Problem Solving Strategies

harambe

06:23

@JohnRennie good morning

Are you free

John Rennie

@harambe morning :-) Yes, I'm free.

harambe

@JohnRennie off doubt.

Do cats die in schrodinger cat experiment

John Rennie

@harambe there isn't a simple answer to that

harambe

It says both alive and dead in Google

John Rennie

The trouble is that explaining the experiment requires explaining quantum mechanics

harambe

06:39

Oh lol. No need xD

harambe

06:50

@JohnRennie imgur.com/a/4fCXKoo

I calculated the answer but only one option is matching

I calculated the energy from qV

Then did 70 % of that successively and calculated the corresponding wavelengths

Only the wavelength 44.1 pm matches one of the answers given

John Rennie

@harambe Which question?

harambe

Q11

John Rennie

So the shortest wavelength is $40 \times 0.7$ keV?

harambe

Didn't understand how the shortest wavelength is 40 times 0.7 keV

John Rennie

The initial electron energy is 40keV. If it converts 70% of that energy into a photon then the photon energy is $40 \times 0.7$ keV.

Which gives $\lambda = 44.3$ pm

harambe

07:00

Yes but shortest wavelength means maximum energy. Shouldn't it be for 0.7 keV

John Rennie

Huh?

Where did 0.7 keV come from?

harambe

@JohnRennie 40 eV

John Rennie

@harambe The electron starts with 40 keV (keV not eV). Yes?

harambe

Yes

John Rennie

And it loses 70% of that at its first collision. So the energy lost is 40 x 0.7 = 28keV. Yes?

harambe

07:10

Yes

John Rennie

So the photon generated in the first collision has an energy of 28keV, and that corresponds to a wavelength of 44.3 pm.

harambe

@JohnRennie I get it. The first collision means first x ray photon

John Rennie

Yes.

So after the first collision the electron has 12keV energy left, and in the second collision it loses 70% of that and generates a second photon.

harambe

@JohnRennie got it. I can do this now

John Rennie

Cool :-)

harambe

07:39

@JohnRennie when an electron gets knocked off from its shell, it becomes a photon

John Rennie

@harambe no, it just becomes a free electron.

Are you asking how the K X-ray photons are generated?

harambe

What is Ek and EL here

John Rennie

Which question?

harambe

@JohnRennie imgur.com/a/8iIhCTf

Abcd

BTW ask some nice question on main to get upload pic privilege of chat.

Energy of electron in Kth shell

And energy of electron Lth shell

harambe

07:44

Okay. Got it

John Rennie

The K alpha photon is generated when the incoming electrons knock out an electron from the 1s orbital, and an electron falls from the 2p into the 1s orbital. So the photon energy is the difference between the 2p energy and 1s energy.

So I assume $E_K$ means the energy of the 1s orbital and $E_L$ means the energy of the 2p orbital.

The K beta photon is generated by an electron falling from the 3p orbital to the 1s orbital.

The L alpha is when a photon is knocked out of the 2s orbital, and the photon is generated by the 3p to 2s transition.

harambe

Okay got it

Abcd

@JohnRennie its explained in terms of shell in the book though...

$K_\alpha : 2 (L)\to 1(K) $

This is what is given in the book.

@JohnRennie Are you free for some concept doubts

John Rennie

@Abcd I'm working at the moment. I'll be about another 15 minutes. Shall I ping you when I'm done?

Abcd

@JohnRennie yes...

@JohnRennie BTW I see what you are doing, you are taking the last subshell with highest $l$ for each $n$. But what about $K_\gamma$ would it be $4s \to 1s$ or $3d \to 1s$

John Rennie

07:51

$\Delta \ell = \pm 1$

Abcd

??

John Rennie

So it would be 4p to 1s.

Abcd

oh

John Rennie

Angular momentum is conserved, and the angular momentum of a photon is $\pm 1$. So during the transition the angular momentum of the atom changes by $\pm 1$.

Abcd

so $K_\delta = 5p \to 1s$

John Rennie

07:53

i.e. p to s or d to p or f to d. But not d to s.

