sin \theta}{\cos \theta} \frac{d}{d\theta}[/math]
but I am not sure yet if it is an eigenfunction, so I will do the calculations but I know how to do that. Thanks a lot.
Merged post follows:
Consecutive posts mergedjust one more question. Its the double derivative question;
[math] \frac {d^2}{dx^2} e^\frac{x^2}{2}[/math]
and this is what I did, frst taking the first derivate:
[math] \frac {d}{dx} e^\frac{x^2}{2} = \frac{1}{2}e^\frac{x^2}{2} [/math]
Now taking the second derivative;
[math] \frac {d^2}{dx^2} (\frac{1}{2}e^\frac{x^2}{2})=\frac{x}{4} e^\frac{x^2}{2} [/math]
could anybody confirm id this is right or not? any help most appreciated!
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well this is why I am slighty confused. The question goes to operate wavefunction by an operator to check if the wavefunction is an eigenfunction of the operator.
The operator is [math] U = \frac{d sin(\theta)}{d(\theta)cos(\theta)} [/math] and the function is [math] f(\theta, \phi) = sin(\theta)sin(\phi) [/math]
Now I know that I will have to use the product rule , but I dont know how, becuase the operator is very strange. Its like its saying that diffentiate the function f by [math] \frac{d sin(\theta)}{d(\theta)cos(\theta)} [/math]
Maybe i havent understood the question properly?
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Hi everyone,
I have been trying to practice calculus for the benfit of my quantum mechanics studies and I am stuck with a problem so thought you people might have some advice to give.Here goes;
[math] \frac{d sin\theta}{d(\theta)cos(\theta)} sin(\theta)sin(\phi)[/math]
Now we have to operate the [math] \frac{d sin(\theta)}{d(\theta)cos(\theta)} [/math] onto function [math] sin(\theta)sin(\phi)[/math].
The operator can expressed into [math] cos^2(\theta)sin^2(\theta) [/math] after diffentiation. Now my question is that can I simply multiply this by [math] sin(\theta)sin(\phi)[/math] or will this be wrong?
It would have been much easier if the expression were [math] \frac{d}{d(\theta)} sin(\theta)sin(\phi) [/math] i.e. to diffentiatie the function?!
I'll be grateful for your help!
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I was just wondering about the reaction of softacids and soft bases. Like generic acid/base reaction give a salt and water/neutral species, does soft acid and soft bases give a salt or they would give an adduct/complex?
Lets say that the soft base is NMe3 and soft acid is a diradical oxgen molecule. Is this similar to the reaction between ammonia and diradical oxygen which goes like this 4NH3 + 5O2 > 4NO + 6H2O?
Second question is more of a thought really. Why do we want to look for electrons when we look at an atom. I am asking this question as a quantum mechanics student. Does position or state of an electron give us all the information about a whole system? I am thinking that once we know which state the electron is in we can determine the reactivity of the system?
any help, thoughts or comments are appreciated.
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you are right hermanntrude. thanks for correcting me!
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hmm, I think we could do the acidbase reaction with it if there is a sp3 hybrid orbital on B and it is empty. But I am not sure if there be one on the B atom?
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hey everyone,
I was reading the inorganic book today and i have got a question. say we have an amineborane adduct lets take (CH3)3CH2N:BH3, can it undergo redox or acidbase reactions? I mean the adduct doesnt act as a base because the empty 2p orbital on B is filled when Nitrogen in (CH3)3CNH2 datively forms a bond with BH3 so I am thinking that it cant have undergo any acidbase reaction. Can anyone tell me if I am thinking rightly here. I will be grateful for any direction.
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Welcome to SFN alan,
hope your stay here is an enjoyable and informative one.
all the best.
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When I was reading a blog today, I saw a post about a sweater which had the complete periodic table on it. The only problem is that it was an ugly colour.
It's my housemate's birthday coming up, and he's doing a PhD in genetics. Does anyone know of any sciencey/geeky gifts online that I could get him along these lines?
Also, if I saw something cool enough, I would probably get something for myself. Oh, and I should probably note that he already has just about every one of those Giant Microbe toys!
Many thanks!
I bought myself a periodic table tshirt which looks really cool, but its not the kind of thing you'd like to wear when going out or as a causual wear. But, if your friend like chemistry stuff i suggest buy him/her a molecular modelling set. hope this helps a bit.
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thank Mr Skeptic, I googled the prices and without the things you stated in your reply it seems the cost can be as much as 300m GBR pounds.
