Problem Solving Strategies

Abcd

06:09

@sammygerbil sorry I had slept by the time you came online...

@JohnRennie Morn?

John Rennie

@Abcd morning :-)

Abcd

@JohnRennie question^^^

@JohnRennie After equating energies I am getting $mc^2 = 0.39 $

@JohnRennie how to decide between options B and D?

John Rennie

@Abcd when you do the energy conservation calculation you are finding the total energy of the neutrino i.e. the sum of its rest mass energy plus the kinetic energy.

Abcd

@JohnRennie OK thats why D. Got it

John Rennie

Cool :-)

tatan

07:36

Can longitudinal vibrations in material medium affect electromagnetic waves like light?

Why is the answer (A) as given below? Can someone explain please?

@JohnRennie Can you help please?

John Rennie

@tatan longitudinal waves in water are density waves. So when a standing wave is formed in the water we get alternating regions where the water is more dense and less dense. The change in density changes the refractive index.

tatan

@JohnRennie "The change in density changes the refractive index"- I didn't find it too obvious... I mean I don't recall reading it in my texts. Am I missing something trivial?

John Rennie

@tatan exactly this phenomenon is responsible for the formation of mirages. OK, that's in air so the density changes can be quite big, but the same basic principle applies to any material.

tatan

07:51

Are you talking about the optical density or the material density? Alternatively, does a change in material density necessarily mean a change in optical density ? @JohnRennie

John Rennie

I'm talking about the density of the material through which the light is passing i.e. kg per cubic metre.

tatan

Thanks ;-)

John Rennie

Light interacts with the electrons in materials. When you change the density of a material (e.g. by compressing it) you change the electron density as well and this will usually mean a change in the strength of the interaction with the light i.e. a change in refractive index.

tatan

I get it.. thanks again ;-)

How does the angle of min deviation vary with the wavelength of incident light? I couldn't find proper explanations in the internet.

I would like to work it out in the case of a thin prism

John Rennie

08:16

The angle of minimum deviation changes with wavelength only because for most materials the refractive index changes with wavelength.

There is no simple way to describe how the refractive index changes with wavelength because it depends on the material being used.

starunique2016

09:39

can anybody suggest a good book for organic seperation techniques.

and the other boring chapters like everyday chemistry and biomolecules

starunique2016

09:56

how to approach

John Rennie

@starunique2016 suppose the velocity of the sliding rod is $v$, then the area swept out per second is $A = v\ell$ and therefore the rate of change of flux linked is $Bv\ell$. OK so far?

starunique2016

ok

yes...after that.

@JohnRennie

John Rennie

10:15

@starunique2016 the induced EMF is equal to the rate of change of flux linked, so the EMF is also equal to $Bv\ell$, and that EMF produces a current through the resistor of $I = V/R = Bv\ell/R$.

starunique2016

yes...all right

John Rennie

And this current means there is a force on the sliding rod of $F = BI\ell$

Can you take it from here?

starunique2016

yes so...B^2Vl/R

can you please go all the way.:)

actually i am really stuck with the terminal speed...is it a really trivival one?

John Rennie

The gravitational force acting down the plane is $mg\sin\theta$.

@starunique2016 The rod is being accelerated down the plane by the gravitational force, and the motion is being resisted by the magnetic force $BI\ell$. At the terminal velocity these two forces are equal, so just set them equal and solve for $v$.

starunique2016

oh sorry,too trivial one

blue_eyed_...

10:39

@JohnRennie, @sammygerbil, Two parallel conductors carrying current in the same direction attract each other. If they are permitted to move towards each other, the forces of attraction do work. From where does the energy come?

John Rennie

If you consider an isolated conductor it has an associated magnetic field and there is energy stored in that magnetic field. That energy is put into field when the current first starts flowing through the conductor. OK so far?

blue_eyed_...

@JohnRennie,yes

John Rennie

@blue_eyed_... when you bring the two isolated conductors together the field is now the combined field of the two conductors, and this combined field stores less energy than the two isolated fields did.

blue_eyed_...

@JohnRennie, why does the combined field store less energy than isolated?

John Rennie

I'm just Googling to see if I can find the derivation. I don't remember it offhand ...

John Rennie

10:59

No hang on ... it's the other way round ...

blue_eyed_...

11:09

@JohnRennie, okay. I'm waiting

John Rennie

Somethings's come up at work and I'll be busy for a bit ...

blue_eyed_...

