Problem Solving Strategies

4:13 AM

@JohnRennie: Hi. It's given for a pure spectrum - "...light of one wavelength should occupy one particular spatial position in the spectrum."; Does it mean a pure spectrum can only be a line and not a band?

Something like the following:

So in reality, will it be highly difficult to see a pure spectrum just because it's so thin?

John Rennie

5:52 AM

No.

Guru Vishnu

@JohnRennie: Good morning sir :-)

John Rennie

Morning :-)

Suppose you choose a wavelength e.g. 456nm.

Guru Vishnu

@JohnRennie Why sir? If it must occupy a unique point in space, then I could think of only one point or a collection of points leading to a line.

John Rennie

If you have a spectrum spread out on a screen, then the question is where along the screen will the light of this wavelength fall.

e.g. if we have the spectrum spread out horizontally and we call the horizontal coordinate $x$, then at what value of $x$ will the 456nm light strike the screen?

@GuruVishnu You remember yesterday we were talking about how a slit of large width will result in a blurred spectrum?

Guru Vishnu

@JohnRennie It's something fixed value. I know it depends on wavelength sir. I had doubts on width of the entire spectrum. Did I explain what I meant, sir?

@JohnRennie Of course yes sir. Now I'm having doubt on the width or the top bottom length of the spectrum.

BTW it's not mandatory to @ username; I'm constantly keeping track of your messages. Sometimes it'll take some time for me to understand or to type a reply :-)

John Rennie

5:57 AM

The vertical height of the spectrum can be anything. Along a vertical line the light has the same wavelength so there is no restriction on the vertical height.

The vertical height of the spectrum on the screen is actually just the vertical height of the slit multiplied by the magnification of the lenses you've used.

Guru Vishnu

@JohnRennie Ok sir. Will that not defy the definition for pure spectrum - "..._light of one wavelength should occupy one particular spatial position in the spectrum._"? Because if we're free to choose any vertical height, then two different points in the same line will have the same colour.

John Rennie

It's poorly worded. It means on spatial position along the spectrum.

Guru Vishnu

@JohnRennie Ok sir. Then I got it. Thank you.

I have some more doubts and I'll ask them after you're done with doubts on Physics. room.

John Rennie

I think you can ask now. Alesha seems happy.

Guru Vishnu

@JohnRennie Ok sir. Thank you.

I was reading about Atomic emission / absorption spectrum from Wikipedia and I had a doubt in that. I'm trying to find the article where I found the doubt sir.

And I'll quote the doubt from that.

John Rennie

6:13 AM

OK

Guru Vishnu

Aha! got it sir.

Source: https://en.wikipedia.org/wiki/Spectral_line

"When a photon has about the right amount of energy to allow a change in the energy state of the system (in the case of an atom this is usually an electron changing orbitals), the photon is absorbed. Then it will be spontaneously re-emitted, either in the same frequency as the original or in a cascade, where the sum of the energies of the photons emitted will be equal to the energy of the one absorbed (assuming the system returns to its original state)"

John Rennie

OK

Guru Vishnu

It suggests a photon of higher energy gets transformed (splitted) into smaller energy photons. But from the following posts on SE, I learnt photon splitting is not possible:

24

Q: Can you split a photon?

I was wondering if a photon is divisible. If you look at a photon as a particle, then you may be able to split it (in theory). Is it possible and how do you split it?

8

Q: What happens when a photon hits a beamsplitter?

Yesterday I read that we can affect the path and the 'form' (particle or wave) of a photon after the fact (Wheeler's delayed choice experiment). Part of what is puzzling me is the beam-splitter. Are the individual photons actually being split into two new photons of lesser energy? This question...

quantum-mechanics photons

I read only answers by user "anna v" as they are easy to understand.

John Rennie

Suppose you have a hydrogen atom, and you fire a single photon at it with the energy equal to the 1s to 3p transition. Then the atom absorbs the photon and jumps to the 3p state. OK so far?

Guru Vishnu

@JohnRennie Yes sir. Shall we use this kind of naming energy levels as I'm quite used to n=1,2,3,4,5,...?

John Rennie

6:18 AM

OK from the n=1 to the n=3 state.

Guru Vishnu

@JohnRennie :-)

John Rennie

Then the atom is going to decay back into the n=1 state, but it can do this in two ways.

