The Bohr Model

  • #1

Main Question or Discussion Point

I am currently taking Physics 30 now and we have just finished the Magnetism unit. The last little bit that we had to learn in the unit was about Albert Einstein's Theory of Relativity. I was just thinking about this....and maybe I'm crazy, but I'll just state my opinion.

I was just thinking that the Bohr Model is....well....kind of impossible. I know that in the Bohr Model we have nucleus in the middle (consisting of protons and neutrons) and if the electrons are always accelerating around the nucleus, they will give off EMR. Because they give off so much energy, wouldn't they just "crash" into the nucleus?

Maybe I'm weird saying this, but it's just something that I thought of....so I was wondering if anyone can help me understand this a bit more. Thanks.
 

Answers and Replies

  • #2
Yes, that's exactly what should happen in that model. The reason why it doesn't is because an electron in an atom isn't really a point particle but is better described as a "spread out" wavefunction, with an energy that has to be one of certain discrete values ("energy levels").

Are you going to take a course on quantum mechanics any time soon?
 
  • #3
dextercioby
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According to both QM and The Standard Model,the electron (in an atom or anywhere else,free or a weird plasma) is percepted as a point particle.The wavefunction and its interesting graphical representation is nothing but a mathematical abstraction of the physical state of the electron.
So i ask you,Mr.loandbehold,
loandbehold said:
Are you going to take a course on quantum mechanics any time soon?
 
  • #4
Claude Bile
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In QM, the electron is described by a wavefunction. The wavefunction is interpreted as being a probability amplitude.

The reason electrons do not radiate while in a fixed orbital is becuase the wavefunctions are invarient with time. Thus the charge density of the electron cloud is in fact, not accelerating and thus no energy is emitted.

When the orbital changes, the wavefunction does vary with time, which is why a transition between orbitals corresponds to an interaction with radiation.

If electrons were point particles, we would expect to see atoms continuously emitting radiation, something that is most definately not observed. Thus this is loose proof that electrons are not point particles - their mass and charge are distributed in space.

Claude.
 
  • #5
Okay, I understand pretty much what is really happening now. But I still think the Bohr Model is kind of impossible....just because electrons give of EMR....and they give off A LOT of energy. This part I still don't really understand. I'm still a high school student, so maybe I'll learn more about this when I hit my post-secondary studies. And I'm also a girl....so I feel kind of weird about this.... :confused:
 
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  • #6
dextercioby
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GodsChild086 said:
But I still think the Bohr Model is kind of impossible....just because electrons give of EMR....and they give off A LOT of energy. This part I still don't really understand. I'm still a high school student, so maybe I'll learn more about this when I hit my post-secondary studies. And I'm also a girl....so I feel kind of weird about this.... :confused:
I honesly think you're confusing Bohr model (and its elliptical orbits extension by Sommerfeld) with the one due to Sir Ernest Rutherford.The so-called "planetary model" That one did fail.Bohr model was essentially correct yielding Balmer formula,but it was incomplete and the dynamics it contained was essentially classical.It remained valid untill the 1926's model of Schroedinger overturned it in the light of the QM.
There ain't no problem to being a girl wanting to learn physics... :rolleyes:
 
  • #7
Oops...yeah I just realized that I did get it mixed up with the Rutherford "planetary model". I had a lack of sleep last night....up late doing Chem (The energy unit is confusing). I'm more awake now that I just read about Rutherford's "planetary model".

*Yawns* and is dying to go back to bed, but she's at school right now having a spare....
 
  • #8
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Actually, Bohr's model also assumes that the electron orbits the nucleus. However, Bohr theorized that at certain discrete levels, the electron would not radiate energy, and therefore would be stable.
 
  • #9
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loandbehold said:
Yes, that's exactly what should happen in that model. The reason why it doesn't is because an electron in an atom isn't really a point particle but is better described as a "spread out" wavefunction, with an energy that has to be one of certain discrete values ("energy levels").

Are you going to take a course on quantum mechanics any time soon?
I am sorry but i don't think that you really know what you are talking about.

