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Ok here is one question I did, but I am not sure of answer. I am asking questions about QM because I am new to this subject area. Here goes the question;

An electron is trapped in a three-dimensional 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?

Now this is what I did.

the energy of electron in a 1-D box is;

En = n^2 *h^2/ 8*m*L^2

inputing the values to get n

n = ((En*8*m*(L)^2)/(h^2))^1/2

n =(9.83)^1/2 = approx. 3

So the energy level that lie below the given energy are n =1,2. And the degenercies are 1 for n =1 and 4 for n=2. Degeneries calculated from n^2.

Is this calculation and result right. Could someone please verify? Sorry had to use excel format for calculation! Thanks a lot for any help!

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No I think you fundamentally messed it up.

An electron is trapped in a three-dimensional 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?

Now this is what I did.

the energy of electron in a 1-D box is;

$E_n = \frac{n^2 h^2}{8 m L^2}$

inputing the values to get n

$n = \sqrt{\frac{E_n 8 m L^2}{h^2}}$

$n =\sqrt{9.83} \approx 3$

So the energy level that lie below the given energy are n =1,2.

Fine till here except that

- I think n=3 would also be fine (or why not?),

- the question does not ask for a particle in 1D,

- you should ask yourself what happens for n=0,

- I did not check the numbers or the term for the energy, only the algebra.

And the degenercies are 1 for n =1 and 4 for n=2. Degeneries calculated from n^2

Where do you get that from? The degeneracy for a 1D particle in a box without additional degrees of freedom is 1 for all energies.

You should look up the 2-dimensional particle in a box and understand things like $H = H_x + H_y \Rightarrow \psi(x,y) = \psi_x(x) \psi_y(y)$ (H being the hamiltonian, psi being the time-independent wave-function and x and y being the respective coordinates) and what that means for the quantum numbers (particularly their number) and the total energy and what $\psi_x(x)$ and $\psi_y(y)$ are.

Sorry had to use excel format for calculation!

???

Edited by timo
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No I think you fundamentally messed it up.

Fine till here except that

- I think n=3 would also be fine (or why not?),

- the question does not ask for a particle in 1D,

- you should ask yourself what happens for n=0,

- I did not check the numbers or the term for the energy, only the algebra.

Where do you get that from? The degeneracy for a 1D particle in a box without additional degrees of freedom is 1 for all energies.

You should look up the 2-dimensional particle in a box and understand things like $H = H_x + H_y \Rightarrow \psi(x,y) = \psi_x(x) \psi_y(y)$ (H being the hamiltonian, psi being the time-independent wave-function and x and y being the respective coordinates) and what that means for the quantum numbers (particularly their number) and the total energy and what $\psi_x(x)$ and $\psi_y(y)$ are.

???

the question asked the degeneracies of states with energy lower than the given one. n=3 is the state with the given energy so lower ones should be n=1,2, right? Also, since its my first year in university and we only been introduced to QM. We have been told that we could use particle in 1-D box model for particle in 2-D or 3-D or N-D boxes. all we have to do is to times the solution of the 1-D model by the appropriate number of dimensions. About the degenericies I think I looked it up from the hydrogen atom spectrum. I understand the hamiltonian operator concept, The kinetic energy is equal to

-h(bar)^2/2m * d^2(psi)/dx^2.

The Kinetic energy is 0 because of no interaction with any other particle (infact there is one one electron inside the box). but I am still not sure about the question. Could you tell me where I was wrong. thanks

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I wonder why you are given questions about 3D particle-in-a-box when it seems you haven't even discussed the 1D case, yet. I'll see if I can construct a short walkthrough for the problem but that'll take a few hours (during which I should better not fall asleep). EDIT: Won't get that done today; will update sometime on the weekend.