Guru Vishnu

@JohnRennie Yes sir. Either directly or through n=2

John Rennie

It can jump directly to n=1 and emit a single photon with the same energy as the original photon that was absorbed.

Or it can decay to the n=2 state and emit a lower energy photon, then the n=2 state decays to the n=1 state and emits a second photon.

Guru Vishnu

Yes sir. Got it. Now we can "jump" to the next point.

John Rennie

6:20 AM

Both emitted photons having an energy less than the original photon that got absorbed.

So one photon got absorbed and two photons got emitted.

Guru Vishnu

@JohnRennie Why sir? Isn't the photon emitted due to n=3 to n=1 same as initial photon energy?

Oh! You're now referring to the second case of an intermediate step (n=3 to 2 to 1)?

If yes, kindly neglect my first message.

John Rennie

I mean both photons emitted in the two step process have a lower energy than the original photon.

Guru Vishnu

@JohnRennie Yes sir. Now one photon in then two photons out. Do you agree with this sir?

John Rennie

Yes

And this is the process that Wikipedia is referring to.

Guru Vishnu

@JohnRennie So one photon splits into two photons. Do you agree sir?

John Rennie

6:23 AM

No

The original photon got absorbed and destroyed.

Two new photons were created.

Guru Vishnu

@JohnRennie Why not? One photon in and two photon out is same as splitting, right sir?

John Rennie

No

Guru Vishnu

@JohnRennie So sad :-)

@JohnRennie Now understood sir. So there is nothing like Law of conservation of photon number?

like we have for conservation of charge.

John Rennie

@GuruVishnu This turns out to be interesting

Guru Vishnu

@JohnRennie Is there any previous question on SE, sir?

@JohnRennie Thank you.

John Rennie

6:26 AM

The number of electrons is a conserved quantity. However its antiparticle, the positron, counts as -1 electrons.

Guru Vishnu

I found it quite difficult to search because this is a new subject for me and I don't have efficient methods of searching.

John Rennie

So suppose an electron and positron come together and annihilate, what is the change in the number of electrons?

Guru Vishnu

@JohnRennie What does annihilate mean? Combine?

John Rennie

@GuruVishnu yes. Do you know that matter and antimatter can combine and turn into energy?

No? Have you never seen a Star Trek movie?

Guru Vishnu

@JohnRennie Sir, I'm sorry. I've heard about antimatter only in some documentaries. I'm afraid I will not be able to understand anything beyond this. A small "yes" or "no" will do it for me for now.

John Rennie

6:29 AM

Annihilation

In particle physics, annihilation is the process that occurs when a subatomic particle collides with its respective antiparticle to produce other particles, such as an electron colliding with a positron to produce two photons. The total energy and momentum of the initial pair are conserved in the process and distributed among a set of other particles in the final state. Antiparticles have exactly opposite additive quantum numbers from particles, so the sums of all quantum numbers of such an original pair are zero. Hence, any set of particles may be produced whose total quantum numbers are also...

@GuruVishnu Yes

Guru Vishnu

@JohnRennie No sir. I have to as it's about space exploration (I guess).

I think I've seen the space ship enterprise from that movie. Am I right?

A big disk kind of space ship

psa

@JohnRennie hey, do you have time at some point for a quick question?

John Rennie

I suspect what I was going to say is a bit pointless if you're not already familiar with the idea of particle-antiparticle annihilation.

But the bottom line is that photons can be created and destroyed, and the number of photons is not conserved.

@psa hi :-)

Guru Vishnu

@JohnRennie Ok sir. No problem. I'll store it for a future discussion with you once I gather some knowledge on Quantum mechanics.

@JohnRennie That's great. Then why not get splitted?

psa

oops

John Rennie

6:32 AM

@GuruVishnu anyhow, the original photon isn't split because it gets destroyed and two new different photons get created.

psa

wrong question

Assuming that Otto von Guericke was able to evacuate just half of the air inside his two famous Magdeburg hemispheres, what would have been the force required to pull them apart? The hemispheres had a diameter of about 50cm.

John Rennie

The energy of the original photon gets split in two, with part going to each new photon, but the original photon gets destroyed not split.

psa

I was assuming it was just $F = PA$, but the fact that only half the air was evacuated is a bit confusing. Wouldn't the pressure vary as more air escaped? Is an integral needed here?

Guru Vishnu

@JohnRennie Ok sir. Thank you. I'll ask after @psa. I think it's a quick question :-)

John Rennie

@psa the force is the force needed for the initial separation i.e. before any air flows in or out of the gap between the spheres.