@JohnRennie, okay. Answer me when you're free

Jack Clerk

11:28

Good morning/night/evening for everyone!

I really need some help in spacetime diagrams as I posted here in my question:

0

Q: Spacetime diagram of Rindler's lenght contraction paradox

I'm struggling to draw properly the spacetime diagram for Rindler's lenght contraction paradox as exposed here: https://aapt.scitation.org/doi/10.1119/1.1937789 How the spacetime diagram draw becomes?

The truth is: I have a small presentation today in the evening, but I do not know how to draw this particular diagram.

small lecture**

Nobody recognizeable

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

Jack Clerk

But this article deals with acceleration.

Nobody recognizeable

@JackClerk these videos maybe helpful to you.

@JackClerk there's no acceleration.

@JohnRennie are you here ?

Spacetime animation solving the train in tunnel paradox

►

@JackClerk this is just a diagram of a similar paradox .

Jack Clerk

11:51

@Nobodyrecognizeable but this paradox is the same as Rindler's ?

Nobody recognizeable

12:04

users.physics.ox.ac.uk/~rtaylor/teaching/train.pdf

@JackClerk heres the paradox. If it fits then that's great if not then you should ask in h bar or physics meta since johnrennie seems to be busy now.

@JackClerk pardon me heres the paradox.

*related question.

Madhuchhanda Mandal

Please help me understand why Fig 3.6 is a reasonable approximation of fig 3.5(c)

harambe

14:10

@sammygerbil are u free

sammy gerbil

@harambe hello

harambe

@sammygerbil I had a doubt in a question

imgur.com/a/glt6mpn

I am getting the answer to be (B) but it's given to be (A)

The cross product of velocity and magnetic field is coming to be positive Y axis in first case and negative Y axis for second

That means it should have force in that direction too

sammy gerbil

14:29

@harambe The cross product v x B should be initially along the +z axis because v is along the +x axis and B is along the +y axis.

harambe

@sammygerbil yes...i just miswrote it

So

@sammygerbil Got it. I got my mistake

sammy gerbil

When the particle reaches x=2x the velocity is along the +z axis and the field is now along the -y axis so the force v x B will be along the +x axis.

Answer A is correct.

harambe

The option has Yaxis to be downward direction.. That's why when I took the cross, it gave the wrong answer maybe

Nobody recognizeable

@sammygerbil if you take ${v_0 \hat{i}}\times {-B_0 \hat{j}}$ then you should get $-vB_0\hat{k}$.

sammy gerbil

14:45

@Nobodyrecognizeable Yes that is correct but how does that affect the answer?

Nobody recognizeable

@sammygerbil its working on negative z axis you said negative x axis ?

sammy gerbil

It x=a the force is in the +z direction, at x=2a the force is in the +x direction.

Nobody recognizeable

@sammygerbil yep fine.

@sammygerbil if you have some free time then ill ask you about the very first question.

sammy gerbil

@Nobodyrecognizeable ok what is your difficulty here?

Nobody recognizeable

@sammygerbil do i have to do it using photons momentum equals $hv/c$ ?

sammy gerbil

14:52

@Nobodyrecognizeable yes

Nobody recognizeable

@sammygerbil can i get the momentum in x axis of photon as $hvcos$\phi /2c$ after the collision?

sammy gerbil

@Nobodyrecognizeable yes

Nobody recognizeable

@sammygerbil can i consider the photon was travelling in x axis initially?

sammy gerbil

@Nobodyrecognizeable yes, why do you think otherwise?

Nobody recognizeable

@sammygerbil should i take the energy conserved?

Actually i was thinking that electron and photon both scattered at different angles so the photon was travelling in x axis or not .

sammy gerbil

15:01

@Nobodyrecognizeable yes, there is nowhere else it can disappear to. The electron has no structure like an atom so it cannot store internal energy. The collision must be elastic.

Nobody recognizeable

Beause classically if i hit anything in x axis it should stay at x axis even after the collision although size of two comes into the matter .

sammy gerbil

We have 3 unknown parameters in the outcome (2 angles + speed of electron). So we need 3 equations (conservation of momentum in x and y directions + conservation of energy). If energy were not conserved we could not get an answer.

@Nobodyrecognizeable In a classical collision eg of 2 billiard balls, one ball which initially travels in x direction does not necessarily continue travelling in x direction afterwards. If it does, the other ball must travel in the x direction afterwards also, otherwise momentum could not be conserved.