The orbits have their specific form because of effective potential energy (at least in the most easy model). I know this is very vague so let me elaborate by taking the simplest configuration possible : The H-atom : 1 positive nucleus and one orbiting electron. Their potential is Coulombic in nature and if you were to plot the effective potential of the Schrödinger-equation as a function of the inter-particle-distance r, you will get a curve that exhibits a minimum at a certain distance between the two particles. This distance is the socalled Bohr-radius for a Hydrogen-atom in the GROUNDSTATE. So when an electron is at this specific distance from a nucleus with one proton, the entire system will be in a state of minimal potential so it is most stable. This is the reason why no collapse takes place. All these calculations are made via time-independent perturbationtheory where the perturbed Hamiltonian expresses the two-particle-interaction...


regards
marlon
 
  • #10
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GodsChild086 said:
I know that in the Bohr Model we have nucleus in the middle (consisting of protons and neutrons) and if the electrons are always accelerating around the nucleus, they will give off EMR. Because they give off so much energy, wouldn't they just "crash" into the nucleus?
Well keep in mind that other interactions also take place in an atom like the spin orbit-coupling and so on...Please read my previous post as to why the electron does NOT crash onto the nucleus...

regards
marlon
 
  • #11
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Moose352 said:
Actually, Bohr's model also assumes that the electron orbits the nucleus. However, Bohr theorized that at certain discrete levels, the electron would not radiate energy, and therefore would be stable.
What ???

Look emission and absorption of photons by an atom occurs when an electron is knocked out of its "position on a shell", leaving an opening which can be filled by another electron taking its place. This corresponds to an excitation-deexcitation process where energy is absorbed in order to knock the electron out and energy is emitted because of the de-excitation. Only certain amounts of energy will be absorbed or emitted and these amounts will correspond to the difference in energy-levels of the particles (electrons in this example) that jump from one level to another in the atom...

Also keep in mind that these energy-levels of an electron only originate from the interaction of this electron with an atomic nucleus in some way...

Correction : i said that energy is absorbed in order to knock an electron out...while this is correct you need to picture it correctly here : for example you can radiate an atom with EM-waves (X-rays,lasers,...) in order to trigger to excitation. So the energy i was talking about comes from an external source like EM-radiation. Also temperature can achieve this...


regards
marlon
 
  • #12
marlon said:
I am sorry but i don't think that you really know what you are talking about.

The orbits have their specific form because of effective potential energy (at least in the most easy model). I know this is very vague so let me elaborate by taking the simplest configuration possible : The H-atom : 1 positive nucleus and one orbiting electron. Their potential is Coulombic in nature and if you were to plot the effective potential of the Schrödinger-equation as a function of the inter-particle-distance r, you will get a curve that exhibits a minimum at a certain distance between the two particles. This distance is the socalled Bohr-radius for a Hydrogen-atom in the GROUNDSTATE. So when an electron is at this specific distance from a nucleus with one proton, the entire system will be in a state of minimal potential so it is most stable. This is the reason why no collapse takes place. All these calculations are made via time-independent perturbationtheory where the perturbed Hamiltonian expresses the two-particle-interaction...
Probably not, but I don't understand what your objection is either.

For a hydrogen atom the way the problem is usually presented is to solve the Schrodinger equation for a Coulomb potential [tex]V(r) \propto -1/r[/tex]. Doing this gives rise to a set of eigenstates and eigenenergies, the lowest of which corresponds to the 1s orbital. In my ignorance, I'd blithely assumed that this was the ground state, and that the reason why this state is stable is due to the fact that there is no lower energy state for it to decay to. Are you saying that this is incorrect?

Of course this is not the whole story since you also have corrections to the Hamiltonian due to things like relativity and spin-spin interactions between the electron and proton. But my understanding was that these will only perturb the eigenstates and eigenvalues slightly, and that you would still have a ground state even if you don't include these corrections.

I don't understand what you mean by the "effective potential of the Schrodinger equation" in this problem. Could you elaborate on this please?
 
  • #13
i thought the only thing which the bohr model failed to do was to explain the line spectra of other atoms than hydrogen. The reason why the electrons don't crash into the nucleus is very simple : uncertainty principle.
 
  • #14
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loandbehold said:
and that the reason why this state is stable is due to the fact that there is no lower energy state for it to decay to. Are you saying that this is incorrect?
No this is wrong,...the ground state is not stable because there is no lower energy possible. The system is stable because of the minimum in potential energy...


And to mysoginfeminists : NO THE UNCERTAINTY IS NOT THE PRIMARY FACTOR HERE...just ask yourself the question how a Bohr-radius is "constructed". What is it? and how does QM gives us this value???

regards
marlon
 
  • #15
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Marlon, can you explain what you mean by "minimum in potential energy"?
 