Abcd

wow nice

John Rennie

08:05

@Abcd I'm free for about half an hour, then I need to work again for quite a long time (couple of hours)

harambe

@JohnRennie one small doubt

imgur.com/a/4fCXKoo

Q12

Can you explain what is happening here

John Rennie

@harambe The K alpha photon corresponds to the transition between the 2p and 1s orbitals. So the energy of this photon (21.3 pm) gives you the energy difference between the 1s and 2p. Yes?

harambe

Yes

John Rennie

And you're told that the energy of the 2p orbital (the question calls it the "L shell") is 11.3keV, so now you can calcuate the energy of the 1s orbital (aka "K shell").

harambe

@JohnRennie a doubt.

John Rennie

08:11

@harambe yes?

harambe

Shells contain many orbitals so how do we know it is between 2p and 1s

ayc

@JohnRennie Ping me when done with harambe

harambe

There is 2s shell in L too

John Rennie

19 mins ago, by John Rennie

Angular momentum is conserved, and the angular momentum of a photon is $\pm 1$. So during the transition the angular momentum of the atom changes by $\pm 1$.

A 2s to 1s transition doesn't change the angular momentum because s orbitals have an angular momentum of zero.

harambe

Okay got it

ayc

08:14

@harambe you done?

harambe

What after calculating energy of 1 s

@ayc just a little bit left

@JohnRennie you here?

John Rennie

@harambe To generate a K alpha line the incoming electrons have to have an energy equal to or greater than the 1s energy.

Otherwise they don't have enough energy to know out a 1s electron.

harambe

Okay

Got it

@ayc done

ayc

@JohnRennie Just wait ...I'll upload pictures of my question

John Rennie

@ayc OK

ayc

08:23

@JohnRennie why is the work done here negative?.....As the gas is undergoing a isothermal compression we know that work has to be done on the gas ....work done on the gas is defined as positive...so here it should be posiitve right........and that formula has a negative right?..i.e..it should be (-2.303nRT)..

John Rennie

@ayc The sign of work is somewhat arbitrary because it depends whether you consider work done on the gas or work done by the gas.

ayc

@JohnRennie My book considers the work done by the gas as negative

John Rennie

Likewise the sign of Q depends on whether you define it as heat flowing into the gas or heat flowing out of the gas.

In this case the internal energy is constant because the internal energy depends on the temperature, so we know $W$ and $Q$ must be equal and opposite.

It looks to me as if this question is defining work done by the gas as positive and heat flowing out of the gas as positive.

So in this case we get a negative $W$ and a positive $Q$.

ayc

@JohnRennie so is the answer given wrong..according to the conventions the book uses?

John Rennie

@ayc well if the work as defined by the book is positive then Q has to be negative. So that is the opposite sign convention to the one used in this question.

Abcd

08:39

@JohnRennie Are you talking about Q absorbed or released?

@ayc are you doing physics or chemistry

John Rennie

@Abcd that's what we're not sure about. Ayc's book seems to use one convention and the question uses the opposite convention.

ayc

@Abcd chemistry

Abcd

Give me 2 minutes to type please

In Chemistry first law is :U = Q+W

In Physics first law is: Q = U+W

In both physics and chemistry Q absorbed by gas is positive

John Rennie

@Abcd I only have a few minutes left if you want to discuss something ...

Abcd

In both Physics and chemistry Q released by gas is negative

In Chemistry work done by gas is Negative

In Physics work done by gas is Positive

That's it.

ayc

08:43

@Abcd In the question since it is a compression ..work is done on the gas....agreed?.....then according to my book work done by the gas is negative..so here work done should be posiitive..and q negative..

Abcd

@JohnRennie the problem is I have checked 10 times and they are not uploading the question paper!! So I am not able to get my questions... We had online exam yesterday and my doubts are from it

John Rennie

OK. I'll get back to work now. I'll be done in around an hour, so ping me after that if you've got the paper.

Abcd

@ayc their work sign is wrong

It should be positive

ayc

@Abcd Yeah..I think so......alright...