Nobody recognizeable

@sammygerbil $hv/c = hvcos\phi /2c + mvcos\theta$

@sammygerbil $hvsin\phi /2c + mvsin\theta$ =0

sammy gerbil

@Nobodyrecognizeable + should be = there

Nobody recognizeable

@sammygerbil $mv^2/2 = h\nu /2$

@sammygerbil please justify that im right or not .

sammy gerbil

15:13

@Nobodyrecognizeable correct so far, assuming this is not a relativistic situation in which $h\nu \approx mc^2$.

However, $mc^2$ appears in the answer, so perhaps this is a relativistic situation?

Nobody recognizeable

@sammygerbil well now how to solve them easily .

sammy gerbil

hyperphysics.phy-astr.gsu.edu/hbase/quantum/compeq.html

Nobody recognizeable

@sammygerbil ok so lets go for relativistic collision problem.

sammy gerbil

Compton scattering

Compton scattering, discovered by Arthur Holly Compton, is the scattering of a photon by a charged particle, usually an electron. It results in a decrease in energy (increase in wavelength) of the photon (which may be an X-ray or gamma ray photon), called the Compton effect. Part of the energy of the photon is transferred to the recoiling electron. Inverse Compton scattering occurs, in which a charged particle transfers part of its energy to a photon. == Introduction == Compton scattering is an example of inelastic scattering of light by a free charged particle, where the wavelength of t...

Nobody recognizeable

@sammygerbil okay after that .

sammy gerbil

15:23

@Nobodyrecognizeable Because this is a MCQ I think you should be able to decide which options are true/false using logical reasoning rather than brute-force calculation.

You can decide options C and D quite easily.

Can you decide between C and D?

Nobody recognizeable

@sammygerbil ok. Lets be clear to you. Im preparing for iitjam an exam ill give after 2.5 years. (After my ug degree) so dont think I'm in any kind of hurry . Rather ill try to solve the question fully.

sammy gerbil

@Nobodyrecognizeable ok. Well please could you look at the derivations which are available online or in your textbook? The chatroom is a good place for discussing doubts, but not for learning material which is already explained very well elsewhere on the internet.

Nobody recognizeable

@sammygerbil ok. Sure . Currently seeking en.wikipedia.org/wiki/… to find out something. So im taking little bit of time.

sammy gerbil

@Nobodyrecognizeable Good. Please try to get to the end of the calculation, rather than asking for confirmation of each step. :)

Nobody recognizeable

@sammygerbil its gonna take a little while so i wanna ask a simple question. So can i?

sammy gerbil

15:32

@Nobodyrecognizeable yes

gateprep

Δ U (nought) and Δ H(nought) are not equal for bomb caloriemeter

is it true

if yes/no why?

sammy gerbil

@gateprep Probably a good one for @JohnRennie. I'm not too hot on thermodynamics.

Nobody recognizeable

sammy gerbil

@gateprep BTW what do you think?

gateprep

They will be equal

Nobody recognizeable

15:35

@sammygerbil 9.46

sammy gerbil

@Nobodyrecognizeable ok what is the difficulty?

Nobody recognizeable

@sammygerbil ok getting back after dinner. You may discuss with @gateprep

sammy gerbil

@gateprep for what reason?

Nobody recognizeable

Bye professor for now.

sammy gerbil

@Nobodyrecognizeable Bye. I shall be online for several hours.

Nobody recognizeable

15:48

@sammygerbil hi im back . Im asking what should be r fram x axis for each case how is it defined ?

sammy gerbil

@Nobodyrecognizeable That was a very quick dinner!

The moment of inertia about z axis through the origin is defined as $I_z=\sum m_i r_i^2$.

Nobody recognizeable

@sammygerbil yes i dont need anymore time to eat 2 breads and vegetable.

sammy gerbil

But we are asked for moment about x axis.

Nobody recognizeable

@sammygerbil oh i think $r_i^2 = y_i^2 + z_i^2$ right ?

sammy gerbil

@Nobodyrecognizeable That is the distance from the origin.

Nobody recognizeable

15:55

@sammygerbil what does distance from x axis mean ?

sammy gerbil

@Nobodyrecognizeable The perpendicular distance. Closest distance.

Nobody recognizeable

3 mins ago, by Nobody recognizeable

@sammygerbil oh i think $r_i^2 = y_i^2 + z_i^2$ right ?

sammy gerbil

@Nobodyrecognizeable Sorry you are correct that is the distance from the x axis.