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Hi there everyone,
its been a long time since i posted;kinda taking a rest from the forums and trying to enjoy the holidays. I was just chatting with someone the other day and he told me that you can actually build a trailer for 300 GBR pounds. Although i was not convinced I couldnt help but ask anyone here if that is possible. lets say its a wooden trailer. is this possible to build it that cheap?
feel free to tell me of your experiences.
Thanks.
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For me when I find what I am looking in life ,I wouldnt complain God if die.
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He was the first ever person I heard to when I started listening to music. Liked him then but never really became a fan, though i was quite unaware about the allegations(I need not mention) leveled against him. I believe anyone who dies deserve some respect if they hadnt really killed anyone or done something of the sort ,humans do have false afterall. God knows him the best so let He be the judge so all I will say is RIP MJ.
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I think even though what you said above may be correct, if the adhesive forces between the glass and liquid are dominant than cohesive forces in the liquid itself then some of the liquid will stick to the wall and make it wet, but if the cohesive forces (attractive forces b/w like molecules) in liquid are dominant than the adhesive forces b/w wall of glass and the liquid as is the case with mercury, surface tension will make the liquid to occupy the minimun surface area and hence will not allow it to stick to the wall.
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So I can use the classical formula for kinetic energy as used in my calculation above? Even if its a quantum mechanics question.
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mercury is a liquid metal (only liquid metal), it has mettalic bonding and surface tension is great in mercury so Cohesive forces in mercury are much more dominant than the adhesive forces with the wall of the glass. so mercury form 'bead' rather than stick to the wall of glass and make it wet.
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what I think is that if we find the energy of electron in 1D box, by assuming that the length of box is [math] 10^{10} m [/math] and n=1 (lowest energy of the system) then we can find the momentum but we we would have to use the classical foruma relating kinetic energy with momentum;
[math] E= \frac {p^2}{2m} [/math]
if the use of the above forumla is acceptable ( i say acceptable because then it will be switching from QM to classical mechanics), then we can find mommentum and then use the de broglie relation to get the wavelegnth?
BTW just did the calculation assuming what i said above is correct, and the de broglie comes out [math] 1.99*10^{10} m[/math]
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hey everyone,
I am doing QM, trying to solve as much problems there are to get a strong hold of the idea. here one question which i cant understand, i think that there is not enough information to solve this one;
Assuming that a chemical bond can be modelled as a onedimensional box,estimate the order of magnitude of the de Broglie wavelength of an electron in a typical chemical bond.
do you think we have enough information in the question to solve this. The de broglie wavelength can be calculated using [math] lambda = \frac {h}{p} [/math] where p is momentum. how do you calculate momentum? I dont get it. Could anyone help. thanks!.
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thanks a lot for verifing that. Just one little question aslo from past papers but a trivial one;
The typical bond lengths of CH, CC and C(triple bond)C bonds are, respectively,110 pm, 150 pm and 120 pm. Use this information to obtain a reasonable length for a 1D box that you can use to model the states of the pi electrons of butadiyne. Butadiyne , HC(tiple bond)CC(triple bond)CH.
My Answer;
there are 2 C(triple)C bonds, 2 CH bonds and 1 CC bond in the molecule. To get the ength of the box we add all the lengths togther;
so 2(110) + (150) + 2(120) = 610 pm.
Do you think this is right or must I have to take the average?
Now I am able to solve most of the problems of particle in a box, QM actually isnt that bad I am starting to like it.
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Well actually i did come up with that expression myself, but it was the book that made the distiction between energy of the system with equal lengths in all direction and one where the length are not equal, so I knew that this expression was for a box of equal legths! I will do the calculation now and see what I can get for an answer. I will post the answer here. Thanks for the help along the way.
Merged post follows:
Consecutive posts mergedright got the answer;
When we plug in the numbers in the expression [math] \frac{H^2}{8mL^2} [/math] for the mass of electron which is [math] me = 9.109*10^{31} [/math] and [math] L= 6.626 * 10^{9} [/math] we get the [math] 1.37 *10^{21} [/math]
we can now write;
[math] E = 1.37*10^{21}{(n^2x + n^2y + n^2z)} [/math]
using different combination of [math] nx, ny, nz [/math];
first for lowest energy of the system [math] nx =1, ny=1, nz=1 [/math]
[math] E = 4.11*10^{21} [/math] which is lower than the given energy, so this combintation is one of the required ones. Becuase we there is only one way to arrange the energy in 111, the degenercy is 1.
for [math] nx=1, ny= 1, nz=2 [/math] , [math] E = 8.22*10{21}[/math] which again has lower energy value than the one given in question. But, now the degenrecy is 3 becuase there are three ways to arrange the energy i.e. 112, 121, 211.
for [math] nx =1, ny=2, nz=1 [/math], [math] E = 1.23*10^{20} [/math]. again this combination meets the condition refered to in the question, and the degenercy is 3.