Edited by timo
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I wonder why you are given questions about 3D particle-in-a-box when it seems you haven't even discussed the 1D case, yet. I'll see if I can construct a short walkthrough for the problem but that'll take a few hours (during which I should better not fall asleep). EDIT: Won't get that done today; will update sometime on the weekend.

well we have been given a couple of lectures on 1-D box; this is what i gathered from them;

when an particle (say electron) is trapped in a box of one-dimension (hence particle in 1-D) of legth x1 =0 and x2=L and it can translate freely along one direction only say x-direction then the probability of finding the electron between the two ends of the box is given by the born interpretation of the wavefunction. also the probability of finding an electron at the walls is zero as wavefunction is zero(potential energy at the walls is infinity) and also outside the box the probability of finding electron is aslo zero. The TI-schrodinger's equation can be formed by solving the hamiltonian operator.

H = E + V (note the sympol '^' is over each letter in the formula).

E is the kinectic energy which is by the virtue of electron transition and V is the potential energy which is due to interaction. Because there is just one particle and it cant interact with anyother particle the potential energy is zero.

So the hamiltonian opertor is H =E and schrodinger's equation can be written as;

H(psi) = E(psi)

- h(bar)^2/2m * d^2(psi)/dx^2 = E(psi)

although we have not been taught to solve the schrodinger's equation however we have been told how to 'keep' the acceptable solutions and 'discard' the unacceptable ones. This is as far as we went. And like I said in my previous message 3-D will be done in more detail next year, we have just for the time being told that for the 3-D we have to multiply the solution to 1-D by 3. but I am not sure if its the same for energy? O yeh, we also covered sketching wavefunctions of 1-D box!. And Thanks a lot for helping!

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Well, a short sketch for the 1D part I had prepared already yesterday, so here it is (note: (E) refers to the time-independent Schrödinger equation, (N) is the normalization criterion).

The 1D particle in a box

A very simple toy example is that of a particle in a 1D box. This system is described by the Hamiltonian

$H(x) = \frac{-\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + V(x), \ V(x) = \left\{ \begin{array}{ccl} 0 &:& 0 \leq x \leq L \\ \infty & : & \text{else} \end{array} \right.$

The wave-function shall be zero where the potential is infinite (which means the particle is inside the box) which due to the continuity of the wave function means that $\psi(0) = \psi(L)=0$. Inside the box (E) for the wave function $\psi(x)$ reads

$\frac{-\hbar^2}{2m} \frac{\partial^2}{\partial x^2} \psi(x) = E \psi(x)$

$\Rightarrow \frac{\partial^2}{\partial x^2} \psi(x) = \frac{-2mE}{\hbar^2} \psi(x)$

$\Rightarrow \psi(x) = A \exp \left( i \sqrt{\frac{2mE}{\hbar^2}} x \right) + B \exp \left( -i \sqrt{\frac{2mE}{\hbar^2}} x \right)$

From boundary condition $\psi(0)=0$ if follows that A=-B which allows rewriting the term as

$\psi(x) = 2A \sin \sqrt{\frac{ 2mE}{\hbar^2}} x$.

Now if A was zero, condition (N) would be violated, so the only way to get the solution in agreement with the other boundary condition $\psi(L)=0$ is to make the Sine zero there (which happens when the argument is a multiple of Pi). This means that

$\sqrt{\frac{ 2mE}{\hbar^2}} L = n \pi \Rightarrow E = n^2 \frac{\hbar^2 \pi^2}{2mL^2}, \ n > 0$.

The solution with n=0 would again violate (N), n<0 gives the same energies. Btw.: Note how the boundary conditions create the "quantization" of energies. Now, writing the possible energies with an index n labeling the remaining degree of freedom (the multiple of Pi in the argument of the Sine), the possible energies for the particle in a 1D box are

$E_n = n^2 \frac{\hbar^2 \pi^2}{2mL^2}$ (1)

For the more-dimensional case the trick is that sometimes (e.g. for the particle in a box) you can rewrite $H(\vec x) = H_x(x) + H_y(y) + H_z(z)$, i.e. cast the Hamiltonian into a sum of sub-Hamiltonians each depending on less coordinates. I'll write a bit more on that over the weekend but today it's my no-physics day so it'll be tomorrow.