Dante

Hi @JohnRennie You here? I had doubt in inorganic chemistry.

Nobody recognizeable

Ok lemme check the answer.

@sammygerbil this is coming out to be 63 while the answer given is 71.

Abcd

16:02

@Dante He doesn't do chemistry man. And he has forgotten most of the stuff of chemistry. He can help in physics related stuff like electrochemistry and solid state though...

sammy gerbil

@Nobodyrecognizeable I get 63 also.

But there should be 2 answers.

Radius of gyration is not 71 either.

Nobody recognizeable

@sammygerbil some writer are miser in terms of giving answers. Well the next question has some correct answers. This is misprinted i think.

Dante

@Abcd Well, could you?

Nobody recognizeable

@sammygerbil is the radius of gyration $\sqrt{6.3}$ ?

Abcd

@Dante please ask in chemistry room.

Dante

16:08

Okay

sammy gerbil

@Nobodyrecognizeable correct

Nobody recognizeable

Just divide moment of inertia by sum of masses and root that.

@sammygerbil by the way have you seen those books which I've given you ?

John Rennie

@Dante I did take an inorganic chemistry course once, back in the mists of time. That was probably before Mendeleev had discovered the periodic table.

10

Dante

Haha

Abcd

@JohnRennie BEFORE the periodic table, really???

John Rennie

16:12

@Abcd I'm English. This is the English idea of a joke.

(no other races laugh for some reason ...)

sammy gerbil

@Nobodyrecognizeable I have come across "Thinking Like a Physicist" before.

Nobody recognizeable

@JohnRennie i forgot vant hoff john rennie got the first noble in chemistry. :p

@sammygerbil for the other one you need djvu viewer.

@sammygerbil do you wanna learn thermodynamics. I ve a problems book on that also .

sammy gerbil

@Nobodyrecognizeable No thank you. I am not in the mood for studying at present.

What was your 2nd book which requires djvu viewer?

Nobody recognizeable

@sammygerbil it was on some mechanics problems basically.

I ve given you problem books actually.

Loop Back

@JohnRennie consider this though experiment. Little quantity of water is poured in a vessel which is then perfectly insulated. The vessel is now kept in anti gravity chamber.(g=0) The water will shape itself into a sphere. Won't this decrease the total entropy

John Rennie

16:22

@LoopBack when the water compacts into a sphere the surface area decreases so the surface energy decreases. This means the water gets slightly hotter as the energy that was at the air-water interface turns into internal energy of the water.

Nobody recognizeable

@JohnRennie so if you make a big planet with water will it turn to a hot star or so ?

John Rennie

@Nobodyrecognizeable no because the change in surface energy is very small. Far too small to do anything dramatic. The point I was making is that when you're trying to account for entropy you have to include everything.

It can be surprisingly difficult to keep track of all the entropy.

For example when an interstellar gas cloud contracts to form a star we go from a huge very disordered dust cloud to a much more ordered small compact object. At first glance this looks as if the energy decreases massively. But only because there is another factor we haven't considered ...

Abcd

@JohnRennie how can water automatically turn into sphere

Nobody recognizeable

@JohnRennie so in case of black hole would we say the total entropy is increasing as it gives away radiations as a lot of particles are hinged together resisting their motion which decreases entropy .

John Rennie

@Abcd surface tension

Abcd

16:29

@JohnRennie ??

Nobody recognizeable

@Abcd because of gravity also.

John Rennie

@Nobodyrecognizeable basically yes. The collapsing dust cloud has to cool to form the star, and the emitted electromagnetic radiation carries away the entropy.

Abcd

Why do we care about stuff like black holes?

John Rennie

@Abcd same reason a balloon is round. The air water interface has a surface energy, so the energy of the water is reduced by forming the shape that has the lowest area to volume ratio i.e. a sphere.

@Abcd it's fun

Abcd

@JohnRennie but useless as well right?

Nobody recognizeable

16:32

@Abcd cause that is somewhere where relativity and quantum mechanics don't agree.

John Rennie

Physics ought to be fun. The trouble with the exam system is that it takes something that should be fun and makes it an ordeal instead.

3

Nobody recognizeable

Can give a lot of things.

Abcd

@JohnRennie But black hole and all have no significance in real world. Why care and waste one's energy?

Nobody recognizeable

And if i say special relativity is useless then it is for our slow moving classical particles but not for the fast ones.