Guru Vishnu

6:35 AM

@psa: Is that a JEE related question. If yes I could give it a try.

psa

not really sure, it's from a 2nd year thermodynamics class but I think it's just meant to be review

John Rennie

@psa do you know how to do the calculation for the Magdeburg spheres?

psa

I don't, no

Guru Vishnu

@psa Then sorry for interrupting.

psa

@GuruVishnu no worries, it might be relevant, I'm just not sure because I've never taken the JEE

John Rennie

6:37 AM

It's done using the principle of virtual work. Well, it can be done in many ways, but virtual work is the easiest.

psa

I only vaguely remember virtual work from Feynman's explanation, and I didn't fully understand it. You'll have to explain that as well or as you go if we take that route

John Rennie

Suppose we pull the spheres apart by an infinitesimal distance $dx$, then the new volume created between the spheres is $dV = Adx$, where $A$ is just the area $A = \pi r^2$. OK so far?

@psa hello?

psa

Hi, sorry

Right, the area is just the area of the centremost cross section?

John Rennie

@psa Yes. It's the area inside the spheres at the mid plane where they separate.

psa

OK, yes

John Rennie

6:47 AM

So the work done is the usual $PdV$ i.e. $dW = (P_{ext} - P_{int})Adx$

psa

Right, I guess we can assume $P_{int} = 0$?

John Rennie

@psa yes, though in your question $P_{int} = 0.5 atm$, which is why I left it in the equation.

psa

ah, I see

John Rennie

Now, there's another way we can calculate the work. If the force pulling the spheres apart if $F$, then the work is just $dW = Fdx$.

psa

So $Fdx = (P_{ext} - P_{int})Adx$

John Rennie

6:50 AM

BOOM! :-)

psa

nice

so then $F = (P_{ext} - P_{int})A = (10135Pa - 50662.5Pa)\pi(0.5m)^2$?

or should I be integrating?

John Rennie

0.5m is the diameter not the radius.

psa

oops

John Rennie

But yes, it is as simple as that.

psa

that's sweet

so that's virtual work then?

i.e. moving a little dx

John Rennie

6:56 AM

Yes. The dx is the virtual displacement.

The technique works in lots of similar problems e.g. calculating the pressure inside a soap bubble.

psa

that's really nice to deal with

I actually use that all the time to set up integrals, I just didn't know that was the essence of virtual work

Nobody recognizeable

@JohnRennie hi.

John Rennie

@Nobodyrecognizeable hi :-)

Nobody recognizeable

@JohnRennie 18 why?

I know du= Tds-Pdv.

Does it signify anything?

John Rennie

7:12 AM

@Nobodyrecognizeable If the gases on the two sides can exchange energy then the hotter gas will heat the colder gas until they are the same temperature. Yes?

Nobody recognizeable

@JohnRennie yes.

John Rennie

So at equilibrium the temperatures are going to be the same.

Nobody recognizeable

@JohnRennie yes.

John Rennie

And if the pressures are different there will be a net force on the movable partition, so the partition will move until the pressures are the same.

Nobody recognizeable

@JohnRennie alright.

John Rennie

7:14 AM

So at equilibrium the pressures and temperatures will be the same on both sides.

Nobody recognizeable

Do you have time for another one?

John Rennie

Give me 15 minutes. I'm working at the moment.

Nobody recognizeable

@JohnRennie OK.

Please ping me when you come back.

John Rennie

7:32 AM

@Nobodyrecognizeable hi. I'm free for about half an hour now.

Nobody recognizeable

@JohnRennie hi

John Rennie

@Nobodyrecognizeable hi

Nobody recognizeable

@JohnRennie is that because we think that heat of resorvoir never changes?

John Rennie

What does the second line of option (d) say - the screenshot has clipped it off.

Nobody recognizeable

@JohnRennie increases.

John Rennie

7:41 AM

OK, so how far have you got with this?

Nobody recognizeable

C

2 mins ago, by Nobody recognizeable

@JohnRennie is that because we think that heat of resorvoir never changes?

@JohnRennie ^^

John Rennie

Entropy is given by $dS = dQ/T$

Nobody recognizeable

@JohnRennie yes.

John Rennie

For the hot reservoir $dQ = -Q_1$. It's negative because heat is being removed.

$T$ has the constant value $T_1$.