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Thank you for your post, its great and helped a lot. Just before you post the other stuff could you check this please;

The question as I undestand it is asking to find the value of n given the energy and the dimension in the three directions (x,y,z). I think we could only use the hamiltonian operator for 1-D box because the dimensions are equal i.e. Lx = Ly =Lz = 6.626 x 10^(-9), so they are dependent of each other. If they were independent, that is to say, their dimensions were diffenent Lx is not = Ly is not = Lz then we could we the hamiltonian operator for the 3-D box with the energy as follows;

Enx,ny,ny = h^2/8m ( (nx/Lx)^2 + (ny/Ly)^2 + (nz/Lz)^2 ).

Am i right till here? it does make sense!. Also if we had different dimensions what value of n do we have to input? thanks very much for your help so far.

side-note;

I am not a physics student; am a chemistry one. I am sure that QM is same for both but I think you can appreciate that chemists dont touch some of the things that are in the realm of physics.

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The question as I undestand it is asking to find the value of n given the energy and the dimension in the three directions (x,y,z).

You are given the total energy and the size of the box in the three directions/dimensions.

I think we could only use the hamiltonian operator for 1-D box because the dimensions are equal i.e. Lx = Ly =Lz = 6.626 x 10^(-9), ...

No, there's not much difference between the lengths being equal or not (there is, but not much).

... so they are dependent of each other.

Hm, you probably mean that if you increase the size of one side then you have to increase the size of the others to keep them being same-length. That is fine but not very exciting. What makes this statement seem a bit odd is that dimension is often understood as the maximum number of independent coordinates.

If their dimensions were diffenent Lx is not = Ly is not = Lz then we could we the hamiltonian operator for the 3-D box with the energy as follows;

Enx,ny,ny = h^2/8m ( (nx/Lx)^2 + (ny/Ly)^2 + (nz/Lz)^2 ).

Yes. Two natural questions:

1) Why?

2) Why would the 3D case with 3 equal-length sides not be Enx,ny,ny = h^2/8m ( (nx/L)^2 + (ny/L)^2 + (nz/L)^2 ), then?

Btw.: Consider leaning TeX. I think learning to use the sfn math commands is a good way to do so. You can make your equations more readable. In "real world" you are likely to use TeX at some point (~100% of the physics student do, dunno for chemists) so learning the TeX math commands (used on sfn) is not a waste of time.

Also if we had different dimensions what value of n do we have to input?

That is pretty much what the question asked from you: You are given a maximum energy -> which combinations (nx,ny,nz) are possible so that the total energy stays below the maximum allowed.

I am not a physics student; am a chemistry one. I am sure that QM is same for both but I think you can appreciate that chemists don't touch some of the things that are in the realm of physics.

I wrote it with chemists in mind. I happen to have a chemistry book (Atkins "Physical Chemistry") at home so I think I approximately know what is demanded from chemists. Also, I once attended a chemistry course that dealt with QM (which is why I have a chemistry book at home). I am not too sure that QM is really the same or at least not taught in the same way to physicists and chemists. Explanations for chemists (and laymen) seem to focus a lot on calculus and the wave function while explanations for physicists are usually a bit more abstract talking about states, vector spaces and algebras.

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Yes. Two natural questions:

1) Why?

2) Why would the 3D case with 3 equal-length 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 3-D 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 1-D 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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The easiest way to see how latex can be used is hitting the quote button on posts that use it and see what the poster typed in. There's a latex tutorial on sfn and also some test-bed too, I think.

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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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Higher dimensions: Seperation of variables

Extending the particle-in-a-box example to two dimensions, the Hamiltonian for this problem reads

$H(x,y) = -\frac{\hbar^2}{2m} \left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} \right) + V(x,y),\ V(x,y) = \left\{ \begin{array}{ccl} 0 &:& 0 \leq x \leq L_x \text{ and } 0 \leq y \leq L_y \\ \infty & : & \text{else} \end{array} \right.$

with $L_x, L_y$ being the lengths of the rectangular box.