John Rennie

@Abcd I have the spare time and energy. One of the few good things about being old :-)

3

Abcd

16:34

@JohnRennie what about youngsters who study it?

Dante

@JohnRennie I wish I get old fast :-/

Nobody recognizeable

@JohnRennie you arent 60 yet . You aren't old yet. @Abcd because they wanna be old studying it:p

Abcd

@Nobodyrecognizeable i dont find these answers practical.

Nobody recognizeable

@Abcd do you know Einstein?

Abcd

@Nobodyrecognizeable yes, who doesnt.

Nobody recognizeable

16:37

@Abcd do you know stephen hawking

Abcd

@Nobodyrecognizeable yes, who doesnt.

Dante

@JohnRennie A doubt in thermodynamics

John Rennie

@Abcd I don't know that much about modern India - I was last there 25 years ago. But I think there is a huge culture difference between India and Europe/USA. In the West if you find something like general relativity fascinating there's a reasonable chance someone will pay you to study it. So you can do what you find fascinating and still eat.

Nobody recognizeable

@Abcd do you know their friends who used to play or stay with them or can you name anybody from 19 th century ?

Dante

John Rennie

16:39

@Nobodyrecognizeable Ramanujan!

Abcd

@JohnRennie that is really amazing about your place. Nothing like that in India. People who do higher studies here generally starve.

Dante

People study here only for a good job lol.

It's more like a competition that education.

Abcd

@Dante just identify all the processes and use the formulas. very simple problem.

John Rennie

@Abcd by the time you're sending your children to college India will be as rich as the West and your children will have the options that today's Indian students don't.

3

Dante

@Abcd Lemme give it a try again.

John Rennie

16:42

@Dante you'll have to catch me tomorrow morning. I'm finished for the day now.

Nobody recognizeable

@JohnRennie @Abcd so for conclusion. You or I or everyone knows the guys who succeeded to science progress then why shouldn't youngter want to see him in the same place?

John Rennie

@Nobodyrecognizeable well, it's nice to eat ...

Abcd

@Nobodyrecognizeable whats the point in being known after your death or even during your lifetime?

One can just stay a "nobody recognisable" and continue their life :)

Nobody recognizeable

@Abcd might be you can get a fraction of fame on being alive also. Einstein did get his nobel and the respect embedded in it.

Abcd

@Nobodyrecognizeable whats the point of fame/reputation etc? Its useless during lifetime as well.

Nobody recognizeable

16:46

@JohnRennie in higher studies good students get monitary help here also. But i have to admit most of them go out for phd or so.

@Abcd well short fame is regarded as a position in society as a job holder big fame is only some guys get.

Dante

@Abcd I don't get it. I'm really bad at thermodynamics :/

Nobody recognizeable

And also everyone has their personal thinking process . And yeah also i can bet competition comes for everyone . So go as you like.

Abcd

@Dante I am busy with maths. You may ask sammygerbil... I find thermodynamics one of the easiest parts of physics though...tastes differ.

Dante

Oh, ok.

Is there anyone else who could help me?

Nobody recognizeable

Since a professor of astrophysics gets 70000 inr a month. I dont think its bad either.

@JohnRennie what do you think.

Dante

16:52

@sammygerbil ?

Or @LoopBack ?

Loop Back

Yes

Dante

Could you help me with that? You seem to be really good at thermodynamics

How do I bump a message in the chat box?

Nobody recognizeable

@sammygerbil @JohnRennie have a nice day professors,goodbye.

Loop Back

@Dante what's you doubt

sammy gerbil

@Nobodyrecognizeable Goodbye.

Dante

16:56

@LoopBack Here

I don't understand how to start the problem.

Since it's a cyclic process, dU must be zero

Isothermal process have dU=0

So sum of dU of adiabatic processes must be 0

That is what I think

But I do not know how to calculate dU for adiabatic process

Loop Back

@Dante answer is D

@Dante If you carefully notice then you will find that the temperature of the source is 1000K and the sink is 250 K

So might be knowing that $$\dfrac{W}{Q_1}=1-\dfrac{T_2}{T_1}$$

Where W is the work done by the gas when it takes heat from the source. $Q_1$ is the heat absorbed from the source.

$T_1 \text{ and } T_2$ are respectively the temperatures of source and the sink.

As internal energy remains constant during the process AB, work done by the gas on it's surrounding is equal to the heat absorbed from the source. So value of $Q_1$ comes out to be 400J.