So $\Delta S = -Q_1/T_1$

(c) is true

Nobody recognizeable

@JohnRennie but the cooler resorvoir takes heat as well.

John Rennie

7:44 AM

The entropy of the hot source decreases and the entropy of the cold sink increases. But option (c) only asks about the entropy of the hot source.

Nobody recognizeable

@JohnRennie overall entropy change is zero. So d.

John Rennie

Yes.

Nobody recognizeable

@JohnRennie OK sorry thats done.

John Rennie

OK

Nobody recognizeable

@JohnRennie don't you agree it should be d.?

ds/dE=1/T

John Rennie

7:53 AM

@Nobodyrecognizeable I agree that dS/dE = 1/T, but that means the lower the gradient on the graph the higher the temperature. Yes?

Nobody recognizeable

@JohnRennie yes

@JohnRennie oh crap sorry.!

John Rennie

:-)

Nobody recognizeable

@JohnRennie g=h-Ts.

We know s is discontinuous then why should g not be?

John Rennie

Both H and S are discontinuous, and the two discontinuities cancel each other out to make G continuous.

Nobody recognizeable

@JohnRennie is it true for everything?

John Rennie

7:59 AM

It's true for all first order phase transitions.

Nobody recognizeable

@JohnRennie ^^

psa

@JohnRennie hey, do you have a sec?

John Rennie

@Nobodyrecognizeable How far have you got with this?

@psa hi :-)

Nobody recognizeable

@JohnRennie lnp=-c/T

psa

I think I'm OK with figuring out $f$ (number of thermodynamically accessible quadratic degrees of freedom), I'm just not sure how I'm supposed to get heat capacity per gram if we're not given the amount of substance present.

Nobody recognizeable

8:08 AM

@JohnRennie now how does the graph appear?

John Rennie

@Nobodyrecognizeable It's either (a) or (c) isn't it?

Nobody recognizeable

@JohnRennie they say c.

@psa one mole substance has or N molecules have molecular weight. Say W g. The 1g has W/N molecules.

John Rennie

@Nobodyrecognizeable When you do the integral you get $\ln P = -\frac{k}{T} + C$ where $C$ is the constant of integration. Yes?

Nobody recognizeable

@JohnRennie yes.

John Rennie

So the question is what is the value of $C$? In graph (a) the constant C is zero and in graph (c) the constant C is positive.

psa

8:12 AM

But we don't have W or N

Nobody recognizeable

@psa well N=6.023*10^23 and W is supposed to be known.

@JohnRennie sorry, yes.

psa

oh I see

$\frac{mol}{g} \cdot N_A = 1/g$

Nobody recognizeable

@JohnRennie are you here?

John Rennie

Suppose we choose the point $P=1N$ so $\ln P = 0$, then we get $C = k/T$, so unless $T = \infty$ that means $C$ is a positive number.

Nobody recognizeable

@JohnRennie yes

John Rennie

8:17 AM

So in general $C$ is going to be a positive number and that means $\ln P$ is going to tend to a constant positive value at large $T$.

Nobody recognizeable

@JohnRennie yes

@JohnRennie Thanks for the help professor have a nice day.

John Rennie

@Nobodyrecognizeable :-)

Knight

@JohnRennie Sir I need to draw a spring, how to draw it? You know some drawing tool software or something?

John Rennie

@Knight I use Google Draw for doing diagrams. I have a drawing of a spring you can copy if you want.

Knight

Yes, sir please give

John Rennie

8:30 AM

This is the link

That's a collection of clipart I use in my drawings. There's a spring in that collection.

Knight

@JohnRennie Thank you so much

John Rennie

You're welcome :-)

psa

@JohnRennie still a bit confused

John Rennie

@psa about the heat capacity question?

psa

yeah

the number of molecules $N = (moles)(N_A)$ but I don't see how that helps

John Rennie

8:40 AM

Let's take the first example, F2.

psa

OK

John Rennie

The molecular weight of F2 is 38, so for one gram N = Na/38.

So $C_v = \tfrac12 \frac{N_a}{38} f k$

psa

ah

John Rennie

$f = 5$ - three translational and two rotational

psa

I see

Nobody recognizeable

8:44 AM

@JohnRennie so we concider low temperature to ignore vibrational ones.

John Rennie

It says room temperature ...

Nobody recognizeable

@JohnRennie oh I see. But is there any standard temperature after which I should consider the vibrational ones?