Defining the two helper-operators

$H_x(x) = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + V_x(x), V_x(x) = \left\{ \begin{array}{ccl} 0 &:& 0 \leq x \leq L_x \\ \infty & : & \text{else} \end{array} \right.$

$H_y(y) = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial y^2} + V_y(y), \ V_y(y) = \left\{ \begin{array}{ccl} 0 &:& 0 \leq y \leq L_y \\ \infty & : & \text{else} \end{array} \right.$

the Hamiltonian can be formally rewritten as $H(x,y) = H_x(x) + H_y(y)$. Now, these helper-operators are simply Hamiltonians for particles in a 1D box. Particularly, the possible energy levels $E_x, E_y$ and the wave-functions $\phi_x(x), \phi_y(y)$ solving $H_x(x)\phi_x(x) = E_x \phi_x(x)$ and $H_y(y)\phi_y(y) = E_y \phi_y(y)$ are known. Because of that, the multi-dimensional problem can be solved very quickly:

Theorem:

Assume an Hamiltonian $H(x,y)$ depending on two coordinates x and y can be rewritten as a sum of two Hamiltonians $H_x(x), H_y(y)$ each depending on only one of the coordinates. Further assume the functions $\phi_x(x), \phi_y(y)$ solve the time-independent Schrödinger equations for the reduces problems $H_x(x) \phi_x(x) = E_x \phi_x(x)$ and $H_y(y) \phi_y(y) = E_y \phi_y(y)$. Then, the wave functions $\psi(x,y) = \phi_x(x) \phi_y(y)$ solves the full problems $H(x,y)\psi(x,y) = E \psi(x,y)$ with $E = E_x + E_y$.

Proof:

$H(x,y) \psi(x,y) = \left( H_x(x) + H_y(y) \right) \phi_x(x) \phi_y(x) = \underbrace{H_x(x)\phi_x(x)}_{=E_x \phi_x(x)} \phi_y(x) + H_y(y) \phi_x(x) \phi_y(y)$

$= E_x \phi_x(x) \phi_y(x) + \phi_x(x) \underbrace{H_y(y) \phi_y(y)}_{=E_y \phi_y(y)} = E_x \psi(x,y) + E_y \psi(x,y)$

$=\underbrace{(E_x + E_y)}_{=:E} \psi(x,y) = E\psi(x,y)$

This separation of variables straightforwardly extends to particles in higher-dimensional rectangular boxes.

So what does that have to do with the original question?

A lot. It tells you why and how you can reduce the question (3D case) to the case you are somewhat familiar with (1D case). Of course, I only explained things up to the point of knowledge that the the question seems to require as being known. Random additional comments:

- An energy state is degenerate exactly if there is more than one combination of (Ex, Ey, Ez) ( or equivalently (nx,ny.nz) ) that leads to the same energy. Note that e.g. (1,2,1) and (2,1,1) are different combinations.

- You originally did some approximation. Do not do this, yet.

- You can probably find all the states fulfilling the max-energy criterion by hand. Start from the state with the lowest energy, then continue looking for the state(s) with the next-lowest energy level until you exceed the maximum allowed.

- The square-case is trivially included in the rectangular case.

- Above mathematical steps may be non-trivial if you are not familiar with QM or differential equations. So don't be afraid if you do not immediately understand a seemingly small step. Sit down and try to understand it. Don't be afraid to ask if you still don't get it after that.

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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;

$E(x,y,z) = Ex + Ey + Ez$

where $Ex = \frac {h^2n^2}{8mL^2x}, Ey = \frac {h^2n^2}{8mL^2y} , Ez = \frac {h^2n^2}{8mL^2z}$

so total energy is;

$Ex,y,z = \frac {h^2}{8m}{(\frac{n^2}{L^2x} +\frac{n^2}{L^2y} + \frac{n^2}{L^2z})}$

so now the schrodingers equation will be;

$H\psi_(x,y,z) = E\psi_(x,y,z)$ and

$-\frac{\hbar^2}{2m} \left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} +\frac{\partial^2}{\partial z^2} \right) = \frac {h^2}{8m}{(\frac{n^2}{L^2x} +\frac{n^2}{L^2y} + \frac{n^2}{L^2z})}$

the total energy provided is $E = 1.35 * 10^{-20}$, so to find the value for n we put E equal to the energy provided.

