Problem Solving Strategies

@JohnRennie when the +ve ones move towards the electrons...both of them gain KE...and finally when they combine with the ions, that KE changes to photons form e's...

John Rennie

It might be simpler to consider a single atom. The recombination of the ion and electron is the reverse of ionisation so the atom has to emit a photon with the energy equal to the ionisation energy to carry away the excess energy.

In a real experiment many of the atoms lose the excess energy by collisions, that is the excited atom collides with some other atom in the container and the energy gets converted into kinetic energy rather than being emitted as a photon.

In fact usually most of the energy gets dissipated this way.

The increased kinetic energy of the colliding atoms is of course just heat, so the material gets hotter.

This is basically how any chemical reaction makes things hot.

user8718165

@JohnRennie I was thinking that photons of infrared region are emitted...so how was heat generated?only light...

@JohnRennie okay sir

John Rennie

I doubt ionisation energies would be in the IR region. They normally have energies in the UV.

user8718165

@JohnRennie okay sir...but hot materials emit infrared radiation...isn't it sir? sorry...if wrong :(

John Rennie

If all the energy is carried away as photons then the material could only get hotter if those photons are reabsorbed and heat the material that way. But in practice only a small fraction of the energy is carried away as photons.

Hot liquids and solids emit radiation by black body radiation. As a general rule hot gases don't emit black body radiation but they may emit photons from rotational transitions. Possibly also vibrational transitions if they are hot enough.

user8718165

9:40 AM

@JohnRennie okay sir...got it. One qn sir, generally in everyday things, temp. doesn't go beyond say 500C so there will be almost no UV radiation...but IR will be there sir

John Rennie

The intensity of the radiation emitted will be described by the black body spectrum.

user8718165

is it correct sir?

@JohnRennie yeah sir...

John Rennie

In principle the spectrum has a tail extending to arbitrarily low wavelengths, but in practice the intensity falls to rapidly with increasing energy that it quickly becomes undetectable.

I'd have to do a calculation to find out what fraction of the radiation is emitted as UV at 500°C, or you could do it.

The equations you need are given here:

Planck's law

Planck's law describes the spectral density of electromagnetic radiation emitted by a black body in thermal equilibrium at a given temperature T, when there is no net flow of matter or energy between the body and its environment.At the end of the 19th century, physicists were unable to explain why the observed spectrum of black body radiation, which by then had been accurately measured, diverged significantly at higher frequencies from that predicted by existing theories. In 1900, Max Planck heuristically derived a formula for the observed spectrum by assuming that a hypothetical electrically charged...

user8718165

:52519934 okay sir...got it...thank you very much...I have a few more qns...

John Rennie

Yes?

user8718165

9:55 AM

@JohnRennie sir can you please tell me a bit about it? I didn't properly get it sir

John Rennie

You'll have to be more specific. What exactly are you asking about?

user8718165

> In a real experiment many of the atoms lose the excess energy by collisions

this one sir

John Rennie

Suppose you have two hydrogen atoms in a gas, and those two atoms collide. In general the collision will be elastic so the total KE before and after the collision is the same. OK so far?

user8718165

@JohnRennie okay sir...

John Rennie

But suppose you give the colliding atoms lots of kinetic energy. Specifically suppose you give them more than 10.2eV of total kinetic energy. This is enough energy to excite the 1s to 2p transition in a hydrogen atom.

So we can now get an inelastic collision.

user8718165

10:03 AM

@JohnRennie okay sir

John Rennie

That is when the atoms collide 10.2eV of the energy of the collision is used to excite one of the atoms from the 1s to 2p state. The total energy is conserved, so now the kinetic energy after the collision is 10.2 eV less than it was before the collision.

OK so far?

user8718165

@JohnRennie okay sir

John Rennie

But we can also get the reverse process. That is suppose we collide a hydrogen atom in the 2p state with another hydrogen atom in the ground state. In this case the excited atom can decay to the 1s state and the 10.2eV gets transferred into kinetic energy. So the kinetic energy after the collision is 10.2eV higher than it was before the collision.

user8718165

@JohnRennie okay sir...got it

John Rennie

We started by talking about ions and electrons. The ion and electron will combine to form a transient state with far too much energy to be stable.