John Rennie

Vibrational modes only start getting excited at hundreds of degrees C

Nobody recognizeable

@JohnRennie for all gases?

@JohnRennie I mean ie there is a Boyle temperature and inversion temperature is there any standard temperature?

For getting the vibrational modes.

John Rennie

No, because it depends on the energy of the vibration. The vibrational modes become important when kT is comparable to the energy of the vibration.

At room temp $kT \approx 1/40$ eV

And vibrational energies span a range of roughly 0.05 to 0.5 eV.

Nobody recognizeable

8:50 AM

@JohnRennie OK I see.

John Rennie

So the lowest energy vibrations will start becoming significant not that far above room temp, but in most cases it needs more like a temperature of 1000K.

We can probably Google for the F-F vibrational energy if you're curious.

Nobody recognizeable

@JohnRennie ok

psa

How many quad. DoF do we get for ammonia? 3 translational, 3 rotational, and if we were to assume the temperature was high enough to activate the vibrational modes, 6 vibrational?

also, the numbers from the CRC handbook for iron actually match 6 degrees of freedom

John Rennie

For complicated molecules the simplest approach is just to say that the total DOFs = 3N, where N is the number of atoms in the molecule

So that's 3 translational, 3 rotational and 3N-6 vibrational

psa

I guess Iron's vibrational modes are actually activated at room temperature

John Rennie

8:58 AM

And as I recall each vibrational DOF has an energy of kT i.e. twice as large as the translational and rotational DOFs.

@psa iron is a solid and that makes life more complicated.

psa

oh interesting

oh right, yes, that's because each term is counted

i.e. each vibrational DOF has two normal modes

John Rennie

For a solid there are no translational DOFs because, well, it's a solid so the atoms can't go anywhere.

And for an elemental solid like iron there are no rotational DOFs because single atoms can't rotate.

Nobody recognizeable

@JohnRennie so in conclusion iron has no internal energy?

psa

They could jiggle around a bit if nothing was frozen out though couldn't they?

John Rennie

So we only get vibrational DOFs due to the vibration of the atoms around their equilibrium position.

And there are three vibrational DOFs, so the heat capacity is 3Nk

psa

9:02 AM

yeah, the way I understand it, every quadratic term is counted, so for vibrational modes if you simplify it to be like a SHM, you get two quadratic terms (I guess you could argue the kinetic terms aren't independent, but from what I understand, experiments have verified that you really do have to count both independently)

John Rennie

Yes. A vibrational mode contributes $kT$ not $\tfrac12kT$

So the molar heat capacity of an elemental solid is approximately $3R$.

Nobody recognizeable

@JohnRennie so it is an exception to equipartition of energy.

John Rennie

@Nobodyrecognizeable huh?

What has equipartition of energy got to do with it?

psa

$\frac{1}{2}kT$ comes from the equipartition theorem

Nobody recognizeable

@JohnRennie equipartition energy states for each degree of freedom has KT/2 energy. While vibrational modes have KT.

psa

9:05 AM

@Nobodyrecognizeable it's not an exception, it's just that you have to count each normal mode, and for vibrations, there's actually two normal modes

John Rennie

The EPT says every available DOF gets an energy $\tfrac12kT$, and each vibrational mode has two DOFs.

psa

if you think of it like a spring, there's two quadratic terms $\frac{1}{2}kx^2 + \frac{1}{2}mv^2$

Nobody recognizeable

@JohnRennie why does vibrational modes have two degrees of freedom?

John Rennie

I can't remember the argument. I would have to Google it, which you could do for yourself.

Nobody recognizeable

@psa but the second one is for translation isn't it?

psa

9:06 AM

it's not

Nobody recognizeable

@JohnRennie np

psa

there's a difference between CoM motion and internal motion here

that distinction is important when you're talking about DoF

the kinetic terms for each translation $\frac{1}{2}mv^2$ are separate for that reason

Nobody recognizeable

@psa so it's internal motion?

psa

yes

it's confusing but it's true, it's also verified by experiment

Nobody recognizeable

google.co.in/url?sa=t&source=web&rct=j&url=https://…

Probably this should clarify.

@psa

psa

9:11 AM

I just read that a few days ago myself

Guru Vishnu

9:38 AM

@JohnRennie: Hi sir.

John Rennie

@GuruVishnu hi :-)

Guru Vishnu

@JohnRennie Are you free now?