Abcd

Because work is done on the gas and it's chemistry

ayc

08:46

@Abcd hmmm

harambe

09:34

@JohnRennie when you get back, can you ping me

John Rennie

09:45

@harambe pingggggg

harambe

@JohnRennie imgur.com/a/c0zk180

Q13

Isn't ionising energy same as the energy to knock out electron from k shell?

John Rennie

@harambe no, the ionisation energy is the energy needed to remove an electron from the outermost orbital

And in argon the outermost orbital is the ... ?

harambe

3p

John Rennie

And the K beta is generated by a transition from ...?

harambe

3p to 1 s

Got it

@JohnRennie Q20

What does energy of atom with no vacancy mean....?

It's orbitals are full, right

John Rennie

10:01

@harambe yes

Effectively you're being given the energies of the 1s, 2p and 3p orbitals.

harambe

@JohnRennie I can't seem to figure energy of 3p

It says 0.530 keV higher than energy of atom with no vacancy

What is the energy of the atom with no vacancy

I am guessing the energy of 1s and 2p is 25.31 eV and 3.56 keV

John Rennie

The energy of an orbital is the energy change if you start with an unionised atom then remove an electron from that orbital.

By "atom with no vacancy" the question just means an unionised atom.

harambe

@JohnRennie energy of unionised atom is zero? Just guessing

John Rennie

Yes, well, kind of. For the purposes of this question you can take the energy of the unionised atom as zero.

ayc

@JohnRennie is latent heat of vapourisation constant for a liquid

John Rennie

10:14

@ayc it's approximately constant. There will be some small change with pressure.

ayc

@JohnRennie Ohk.Thank you

John Rennie

8

Q: What is the effect of an increase in pressure on latent heat of vaporization?

What is latent heat of vaporization ($L_v$) in the first place? Wikipedia seems to indicate that it is the energy used in overcoming intermolecular interactions, without taking into account at all any work done to push back the atmosphere to allow for an increase in volume when a liquid boils. I...

harambe

10:27

@JohnRennie Q21

The energies of X rays are

58.2keV for K alpha, 67.2 keV for K beta and 9 eV for L alpha

For no enmiting of X rays the energy should be less than 9 eV

So Voltage should be 9 V

What's wrong in this logic

John Rennie

Tungsten L-alpha is 9keV not 9eV

harambe

Yeah. A typo

John Rennie

The minimum electron energy is the energy of the 2p orbital, because you have to eject a 2p electron to get a L line. So you have to work out the energy of the 2p orbital from the info you're given.

harambe

Didn't get this

Why 2p?

John Rennie

Oops, 2s I meant

harambe

10:35

@JohnRennie okay so x ray is created due to removal of 1 s or 2s

1 s has more energy so we go for 2 s?

John Rennie

Yes, the incoming electron knocks out the 1s or 2s electron, and the X-ray photon is generated when one of the other electrons drops into the hole.

harambe

Oops 1 s has less energy

John Rennie

The L lines are always lower energy than the K lines, so if you're being asked for the maximum energy for no K or L line then use the L line as your benchmark i.e. the energy needed to knock out a 2s electron.

@harambe careful! The orbital energies are negative so 1s has less energy than 2s in the sense that it is more negative.

The magnitude of the 1s energy is a lot bigger than the magnitude of the 2s energy.

harambe

Got it but one doubt

John Rennie

Yes?

harambe

10:40

@JohnRennie imgur.com/a/hDZiFjw

John Rennie

Yes?

harambe

Ignore it. I thought you were comparing Beta and alpha

@JohnRennie the potential difference would make the electrons hit target and this would realease photons. The coming electrons have to have energy equal to 1s or 2s to remove electrons in 1s or 2s.So why exactly the qV should be equal to energy of 1s?