  • #16
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Here this is an easy course on quantum physics I taught to one of my friends via E-mail. The bohr model is old, and outdated, since then, a new model was made.

Electrons are attracted to protons, but repell electrons. So, instead of all the electrons being bunched up right next to the nucleas, they orbit around the nucleas in shells. These shells can sometimes contain sub-shells. For example, the first shell contains only one sub-shell. As an electron gets further away from it's atom, it must have more "quantum energy." Electrons want to get as close to the nucleas as possible, but according to quantum physics, no to electrons can have the same "quantum energy." So, they orbit in shells. The electrons orbit in orbitals. The sub-shells have orbitals. For example, the 1 shell has an S orbital. Because it's an s orbital and it's the first shell it's labelled 1S. For 1-First shell-, S-S orbital. An S orbital has the shape of a sphere. An orbital wants to fill it's self. Alright, so why would the atom want to have 8 electrons in it's outer most shell, good question. The second shell has two sub-shells. One sub-shell has an S orbital, and the second has three P orbitals. The reason it has three is because they can arrange themselves according to X,Y,Z. Each orbital has only two electrons, because no two electrons can have the same "quantum energy." So, for the valence shell of an atom with two shells, one S orbital and three P orbitals. Two electrons an orbital adds to...8. Hydogen, on the other hand, only has one shell. So, to fill it's valence shell, it only needs two electrons. It already has one - Hydogen = one proton, one electron - so, it only needs to bond with one atom to fill itself. Carbon, on the other hand, has two shells, so it needs 8 to fill it's valence shell. So....

H
H C H Methane!!! CH4.
H

If you were to count it up everyone's filled. The carbon atom has 6 electrons. 2 in it's first shell, and 4 in it's valence shell. It needs 8 in it's valence shell. So, it shares one with hydrogen, and the hydrogen shares one of the carbons. This gives the carbon an extra electron, and the hydrogen it's desired two. The carbon, then, bonds with three more to add to 8.

HOH Water!!! H20. Oxygen has six valence electrons, meaning it needs 2 to gain, which it does with 2 hydrogen molecules.

O=O Oxygen!!! O2.

You're probably wondering, why is there an equals sign between the Oxygen molecules?
This indicated a double bond. Oxygen has six valence electrons, when it bonds with another oxygen, it gets 7. That's not the desired 8. So, it makes a double bond, and they share two electrons each. Which adds to 8.

O
O O Ozone!!! O3. Each one of these atoms share with each other, making 8.

That's covelant bonding!!!
This "quantum energy I told you about is somewhat true. What's really true is that there are four "quantum numbers" that cannot match.
The first is N.
N is the energy of an electron. For example, an electron in the first shell would have an N of 1. An electron in the second shell would have an N of 2. An electron in the third shell would have an N of 3.
N=1, means it's in the first shell.
The second is L. It's actually a greek cursive L kind of like this. l. Okay. This sign is the orbital. L = N - 1. That's the equasion. So, if N = 1, then, L = 0. 0 is an S orbital.
If N = 2, L can equal either 0 or 1. If it is 1, that's a P orbital. If N = 3, then that can be either 0,1 or 2. An S,P or...a D orbital.
Now, the third quantum number is M. It is the orientation of the orbitals, you know XYZ.
M can equal anything between -L and +L. For example if L is 1, then M can equal -1,0,1.
This is 3 different ways of arranging the P orbital.
Now the final one is Ms. For Spin. The spin of the electron can equal - 1/2 or 1/2.

Okay, so let's look at the possible arrangements of some electrons.

N L M Ms
1 0 0 -1/2
1 0 0 1/2 First shell, only can have two electrons.

2 0 0 -1/2
2 0 0 1/2
2 1 -1 -1/2
2 1 -1 1/2
2 1 0 -1/2
2 1 0 1/2
2 1 1 -1/2
2 1 1 1/2 Second shell, eight electrons, but none of them, nor the one's in the first shell have the same 4 quantum numbers.

HOPE YOU UNDERSTAND. IT TOOK ME A WHILE TO WRITE, I'D HATE TO LOSE IT AT THE LAST MOMENT, LIKE THE POWER SHUT DOWN OR SOMETHING. IF YOU UNDERSTAND THIS, YOU WILL UNDERSTAND THE REST.
HERE'S SOME SITES.

http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch6/quantum.html

http://lectureonline.cl.msu.edu/~mmp/period/electron.htm I want a particualar question from this site from you.

P.S. I'm a Christian!!!
 