So the answer to question;
An electron is trapped in a threedimensional box with edge lengths Lx=Ly=Lz= 6.626 x 10^(9) m. Among the energy levels lying below
E = 1.35 x10^(20) J, how many are degenerate? What is the degeneracy of each one of them?
There are 3 degenerate states and their degenercies are 1,3 and 3.
could you verify it please?
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Is that the same n or is it three different n?
right, now, we want to find 'n' and it is not neccessary that they are same. They will be same when we know that the translation of electron at a particular direction has the same energy as the other direction e.g.
[math] nx = ny [/math]
all we know is that the lengths are equal so the energy of the system can be equal to;
[math] E = \frac {h^2}{8mL^2}{(n^2x + n^2y + n^2z)} [/math]
I got the above expression from a physical chemistry book which says that when the lengths a=b=c then the above expression is applicable for energy of the system.
we could use [math] n=1 [/math] for [math] nx,ny,nz[/math] to get the lowest energy of the system and see if that energy equals the energy given in the question, if it is lower than the provided energy we have one state which has lower energy than asked in the question. how am i doing uptil here?
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Ok here goes;
the question is about 'a particle in 3D box'. The hamitonian operator for it is;
now because the total enegy is equal to sum of energies in the 3 directions we have;
[math]
E(x,y,z) = Ex + Ey + Ez [/math]
where [math] Ex = \frac {h^2n^2}{8mL^2x}, Ey = \frac {h^2n^2}{8mL^2y} , Ez = \frac {h^2n^2}{8mL^2z} [/math]
so total energy is;
[math]
Ex,y,z = \frac {h^2}{8m}{(\frac{n^2}{L^2x} +\frac{n^2}{L^2y} + \frac{n^2}{L^2z})} [/math]
so now the schrodingers equation will be;
[math] H\psi_(x,y,z) = E\psi_(x,y,z)[/math] and
[math] \frac{\hbar^2}{2m} \left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} +\frac{\partial^2}{\partial z^2} \right) = \frac {h^2}{8m}{(\frac{n^2}{L^2x} +\frac{n^2}{L^2y} + \frac{n^2}{L^2z})} [/math]
the total energy provided is [math] E = 1.35 * 10^{20} [/math], so to find the value for n we put E equal to the energy provided.
[math] \frac {h^2}{8m}{(\frac{n^2}{L^2x} +\frac{n^2}{L^2y} + \frac{n^2}{L^2z})} = 1.35*10^{20} J[/math]
by what we do after I cant seem to figure out, we cant possible take the LCM of the terms in bracket on the left hand side of the equation and take the denominator to the other side of the equation?
note; by trying to use LaTex, i can appreciate now that it takes ages and is a monotonious work to write all that stuff dowm. grateful for your help!.
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thanks I got it. I might not be using it immediately; takes a bit of time. If you have ever worked on maple then I think its very easy to use. Except that a few commands will need some learning!. sorry it will be really frustrating for you I am sure, but what do you think of my reasons to use the expressions and in your opinion whats the answer!
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Yes. Two natural questions:
1) Why?
2) Why would the 3D case with 3 equallength sides not be Enx,ny,ny = h^2/8m ( (nx/L)^2 + (ny/L)^2 + (nz/L)^2 ), then?
.
1).well, we are considering a particle in a 3D box with different lengths. So the particle will have different energies in different directions.
2). Well i tried to solve it with that expression and am reduced to;
E8m/h^2 = ((nx+ny+nz/Lx,y,z,^2))
here the algebra seems a bit complicated because on the right hand of equation you have a numerator which is a sum and demoniator cant possibly be taken to the left or can it? Any way, even if you break that rule you still get the same answer i.e. n= approx 3.
So i guess we could use the energy forumla from 1D box?.
I too have atkins 'physical chemistry', although it is recommended for next year I am trying to use it so solve that problem.
How can I use the LaTex, are there any instructions here in sfn. Thanks a lot.
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Quantum Physics
in Homework Help
Posted
dont worry, pay attention to word 'Quantum', look it up in dicitionary if you dont already know. I am a chemistry student and I never did physic before starting university and trust me you do a lot of quantum mechanics in physcial chemistry. Why, well atoms, molecules are very very tiny and our everyday observation give us a misleading picture in general so we use quantum mechanics to help us understand what is possible and what isnt even if the impossible may actually be possible in the real 'quantum' world.
hope this helps you to think about it, also there are some books that will also help you with your essay. maybe google 'quantum mechanics and real world' who know something interesting might show up.
good luck.