Now the usefull work done by the done by the gas equals 300J.(this is obtained from the above equation).

So the heat rejected to the sink will simply be given by $$Q_2=Q_1-W$$ $$\therefore Q_2=100J$$.

$Q_2 $is the heat given to the sink, or heat given off in the process CD

Note:- According to second law of thermodynamics, the entire heat $Q_1$ given to the working gas by the source cannot be used to do work.

Abcd

20:45

@sammygerbil Are you there? ...I have a few problems.

sammy gerbil

21:07

@Abcd hello

Abcd

@sammygerbil A parallel beam of uniform monochromatic light of wavelength 2640 A has intensity of 200 W/m^2 . The number of photons in 1mm^3 of this radiation is?

sammy gerbil

@Abcd Any ideas?

Abcd

@sammygerbil $E_1 = hc/\lambda, I = E/At$

sammy gerbil

@Abcd What is $t$? How would you define or calculate that?

Abcd

21:22

@sammygerbil time, 1 second.

@sammygerbil Please reply I have to do other questions as well.

sammy gerbil

@Abcd 200 W = 200 J/s. How many photons per second?

Abcd

@sammygerbil damn it! I messed up intensity and power

sammy gerbil

The photons are travelling at speed $c$. So in 1s the photons in a box of length of $c$ metres are incident on a 1m^2 area. The energy calculation gives you the number of photons, divide by the volume of this box to get density.

Abcd

@sammygerbil 26.6 * 10^19 ? Right?

sammy gerbil

@Abcd Probably. Do you really need me to check? Does the question have an answer which you can check?

Abcd

21:35

got it thanks.

@sammygerbil Please see part (b) w/ attempt^^^

sammy gerbil

@Abcd If the source is further away, do you expect the current to be smaller or larger?

Abcd

@sammygerbil even I expect it to be smaller but i have no clue why the equations are suggesting otherwise

sammy gerbil

Saturation current depends on the number of photons reaching the photoelectric cell per second.

Abcd

@sammygerbil In my attempt n denotes number of photons and A denotes area.

sammy gerbil

@Abcd The total number of photons crossing a spherical surface is the same at any radius.

Abcd

21:45

@sammygerbil whats wrong in my attempt?

$I = E/At$

E = $n h\nu$

sammy gerbil

You should have $i \propto 1/d^2$ not $i \propto d^2$

Abcd

But please tell how to derive it.

Why has my attempt gone wrong?

Whichformula have I misused?

sammy gerbil

If $N$ is the total number of photons emitted by the source per second, and $A$ is the fixed area of the detector, then the number of photons reaching the detector per second is $$n=\frac{A}{4\pi r^2}N$$

Saturation current is proportional to $n$, so $i \propto 1/r^2$ where $r$ is distance between source and detector.

Abcd

@sammygerbil Why $A/ 4\pi r^2$?

sammy gerbil

@Abcd $N$ photons cross a spherical surface of area $4\pi r^2$. The fraction of this area which is covered by the detector is $\frac{A}{4\pi r^2}$.

Abcd

21:53

@sammygerbil got it thanks.

sammy gerbil

So the fraction falling on the detector is $n=\frac{A}{4\pi r^2}N$.

Abcd

@sammygerbil so answer should be 18/ 9 ?

Yes thats correct.

sammy gerbil

@Abcd Yes.

Abcd

yesterday, by Abcd

16 mins ago, by Abcd

@JohnRennie potential difference across X ray tube is increased. Will intensity increase or decrease?

yesterday, by Abcd

@JohnRennie why does overall intensity increase?

@sammygerbil Please tell the reason.

sammy gerbil

@Abcd The PD is that between the cathode (where electrons are produced by heating) and the anode (to which they are attracted and accelerate). Increasing PD increases the KE of electrons arriving at the anode. This increases the energy and intensity of X rays produced by bombardment.

Abcd

21:58

@sammygerbil didnt get last sentence.

sammy gerbil

Higher energy electrons can knock out electrons which are more tightly bound to atoms.

Other electrons fall back into the vacant energy level and emit an X photon.

Another way intensity can increase is when the current from the cathode has not reached saturation.

Then the increased PD drags more electrons away from the cathode, and they reach the anode quicker because they are moving faster. So there are more X photons generated per second.

@Abcd So there are 2 ways in which increased PD can increase intensity of X rays : (1) increase in energy of each X photon emitted, (2) increase in the number of photons emitted per second.