  • #17
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Moose352 said:
Marlon, can you explain what you mean by "minimum in potential energy"?

Well you know that the harmonic oscillator has a parabolic potential. The minimum in potential is at the bottom of this curve, That's all...

In this case, if you look at certain potential values while varying the distance between the two particles you will see the following evolution. Let's start by bringing them together from a great distance...So in the beginning the inter-particle distance (=r) is very great.

1) The potential lowers as r lowers.
2) When r = Bohr-radius, V is minimal (ie : lowest possible value)
3) if r is smaller then the Bohr-radius the V will increase rapidely...

Don't mind too much about the concept of this effective potential. It has the same meaning as the real potential involved in the interaction, but it is used in order to make calculations easier. Basically the effective potential equals the ordinary potential + some extra terms that describe other interactions...


regards
marlon
 
  • #18
marlon said:
No this is wrong,...the ground state is not stable because there is no lower energy possible. The system is stable because of the minimum in potential energy...
I really don't understand what you mean. For the electron to decay and lose energy, there must be a lower energy state for it to go to. So, if the system is in it's ground state, then by definition it is in its lowest energy state, and therefore it can't decay. What am I missing here?

And could you write down the form of your effective potential, please? You say that it has a minimum, so there must be repulsive terms at small distances that counteract the Coulomb potential. What is the physical origin of these terms, exactly?
 
  • #19
marlon said:
Well you know that the harmonic oscillator has a parabolic potential. The minimum in potential is at the bottom of this curve, That's all...

In this case, if you look at certain potential values while varying the distance between the two particles you will see the following evolution. Let's start by bringing them together from a great distance...So in the beginning the inter-particle distance (=r) is very great.

1) The potential lowers as r lowers.
2) When r = Bohr-radius, V is minimal (ie : lowest possible value)
3) if r is smaller then the Bohr-radius the V will increase rapidely...

Don't mind too much about the concept of this effective potential. It has the same meaning as the real potential involved in the interaction, but it is used in order to make calculations easier. Basically the effective potential equals the ordinary potential + some extra terms that describe other interactions...
Huh?!!...I thought that the Bohr radius just drops out when you solve the Schrodinger equation with the Coulomb potential. Why do you need "other interactions" in there as well? Just write down the usual time independent Schrodinger equation for the hydrogen atom:

[tex]-\frac{\hbar^2}{2\mu} \nabla^2 \psi - \frac{ke^2}{r} \psi = E \psi[/tex]

where [tex]\mu[/tex] is the reduced mass, and [tex]k=1/(4\pi \epsilon_0)[/tex]. Now I'll solve for the ground state wavefunction. This is going to be spherically symmetric (i.e. I don't have to worry about anything other than the radial part of the wavefunction), and I'm going to assume that it has a exponential form. So, I can write it as [tex]\psi(r)={\rm e}^{-r/R}[/tex] (ignoring the normalization), where R is a variable. Substituting this into the Schrodinger equation gives:

[tex]\left( \frac{2E}{k e^2 a_0}+\frac{1}{R^2} \right) + \frac{2}{r} \left (\frac{1}{R}-\frac{1}{a_0} \right)=0[/tex]

where [tex]a_0=\hbar^2/k\mu e^2[/tex] is the Bohr radius. This equation is solved by taking [tex]R=a_0[/tex] and [tex]E=-k e^2/(2a_0)[/tex]. So the ground state wavefunction is proportional to [tex]e^{-r/a_0}[/tex] while the energy is [tex]-k^2 e^4 \mu/(2\hbar^2)[/tex].

What's the problem with that?
 
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  • #20
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loandbehold said:
I really don't understand what you mean. For the electron to decay and lose energy, there must be a lower energy state for it to go to. So, if the system is in it's ground state, then by definition it is in its lowest energy state, and therefore it can't decay. What am I missing here?
Yes this is true but that was not what i meant. I mean that higher-energy levels are also stable when they are in a potential-minimum that corresponds to this higher-energy-level...

Again all you say is very true...But electrons at higher energy-levels will remain at their level when we do nothing with the atom. So they will not decay...they will stay in that specific state...


And could you write down the form of your effective potential, please? You say that it has a minimum, so there must be repulsive terms at small distances that counteract the Coulomb potential. What is the physical origin of these terms, exactly?
Indeed, here's the expression : [tex]V_{eff}(r) = \frac{-Ze^2}{4{\pi}{\epsilon_0}r} + \frac{l(l+1)\hbar^2}{2{\mu}r^2}[/tex]

mu is the reduced mass.
the term with the l-quantumnumbers is the repulsive centrifugal barrier term.

