@JohnRennie: Hi sir. Good morning :-)

@GuruVishnu hi :-)

It seems you woke up 30 minutes earlier than me.

It's 5:15 a.m. there, right?

The clocks have just gone forward i.e. the UK has switched to British Summer Time, so I'm up an hour earlier than usual.

I see.

Are you free now, sir?

Yes, though I need a couple of minutes to make the first coffee of the day. This is essential if you want coherent answers from me :-)

Fine sir :-)

OK, coffee is made. What do you want to ask?

Ok sir.

Is it possible to produce $L_\alpha$ X-rays without producing $K_\alpha$ X-rays? I think yes, if we could create a vacancy in the $L$ shell without creating any in the $K$ shell. However, I don't know whether it's possible?

I'll have to remind myself what the mechanism for the L line is. Give me a moment ...

May I explain sir?

@JohnRennie The characteristic X-ray lines are due to electrons jumping from a higher energy level to the lower one to fill the vacancy.

Yes, I remember that much :-)

So the L line is when an electron is ejected from the 2s orbital and an electron drops into it from a higher orbital.

Fine :-) We call the lines as $\alpha, \beta, \gamma$ and so on depending upon whether it's the next higher or the second next higher orbital.

@JohnRennie Yes sir. For example, $L_\alpha$ means a transition from $n=3$ to $n=2$.

$L_\beta$ means a transition from $n=4$ to $n=2$.

I think it would be hard to find a way to eject L electrons without also ejecting K electrons.

I guess if you tuned the energy of the incident electron beam really carefully you might do it.

Ok sir. Thank you for the information. However, I don't know how to tune it properly without removing a K electron.

It always has energy less than that of a L electron. And so anything which could excite L can also excite a K electron.

It takes more energy to eject a K electron that to eject an L electron.

So in principle if you hit the material with an electron beam of an energy just less than the K edge energy you would only see an L line.

Ok sir. In my previous message, I forgot that the excited orbitals are inverted and the reference energy level is shifted.

Now I can understand this. Thank you sir.

@GuruVishnu in practice if you did this the X-ray intensity would be very low because we need electrons much more energetic than the line energy to get a decent probability of ionisation.

I didn't know that. I thought if the intensity is low, we need to increase the current in the filament circuit so that more electrons can be emitted due to thermionic emission. I thought energy is unrelated to intensity.

Well yes, intensity is proportional to the number of electrons hitting the target, but the ionisation probability is also strongly dependent on the electron energy.

Ok sir.

So if you keep the beam current the same but increase the energy e.g. from below the K energy to well above it you'll see the K intensity rise smoothly as the energy is increased.

The intensity doesn't switch from zero just below the edge energy to maximum intensity just above it.

Fine sir. It's intuitive and is similar to photoelectric effect. It seems, X-ray emission is an inverse of photoelectric effect. Is this idea correct?

Photoelectric effect: Light $\to$ electrons

X-ray emission: Electron $\to$ light

*light=electromagnetic radiation

Not really.

X-ray emission is a two step process while photoelectron emission is a one step process

@JohnRennie Yes sir. I'm able to see the difference. Maybe we could say it's only applicable for continuous X-rays and not for characteristic.

The continuous background is due to bremsstrahlung radiation. Whenever you accelerate or decelerate an electron it emits EM waves.

I haven't studied about bremsstrahlung (what a difficult spelling :-) ) radiation. However, now I can understand why EM waves are produced in a continuous fashion. Just now realised it's due to acceleration of electrons. Earlier, I considered the nature to be beyond my current scope - something related to advanced QM.

Once you're free could you reply to the following, sir?

> In Moseley's law for characteristic X-rays $\sqrt\nu=a(Z-b)$, $a$ and $b$ are independent of the material.

I'm unable to see how $a$ and $b$ are independent of the material. I think both of these two constants depend upon the initial and the final orbitals.

Could you explain the above statement, sir?

Modeley's law gives the energy of the K alpha line, so the start and end orbitals are always the same i.e. it's always the $2p \to 1s$ transition.

Is that only for K alpha line? I thought it's applicable to all kinds of transitions.

The K alpha line occurs when the incident electron ejects electrons from the 1s rbtal and the 2p electrons falls down into the empty 1s.

So the K alpha line is always the energy difference between the 1s and 2p orbitals.

Yes sir. I know that. But isn't Moseley's law applicable for all transitions?

No

I see. I haven't learnt about that yet.

Ah, wait.

There is a Moseley type relationship for each line, but the constants are different for each line.

Ok sir. We can continue after your discussion on CodeClub. No problem.

So for the K alpha $A$ and $b$ will have some values, for the L alpha $A$ and $b$ will have different values, and so on.

Ok sir. So finally $a$ and $b$ are independent of material but dependant on the initial and final states.

Yes

But note this is an approximate law. It approximates the energies of the lines. It isn't an exact equation.

Oh. Is it something like the Rydberg's empirical formula?

Yes.

Ok sir. Thanks for the new info. I thought it was as accurate as $F=ma$

It seems things get fuzzier for all formulas of particle physics.

The energy levels in multielectron atoms are complicated and cannot be described by any simple equation. But there are approximate equations like Rydberg's and Modeley's equations that give reasonable results for some special cases.

If you go on to do physics at university you'll learn the gory details :-)

:-)

I concluded the answer is (d)$>I_0$ and the answer is correct.

May I know whether I should know anything more about X-ray transmission through aluminium and metals in general?

For example, will the transmitted intensity increase forever with increase in incident intensity? Or will it saturate beyond a particular value?

Then, will there be any frequencies in the X-ray spectrum which would be shielded like a crossed polaroid?

Increasing the temperature of the filament increases the number f electrons so it increases the beam current, but it doesn't change the energy of the electrons.

It increases the X-ray intensity simply because more electrons means more atoms in the target get ionised.

@JohnRennie Yes sir. I understand these points. I must have been a bit specific. My question isn't about the filament. But effect on the intensity of the transmitted beam due to a foil placed in the path of X-ray beam.

Only a tiny fraction of the atoms in the target get ionised so the number of atoms ionised is proportional to the beam current. You would melt the target long before you approached any saturation point.

I think we need not worry about the working of the Coolidge tube for the moment. Just assume we can play with any values of intensity of X-rays emitted by it.

I need to work for a few minutes ...

Ok sir.

@GuruVishnu that was a long few minutes! Anyhow I'm around for a few hours now.

@JohnRennie Hi sir. Is this a consequence of time-dilation? :-)

@GuruVishnu :-)

desmos.com/calculator/i9badao9q6

Please ignore these sir. I'll ask my doubt afterwards.

OK :-)