$\frac {h^2}{8m}{(\frac{n^2}{L^2x} +\frac{n^2}{L^2y} + \frac{n^2}{L^2z})} = 1.35*10^{-20} J$

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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now because the total enegy is equal to sum of energies in the 3 directions we have;

$E(x,y,z) = Ex + Ey + Ez$

where $Ex = \frac {h^2n^2}{8mL^2x}, Ey = \frac {h^2n^2}{8mL^2y} , Ez = \frac {h^2n^2}{8mL^2z}$

Is that the same n or is it three different n?

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!.

You get faster with experience plus you have copy&paste which is a huge advantage over using pen&paper. In fact I sometimes to tedious calculation on latex rather than paper when they involve a lot of repeating the same stuff because I am a) sometimes a bit faster then and b) less prone to errors when copying something from the previous line to the new one.

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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.

$nx = ny$

all we know is that the lengths are equal so the energy of the system can be equal to;

$E = \frac {h^2}{8mL^2}{(n^2x + n^2y + n^2z)}$

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 $n=1$ for $nx,ny,nz$ 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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right, now, we want to find 'n' and it is not neccessary that they are same.

Yes.

all we know is that the lengths are equal so the energy of the system can be equal to;

$E = \frac {h^2}{8mL^2}{(n^2x + n^2y + n^2z)}$

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.

You should by now have been able to come up with that yourself. It's merely the sum of energies of three degrees of freedom (x, y and z) in a 1D box.

we could use $n=1$ for $nx,ny,nz$ ...

Let's just call it nx=1, ny=1, nz=1. You mean the right thing, but there simply is no variable called "n", anywhere.

..to get the lowest energy of the system and see if that energy equals the energy given in the question, ...

More generally: How it relates to the given energy. It's quite improbable that a quantized energy will exactly equal a given one.

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?

Yes. Now try to find all states whose energy is lower than the given. Then think about the degeneracy.

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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 merged

right got the answer;

When we plug in the numbers in the expression $\frac{H^2}{8mL^2}$ for the mass of electron which is $me = 9.109*10^{-31}$ and $L= 6.626 * 10^{-9}$ we get the $1.37 *10^{-21}$

we can now write;

$E = 1.37*10^{-21}{(n^2x + n^2y + n^2z)}$

using different combination of $nx, ny, nz$;

first for lowest energy of the system $nx =1, ny=1, nz=1$

$E = 4.11*10^{-21}$ 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 $nx=1, ny= 1, nz=2$ , $E = 8.22*10{-21}$ 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 $nx =1, ny=2, nz=1$, $E = 1.23*10^{-20}$. 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 three-dimensional 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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When we plug in the numbers in the expression $\frac{H^2}{8mL^2}$ for the mass of electron which is $me = 9.109*10^{-31}$ and $L= 6.626 * 10^{-9}$ we get the $1.37 *10^{-21}$

That should carry some unit, probably Joule.

for $nx =1, ny=2, nz=1$, $E = 1.23*10^{-20}$. again this combination meets the condition refered to in the question, and the degenercy is 3.

"nx=1, ny=2, nz=1" probably is a typo and should read "nx=1, ny=2, nz=2"?

There are 3 degenerate states and their degenercies are 1,3 and 3.

A degenerancy of 1 is usually not referred to as degenerate.

I did not check the numbers. Technically, a proof that all other combinations of nx,ny,nz would exceed the maximum energy is missing. Other than that and the (minor) comments above it looks fine to me.

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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 C-H, C-C 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 1-D box that you can use to model the states of the pi electrons of butadiyne. Butadiyne , H-C(tiple bond)C-C(triple bond)C-H.

there are 2 C(triple)C bonds, 2 C-H bonds and 1 C-C 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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