Keep in mind that this is the radial solution to the Schrödinger equation and the l-quantumnumbers come from the parity-symmetry-properties of the spherical harmonics, which contain the angular degrees of freedom of the system's wavefunction. These spherical harmonics give rise to this extra repulsive term.

Now, the effect of this centrifugal barrier is to reduce the depth of the potential well. This effect will also increase as l increases. For l > 2 no bound states (like an electron orbiting some proton) will be found because the barrier-term becomes more and more repulsive... Keep in mind that this summary is valid for an attractive interaction between a proton and an electron. For l = 1 you will clearly see that V_eff exhibits a minimal value when r has a certain value. This corresponds to the solution we are looking for....

regards
marlon
 
  • #21
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loandbehold said:
What's the problem with that?
You are forgetting the influence of the angular parts of the system's wavefunction. You are not using spherical harmonics and what about the centrifugal barrier term in the potential generated by L² acting on the spherical harmonics????Where is that ????

Basically your picture is not complete enough. Indeed when a potential is central in nature (this means that the potential only depends on the magnitude r of the position-vector )you will only need to solve for r-dependent solutions yet you cannot rule out the influence of the angular degrees of freedom like i stated before...


regards
marlon
 
  • #22
marlon said:
You are forgetting the influence of the angular parts of the system's wavefunction. You are not using spherical harmonics and what about the centrifugal barrier term in the potential generated by L² acting on the spherical harmonics????Where is that ????

Basically your picture is not complete enough. Indeed when a potential is central in nature (this means that the potential only depends on the magnitude r of the position-vector )you will only need to solve for r-dependent solutions yet you cannot rule out the influence of the angular degrees of freedom like i stated before...
I wasn't "forgetting". I neglected the angular parts of the wavefunction specifically because, as I stated in my post, I was solving for the ground state, where the wavefunction is spherically symmetric. So the centrifugal barrier term isn't there. This can be seen by setting l=0 in the effective potential you wrote down earlier. I know that to be more general you need to also include the angular degrees of freedom, but when l=0, as is the case for all of the s-orbitals including the ground state, then you can simply solve the radial Schrodinger equation with a Coulomb potential.

And that is my point. You seem to think that it is important that you have a minimum in the potential at some r in order to understand why the electron is stable. But for the ground state you don't have such a minimum- you just have the normal 1/r Coulomb potential. So how does that fit in with your argument?
 
  • #23
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loandbehold said:
I wasn't "forgetting". I neglected the angular parts of the wavefunction specifically because, as I stated in my post, I was solving for the ground state, where the wavefunction is spherically symmetric. So the centrifugal barrier term isn't there. This can be seen by setting l=0 in the effective potential you wrote down earlier. I know that to be more general you need to also include the angular degrees of freedom, but when l=0, as is the case for all of the s-orbitals including the ground state, then you can simply solve the radial Schrodinger equation with a Coulomb potential.

And that is my point. You seem to think that it is important that you have a minimum in the potential at some r in order to understand why the electron is stable. But for the ground state you don't have such a minimum- you just have the normal 1/r Coulomb potential. So how does that fit in with your argument?

You are right on the l=0-thing here. But if l = 0 you will see that the potential will evolve towards negative infinity when r = 1 (r is given in units of Bohr radius). Basically this means that when r is equal to the Bohr-radius, the potential evolves towards minus infinity...What better absolute minimum can you possibly come up with??? See my point ???

regards
marlon
 
  • #24
marlon said:
You are right on the l=0-thing here. But if l = 0 you will see that the potential will evolve towards negative infinity when r = 1 (r is given in units of Bohr radius). Basically this means that when r is equal to the Bohr-radius, the potential evolves towards minus infinity...What better absolute minimum can you possibly come up with??? See my point ???
Not really. Surely if l=0 then you just have the -1/r term, which diverges to minus infinity at r=0, not at r=1.
 
  • #25
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loandbehold said:
Not really. Surely if l=0 then you just have the -1/r term, which diverges to minus infinity at r=0, not at r=1.

Well, yes, ofcourse :rolleyes: , but you still don't get the point do you???

The strong decrease in potential energy means what do you think ??? There is no solution for r = 0 because it is not physical so not worthy of mentioning here...

regards
marlon
 

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