That transient state could emit a photon and decay into the (stable) ground state.

But it can also collide with another atom and the energy could be converted into the kinetic energy of the two colliding atoms.

Just as I described in the case of colliding hydrogen atoms.

user8718165

10:11 AM

@JohnRennie sorry for waiting sir...my messages are getting timed out :(

John Rennie

And that increased kinetic energy is of course just heat. So the ions and electrons recombine and get the gas gets hotter.

user8718165

@JohnRennie okay sir...got it...is that thing a phonon sir?

John Rennie

We don't normally get phonons in gases.

user8718165

@JohnRennie okay sir...the process you described happens only in gasses...or also in solids and liquids?

John Rennie

@user8718165 You know how a diatomic molecule has distinct vibrational energy levels?

user8718165

10:16 AM

@JohnRennie equipartition theorem sir? Otherwise I don't

John Rennie

No, not the equipartition theorem.

user8718165

@JohnRennie sorry...I don't know

John Rennie

The energy levels of a quantum harmonic oscillator are quantised. The energy of the ground state is $\tfrac12 h\nu$, then the first excited state is $\tfrac32 h\nu$, the second $\tfrac52 h\nu$, and so on.

The energies of the states are separated by $h\nu$.

user8718165

@JohnRennie okay sir

John Rennie

That's why the oscillator absorbs and emits energy in photons of energy $h\nu$.

user8718165

10:19 AM

@JohnRennie okay sir

John Rennie

Now suppose you take a chunk of metal e.g. a tuning fork. It too has a characteristic vibrational frequency. Obviously for tuning forks they are designed to have a specific frequency so you can use them for tuning pianos.

user8718165

@JohnRennie okay :)

John Rennie

In principle these vibrational modes are also quantised and absorb and emit energy in quanta of $h\nu$, though in practice the energy of these quanta is far too small every to be detected experimentally.

Anyhow, we call these quanta phonons.

The phonons are the quanta of the macroscopic oscillations of the metal, or liquid, or whatever is oscillating.

But gases don't have these sorts of oscillations because the atoms are moving relative to each other rather than having approximately fixed positions as they do ina solid.

So that's what phonons are.

user8718165

@JohnRennie got it sir

@JohnRennie Got it now :)

@JohnRennie sir just 1 last qn...

John Rennie

Yes?

user8718165

10:31 AM

@JohnRennie sir, this one, how will the e give off its energy as the KE of the atoms? I mean there has to be something which converts that energy to KE...does the e lose energy just by the atoms colliding?

I'm sorry if you mentioned it before and I'm asking again

John Rennie

I don't think there's a simple answer to that.

Consider the reverse process i.e. two ground state atoms colliding and the energy going into an excited state.

And take a classical view of the process.

As the atoms collide the electrons in the atoms can come close together, and sometimes very close together. That means there is a strong force between the electrons, and that force can do work by pushing one of the electrons into a higher orbit.

The de-excitation is just the reverse of this.

This isn't really what happens because the electrons aren't point particles and the 1s, 2s, etc aren't orbits. The real calculation would be complicated. However the overall process ends up the same.

@user8718165 yes

user8718165

@JohnRennie Sir you learnt thiese things in Cambridge sir?

John Rennie

No, I didn't take the specialised quantum physics courses at university. I was more interested in quantum chemistry.

This is stuff I've learned from hanging around on the Physics Stack Exchange :-)

user8718165

@JohnRennie Sir...I thought these were written in some very advanced sort of book

John Rennie

@user8718165 I'm sure they are. It's just that I haven't read that book :-)

user8718165

10:43 AM

@JohnRennie sir could you please refer me some books like that? these things are not written in ordinary books... I'll go buy them after exams are finished :-)

John Rennie

I can't I'm afraid. I'm out of touch with science textbooks and don't know what books would be good.

user8718165

@JohnRennie okay sir...no worries...for how long will you be there? :)

John Rennie

I'm here for a couple hours more

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Problem Solving Strategies