John Rennie

Yes

Guru Vishnu

I've already received an answer from a good user. But do you have anything more to say about this question sir - physics.stackexchange.com/q/526618/238167

Before I proceed to accept the existing answer.

I asked that question when you were busy here sir.

John Rennie

In practice lines from different elements only very rarely overlap.

So given that know the lines for e.g. sodium, we can go through our spectrum and remove every line for sodium.

Guru Vishnu

9:41 AM

@JohnRennie Ok sir. The main problem I thought was - how to find whether one particular element is due to one particular element.

@JohnRennie Ok sir. Now understood.

@JohnRennie: But what if there exists some other element which looks the same as that of the combination?

John Rennie

Suppose you find a line that had the same wavelength as a sodium line, then it's 99% probable that it's due to sodium.

(because elemental lines almost never overlap)

If you find another or two more lines that match sodium then you are almost certain that sodium is there.

Guru Vishnu

@JohnRennie Is there any simple reason behind this fact sir?

John Rennie

The positions of lines is effectively random. It's actually related to the electronic structure of the atom, but that's so complicated that there is little or no correlation between the lines in different atoms.

And the width of a line is small compared to the gaps between the lines.

So just by random chance if you scatter lines at random it's unlikely two lines will overlap.

Guru Vishnu

@JohnRennie: Thank you sir. Will you copy your messages into an answer, so that I could cast my votes. I think this explanation is also great.

John Rennie

It does happen, and offhand I don't know what percentage of lines overlap, but it happens rarely enough that it's easy to separate the lines from different elements.

@GuruVishnu I wouldn't copy the answer as it is.

It's too vague. You would need to quantify what fraction of lines overlap.

Guru Vishnu

9:47 AM

@JohnRennie No sir. I didn't ask you to copy the existing answer. But the messages in the chatroom.

If it's not possible, no problem

John Rennie

I don't know what that fraction is, but you could probably Google for it.

Guru Vishnu

Is you explanation same for the following message I sent previously, sir?

5 mins ago, by Guru Vishnu

@JohnRennie: But what if there exists some other element which looks the same as that of the combination?

John Rennie

Yes, basically no two elements look the same.

Guru Vishnu

@JohnRennie Ok sir. Could you please tell what should I reply to the following comment: I think I'm in a small trouble.

What is observed is not the emission spectrum but the absorption lines, Fraunhofer lines. Atomic lines are narrow. — Pieter 6 mins ago

As far as I know Fraunhofer lines are dark lines observed from sun light.

I also learnt about it from Wiki.

John Rennie

Correct. It happens because light from the Sun is absorbed by atoms in the corona.

The Sun produces mainly black body radiation i.e. a continuous range of wavelengths.

Guru Vishnu

9:52 AM

@JohnRennie Yes sir. No problem in understanding that :-) The user thought my image:

is scientifically accurate.

But it isn't. How and what should I reply?

This is something which I don't much about.

John Rennie

Niels's answer looks fine to me. I would just accept it.

Guru Vishnu

@JohnRennie Yes sir. It's good. Usually, I give some time before answer arrives and then accepting. I'll give it some more thought and ask any doubts if any. I used to imagine what if even more answers come in? Will they stop writing just because I accepted one already? The ultimate goal of mine is to learn more interesting stuff.

John Rennie

Admittedly he though you meant emission lines, but emission and absorption lines have the same wavelengths. Which you get just depends on what you're looking at.

As a general rule a question stops attracting answers as soon as an answer is accepted.

Guru Vishnu

@JohnRennie Why did Mr.Fraunhofer come in?

John Rennie

Presumably Fraunhofer was the first to observe the dark lines.

Wikipedia says:

> In 1802, the English chemist William Hyde Wollaston[2] was the first person to note the appearance of a number of dark features in the solar spectrum.[3] In 1814, Fraunhofer independently rediscovered the lines and began to systematically study and measure the wavelengths where these features are observed.

Stigler's law strikes again :-)

Guru Vishnu

9:58 AM

:-)
Or may be as per "_Atomic lines are narrow._" :: Are the lines in my image very fat?

John Rennie

Though it seems Fraunhofer studied the lines in much more depth than Wollaston so naming the lines after him is probably fair.

@GuruVishnu I doubt anyone will take your spectrum that literally.

It's obviously just meant as an illustration.

Guru Vishnu

@JohnRennie Unfortunately, he did :-)

@JohnRennie Yes sir.

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