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What bugs my is that Bohr derives the energy from very few assumptions and sets up the solution through a natural force balance. Why is it that a faulty model can deduce the energy levels?
Bohr's model looks weird in the context of proper quantum mechanics, but it's a lot more solid than we usually give it credit for. A lot of features of it must work, ...
10
Bohr's model actually is part of the "old quantum theory" prior to the advent of wave and matrix mechanics that is based on the concept of classical mechanics that the motion of particles (electrons) can be described in phase space with general coordinates $(q_i,p_i)$. Under this assumption orbits, in particular periodical orbits exist. The essential ...
8
Bohr's guess was that orbital angular momentum is quantized in multiples of the quantum of action, and that is correct.
Bohr assumed circular orbits and that is wrong. But one feature of the hydrogen atom is that the $1/r^2$ interaction leaves all states with the same principal quantum number degenerate: "circular" orbitals where $\ell = n-1$ have the same ...
7
When a photon with the appropriate energy hits an atom, the electron will make a transition from the ground state to a excited state. This will make the potential energy of the atom higher.
This is mostly correct, but it's not the potential energy of the atom which is higher, it's its internal energy
Also, momentum is conserved,
this is correct, but
...
6
Two-photon emission does exist, or else the 2s state of hydrogen would be stable. You can get a pretty decent estimate for this kind of rate without any fancy math or physics, just using the energy-time uncertainty principle. The typical rate of emission for a photon, when not forbidden by parity, is $R \sim 10^9\ \text{s}^{-1}$. We can think of the two-...
4
A reasonable answer is given in
A.R.P. Rau, "The negative ion of hydrogen". J. Astroph. Astron. 17, 113 (1996)
where Rau explains as follows:
Of particular interest among the $Ν = 2$ states is the lowest one of $^3 P^e$ symmetry, described in independent-electron terms as $2p^2$. This is bound below the $\mathrm H(N = 2)$ threshold with about $9.6 \:\rm ...
4
The Bohr model is far from exact, and the electrons don't strictly have 'radius of orbit' as they have a distribution function across space, but basically your intuition is not worng: as $n\to \infty$, the available angular momentum levels also grow, since $L<n$ (so for $n=15$, for example, the quantum angular momentum number can grow up to $14$). As $L$ ...
4
I am a bit confused because if that's the case, the atom will gain kinetic energy.
In the way you've set it up (atom is initially at rest in our reference frame), then that is correct.
Doesn’t that violate the conservation of energy?
That depends on the energy inputs and outputs. We haven't described them completely yet.
Since the energy is already ...
2
When the electron returns back to it’s ground state, a photon is emitted in a random direction
Not exactly. If it's a spontaneous emission - then YES. If it's a stimulated emission - then NO, photon is emitted in the same direction of incident photon which has hit the atom. LASER or MASER works according to this principle. Btw, stimulated emission was ...
2
I am studying Beiser's Modern Physics and according to that in the Bohr Model the wave the probability density of ...
This is a misunderstanding of the text. There are no probabilities, no waves, and no orbitals in the Bohr model. Your question makes perfect sense, but it is inscribed within full-grown QM ─ it has nothing to do with the Bohr model.
the ...
2
Hydrogen is an obvious counter-example.
Indeed it is. The claim as stated is false.
So I am trying to work out what he could have meant?
You'll have to ask him. There's no way for us to read his mind.
2
In the Bohr model, the radius of the $n^\text{th}$ orbit is
$$r_n=n^2a_0,$$
where $a_0$ is the Bohr radius.
Using hydrogenic wavefunctions from the Schrodinger equation, the expectation value of the radius in the orbital with quantum numbers $n,l,m$ is
$$\langle r\rangle=\frac{3n^2-l(l+1)}{2}a_0.$$
I didn’t look up what it would be using wavefunctions ...
2
There's not really any such thing as heating an individual atom. When you heat a gas like in a candle flame, the heat is the random motion of all the different atoms. The randomness is what makes us call it heat. If there's only one atom, there's no way to say if its motion is random or not.
And they reflect other forms of light that don't supply energy ...
1
Don't forget the explicit form of the eigenfunctions of angular momentum:
$$Y_{l, m}(\theta, \phi) = \Theta(\theta)\textrm{e}^{im\phi}$$
which ensures that the wavefunctions are single valued under a full rotation $\phi \rightarrow \phi + 2\pi$. Here $\Theta(\theta)$ is the non-trivial $\theta$-dependence. Can you see how your second equation arises if you ...
1
Radium that was used for luminous indicators was combined with a phosphor. The radiation excited the phosphor and that was what glowed (usually green).
If you have enough radium in the dark, it can ionize the nitrogen in the air. As the nitrogen recombines with electrons, various emissions are possible, but in the visible band, blue is strong. You'll see ...
1
Upon reading the question carefully, I believe that the problems of the OP have nothing to do with the quantum nature of the interaction, but simply with the understanding of how modes work. To see this, let us simply write the interaction term in a different form which is in fact also mentioned in the question. Putting in the relevant functional dependences
...
1
”… Having such a good prediction one would expect that there exists an extension or modification to it.”
QED is an extension or modification to Bohr’s model and there exists an extension or modification to it. The sequence of modifications (extensions) is as follow:
QED -> Schrodinger’s model -> De Broglie’s model -> Bohr’s model.
I believe it is “obvious”...
1
Bohr said that an electron would revolve in certain stationary orbits and he gave its mathematical interpretation.
While in quantum mechanics we deal with probability of finding an electron around the nucleus so we can't derive the Bohr's postulates as Bohr said that "electrons revolve" in circular orbits while QED says "electrons may be found" in region ...
1
Another form of two photon emission is two photon excited fluorescence.
Two photon absorption is the absorption of two photons by a molecule to excite from ground state to excited state.
Two-photon absorption can lead to two-photon-excited fluorescence where the excited state produced by TPA decays by spontaneous emission to a lower energy state.
First, ...
1
Then you learn, that the Hydrogen negative ion with an extra electron is stable. Ok, so you have to accept that it is not so simple, and it is all QM.
You don't have to go to QM to have a stable system with two electrons and one proton. If you want to think within the classical electrostatic planetary model of the atom (ignoring for now that it doesn't work ...
1
A lot of work has been done since the 1984 review by Moriya and Takahashi, both theory and experiments.
On the theoretical side, calculations have been refined a lot. The effect of the on-site $dd$ correlation energy $U$ was taken into account first by LSDA+U, then by LSD+DMFT (a dynamic mean field theory), for example this paper from 2001 about high ...
1
Given the index $i$, first calculate $n$ using
$${n=}\left\lfloor\frac{1}{2}+\frac{(-108+108i+\sqrt{3}\sqrt{3887-7776i+3888i^2})^{1/3}}{2\times 3^{2/3}}+\frac{1}{2\times 3^{1/3}(-108+108i+\sqrt{3}\sqrt{3887-7776i+3888i^2})^{1/3}}\right\rfloor.$$
Then calculate $l$ using
$$l=\left\lfloor\sqrt{i-\frac{2n^3-3n^2+n}{6}-1}\right\rfloor.$$
Finally, calculate $...
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