John Rennie

The potential difference determines the energy of the incoming electrons. e.g. if the potential difference is 40kV then the energy of the incoming electrons is 40keV.

harambe

Okay

John Rennie

So suppose the energy of the 2s orbital is 14keV (I just made up that number)

Then to avoid generation of L photons the incoming electrons need to have an energy less than 14keV so they can't knock out a 2s electron. Yes?

harambe

10:48

Yes

John Rennie

And that means the potential difference has to be less than or equal to 14 kV.

harambe

@JohnRennie are electrons from filament absobed in the target

John Rennie

@harambe yes, the incoming electrons will eventually come to rest in the target, though this usually takes lots of collisions.

In X-ray tubes the target is earthed to stop a negative charge building up on it.

ayc

@JohnRennie Can I think of heat of vapourisation like this: I have a container filled with some amount of water.If I wish to vapourize the water then I have to add some energy to break intermolecular bonds(H-bonds) and lets say volume was v.If pressure was more than before then I will have to add more energy than before to make it reach volume V..Now if I want the volume to be 2V,pressure is same as initial case, then I have to add more energy....

John Rennie

@ayc kind of. The free energies of the molecules in both the vapour and the solid depend on the pressure so it will actually be quite complicated to calculate how the enthalpy changes. There is an equation that describes this. From memory it's the Gibbs-Duhem equation, though I wouldn't swear to this.

harambe

10:56

So in a way I can think that the electron will remove an electron from the 1s or 2s before coming to rest if they have energies higher than 1s or 2s?

John Rennie

@harambe yes

ayc

@JohnRennie I'll keep that in mind.....the equation looks a bit complicated ..I'll have a look at it after finishing the chapter....anyways thank you for your time!

harambe

@JohnRennie is there a time interval for the X rays to be produced in target after electron strikes it

John Rennie

@harambe I suppose so, but electrons with energies of tens of keV are moving at a significant fraction of the speed of light, so it's not going to take very long :-)

Though I suppose the time is probably the lifetime of the ion rather than the travel time of the electron i.e. the time taken for a 2p electron to drop into the 1s orbital. That's still going to be a pretty short time. Picoseconds or femtoseconds I'd guess.

harambe

Okay got it

Another thing. Heat is produced in process of X rays. Why

John Rennie

11:06

Not all the energy of the incoming electrons is released as photons. In fact only a small fraction of the energy is released as photons. Most of the energy gets converted into heat.

So the target gets mighty hot!

harambe

Got it

harambe

11:17

@JohnRennie imgur.com/a/LbmRcAf

Q23

Don't you think they should have given time here?

John Rennie

They have! They have given you the power, 200W, which is energy per unit time.

harambe

@JohnRennie what exactly does unit time mean

John Rennie

Well, what is the definition of the Watt i.e. what does one Watt mean?

harambe

One Joule energy per sec

John Rennie

Right, so 200W is 200 joules per second. So you divide 200 by the energy of an electron (in joules) and that gives you the number of electrons per second.

harambe

11:23

Okay

That's a cool way to think

John Rennie

@harambe The whole point of doing all these annoying questions is that you get used to the right way to think about them. I only know the right way to approach them because I went through the pain too (a long time ago! :-).

harambe

@JohnRennie yea. I can see that xD

One doubt I had regarding this

John Rennie

Yes?

harambe

Here the total energy of neV is used for heat while in the numericals until now I have used every for X rays generation. Is the number of electron so close to 1 for X rays production

John Rennie

The probability of an electron knocking out a core electron is low, so maybe only one ina hundred electrons knocks out a core electron and thereby causes an X-ray photon.

If an electron does knock out a core electron then it transfers most of its energy to the ion and therefore eventually to the X-ray photon. In this sense the process is very efficient.

harambe

11:38

Lol every is the autocorrect of ev

John Rennie

But the other 99 electrons don't knock out an electron and all their energy ends up as heat (less a small amount of energy that goes into the continuous X-ray background).

ayc

@JohnRennie Could you explain the negative heat capacity:en.wikipedia.org/wiki/… in your own words

John Rennie

So overall the process is very inefficient because only one photon gets created for every 100 electrons. That's why the question is asking you to ignore that 1% and estimate the heating.

@ayc My own words:

14

Q: Explanation for negative specific heat capacities in stars?

I've just found out that a negative specific heat capacity is possible. But I have been trying to find an explanation for this with no success. Negative heat capacity would mean that when a system loses energy, its temperature increases. How is that possible in the case of a star? Musn't there b...

thermodynamics astrophysics stars

ayc

@JohnRennie you already did.....nice

John Rennie

Also

7

Q: Can a black hole have negative temperature?

Stephen Hawking said that black hole also have temperature and it is related to its mass so in other words a black hole can also be shown to have a negative temperature! I know that nothing is colder than absolute zero and negative temperature are hotter than all positive temperature including in...

black-holes temperature

harambe

11:44

@JohnRennie doubt

Why is pluto not considered a planet?

Is there any reason for this down grade

John Rennie

13

Q: What is the current status of Pluto?

Pluto has been designated a planet in our solar system for years (ever since it was discovered in the last century), but in 2006 it was demoted. What caused this decision? And is there a chance that it could be reversed? Edit: well, http://www.dailygalaxy.com/my_weblog/2017/03/nasas-new-horizon...

planets definition asteroids

harambe

Okay that makes little bit of sense to me

@JohnRennie last question for today

imgur.com/a/d7Ci2B8

Q25

Can you explain what is happening here

tatan

@JohnRennie Hi

John Rennie

@harambe conservation of momentum. Remember that the momentum of a photon is given by $p = h/\lambda$.

@tatan morning :-)

harambe

Yea. De broglie wavelength

tatan

11:54

Can you help me with a bit of current electricity?

harambe

Is the collision elastic

Also by atom what is meant here

John Rennie

The idea is you start with a free iron ion floating in space so its momentum is zero. This ion emits a K-alpha ray that zooms off in one direction, so to conserve momentum the ion has to move in the other direction.

tatan

I am faced with the question if drift velocity changes with non-uniform cross section of a conductor. I was a bit confused. Searching yielded these -physics.stackexchange.com/questions/138458/… and physics.stackexchange.com/questions/145926/… Now, I am still confused. Can you explain please?

@JohnRennie Please help once you are done with Harambe ;-)

John Rennie

Call the mass of the atom $m$, and the velocity $v$. Then to a first approximation we can just say that for conservation of momentum $mv = h/\lambda$.

To be more precise we'd need to take into account that the KE of the atom has to come from the energy f the photon i.e. the photon is slightly red shifted, but I'd guess the question doesn't expect you to worry about that.

@tatan hi, I think harambe and I are done

tatan

Hi

I have been given that a steady current I flows through a non uniform cross section of a metallic conductor. Will the drift velocity change?

John Rennie

12:08

@tatan Imagine pumping water through a pipe that varies in width. The water is going to speed up in the narrow sections of the pipe and slow down in the wide sections of the pipe. Yes?

tatan

Yes

That implies drift velocity changes. does it?

John Rennie

The electrons in a conductor behave much like a fluid flowing in a pipe. The electron drift velocity corresponds to the water velocity.

So, yes, the drift velocity is going to change.

tatan

Okay. But I have one doubt

John Rennie

@tatan Yes?

tatan

In all proofs of current electricity like i=neAv_d or v_d=\frac{eEt}{m} etc. one thing we assumed at the beginning is that drift velocity remains constant. Were we making some specific assumptions there?

I mean in the same question, there is no doubt that current density changes. But if we are asked about electric field in the same situation can I use the equation $\J=\rho E$? Because in deriving $J=\rho\times E$ the first thing we assumed was that $v_d$ was constant

So it shouldn't be valid here right?

John Rennie

12:17

If the conductor gets narrower that means its resistance increases (NB its resistance not its resistivity) and that means the voltage drop across the narrow bits is going to be higher than across the wider bits. That means the field $E$ will be greater in the narrower bits, and that's why the current density is higher there.

tatan

Ok... but I suppose none of the equations like I=neAv_d or v_d=\frac{eEt}{m} are valid, is it?

John Rennie

The equation $v_d=\frac{eEt}{m}$ is valid as long as you note that $E$ varies along the conductor.

And $I=neAv_d$ is valid as long as you note that $A$ varies along the conductor.

tatan

But, how do I derive $v_d=\frac{qEt}{m}$ in such a case. Normally what I do is to write Newtons Law for the conductor $m\frac{dv}{dt}=qE-\frac{mv_d}{t}$ in which the LHS terms goes to $0$ (as a result v_d is constant). However we can't do that in a conductor with continuous variation of cross section. Can we?

John Rennie

If you know the overall resistance, and therefore overall current, you know the number of electrons per second flowing along the conductor. The total number of electrons per socond is constant all along the wire, so you can use this to determine the drift velocity from the cross sectional area.

tatan

12:36

ok... thanks!

harambe

@JohnRennie is your irodov from your college days

John Rennie

12:55

@harambe no. At college we used a similar book but published the university (Cambridge). I forget how I acquired my copy of Irodov.

Sid

@JohnRennie Is it standard that all books used by students are published by Cambridge?

harambe

13:58

@Sid is dearrangement in pnc in mains? Is it okay to skip it

Sid

@harambe I am sorry, I am completely(and thankfully) out of touch with JEE syllabus

So, you would need to ask some of your batchmates about that

harambe

Okay.

Abcd

14:29

PnC has no limit or syllabus. And Derangement is a 10-20 minute topic with 1 formula. So its not OK (read: advisable) to skip it.

@sammygerbil Please ping when you are free for problems.

harambe

14:56

Yea. Thsmat makes sense

John Rennie

16:18

@Sid it wasn't a university policy, but with Cambridge having so many leading scientists working there it was inevitable that books by university staff were on our reading list.

This is the book though those are silly prices on Amazon. It must now be a collectors item. Maybe I should sell my copy :-)

Dante

@JohnRennie Hello! Are you free for a simple communication problem?

communication systems*

John Rennie

@Dante yes, if it's quick.

Dante

It is I suppose

Lemme upload the pic

First one

John Rennie

I have to admit I don't know what a modulation index is.

Dante

It's ratio of amplitude of message wave to that of carrier wave

John Rennie

16:30

Are we looking at question (1)

Dante

Yeah

John Rennie

That's frequency modulated not amplitude modulated.

Dante

Ah damn, I had studied only basics from NCERT which had only amp mod

Sorry for wasting your time

I'll have to study this concept first

John Rennie

@Dante well the question does say A amplitude modulated wave but then the wave they give is frequency modulated. And as far as I can see none of the options they give is correct.

Dante

Yes, it didn't seem like an amplitude modulated wave to me too, amplitude is constant and frequency is complex here.

@JohnRennie hold on for 1 min I might have another simple doubt

John Rennie

16:37

@Dante OK ...

Dante

@JohnRennie second one

John Rennie

If the modulation is 75% then presumably the wave is something like $(1 + 0.75\sin(\omega_1 t))\sin(\omega_2 t)$. Does that look a sensible equation for the AM wave?

Dante

Yes

John Rennie

Where $\omega_2$ is the carrier frequency and $\omega_1$ the modulating frequency.

Dante

yea

John Rennie

16:44

So to find the power you need to square that function and then integrate it ...

Dante

@John what's radiated power?

John Rennie

I think that just means the power carried by the wave, which is proportional to the amplitude squared.

Dante

Hmm, I see..

Thank you!

Sid

17:24

@JohnRennie 36 Euros.. That doesn't seem that much expensive. Collectors' items would cost a bit more, no?

John Rennie

@Sid it's only a tiny book. 56 pages including the index. My copy cost me £1.95 in 1980.

Sid

@JohnRennie Accounting inflation, that would cost what in current times?

John Rennie

I suppose these days you might £5 for something like that.

Sid

Ah. Then, you should sell it assuming you aren't too attached to it. You will be making a nice profit

harambe

23:00

@sammygerbil Hi .If you are free can you tell me if there an intuitive way of understanding power

Transcript for

Problem Solving Strategies