Abcd

@sammygerbil i think you mean electron

sammy gerbil

22:14

@Abcd No I do mean the number of photons emitted by the anode. Which of course equals the number of electrons reaching the anode.

The intensity of any beam of EMR depends on the energy of each photon and the density of photons.

Abcd

@sammygerbil what happens to electron after striking another electron during x ray emission.

sammy gerbil

@Abcd It loses energy and wanders around inside the metal anode. It might fall into an atom which has a vacancy in its energy levels. Otherwise it is conducted back to the cathode. The anode is connected back to the cathode via a power supply, so these electrons are dragged back to the cathode where they can be 'evaporated' again and be accelerated towards the anode etc.

Abcd

@sammygerbil nice

sammy gerbil

@Abcd The bombarding electron ejects a bound electron from the atom which it strikes. An X photon is not emitted until a free electron in the metal falls into the vacant energy level.

Abcd

@sammygerbil y

sammy gerbil

22:36

@Abcd I have missed out the most important process for X ray generation : braking radiation. Most of the X rays are produced when the fast electrons are rapidly decelerated in the anode. The X rays which come from the ejection of electrons from atoms is a minor process, but it is responsible for the spikes in the energy spectrum.

See radiologycafe.com/radiology-trainees/frcr-physics-notes/…

Abcd

@sammygerbil y = why.

sammy gerbil

@Abcd Apart from braking radiation, photons are emitted by atoms when an electron falls to a lower energy level.

Abcd

@sammygerbil Question with attempt^^^^. Answer given is 2.48, my answer is 2.4

sammy gerbil

@Abcd Isn't that close enough?

The data is given to only 2 decimal places, so the answer should be rounded to 2dp.

Abcd

@sammygerbil See the values given are so precise and neat...so there must be something that I am missing which is why the actual answer is 2.48

sammy gerbil

22:51

The values given don't seem precise and neat to me. Correct mass of electron is 0.511MeV. Taking this as 0.5MeV is very approximate indeed.

Dante

@LoopBack Thanks a lot!

sammy gerbil

I don't see how they get 2.48 using the figures given.

(Actually electron mass is given to only 1dp accuracy, so even 2dp in the answer may not be justified.)

Abcd

23:08

@sammygerbil doubt

sammy gerbil

@Abcd yes?

Abcd

@sammygerbil first part unfortunately.

@sammygerbil Attempt is:
$N = \dfrac{99}{2}\times 10^{23}$
$\implies n = N \times \dfrac{1}{4}\left(\dfrac{r_{atom}}{r_{distance}}\right)^2$

sammy gerbil

@Abcd Can you explain your calculation? I don't understand it.

Abcd

@sammygerbil N is total number of photons emiited in second.

Area of atom is $\pi r^2$

surface area of full sphere around the point source is $4\pi r^2$

1 hour ago, by sammy gerbil

If $N$ is the total number of photons emitted by the source per second, and $A$ is the fixed area of the detector, then the number of photons reaching the detector per second is $$n=\frac{A}{4\pi r^2}N$$

sammy gerbil

@Abcd ok. How are you getting $N$?

Abcd

23:16

@sammygerbil $N = \dfrac{P\lambda}{hc}$

Dante

@sammygerbil Hi! I'll drop a question here, please reply whenever you're done with Abcd.

sammy gerbil

Your method is correct. I assume you have calculated $N$ correctly.

@Dante ok

Abcd

@sammygerbil Answer is 5/16 :(

sammy gerbil

@Abcd $n=5/16$ per second? That cannot be correct.

What answer do you get for $n$?

Abcd

@sammygerbil why not, an atom is small so only fractional of photons might be striking it.

@sammygerbil very large

sammy gerbil

23:23

@Abcd Good point. What is your answer?

Abcd

@sammygerbil $= 25\times 99000/8$

sammy gerbil

@Abcd Check your calculation for $N$.

Abcd

@sammygerbil i have checked it thrice already

Dante

@sammygerbil How do I make out when I do I have to apply Kirchoff's law for a single loop and when do I have to find potential difference between two same points using different paths to get the unknown?

sammy gerbil

@Abcd I get N = 5 x 10^18 per second.

Dante

23:33

Look at problem 18 and 19, they've done both in different way.

Abcd

@sammygerbil I might be making some error because i am very tired atm.

sammy gerbil

@Dante Some problems can be solved by different methods. You can choose whichever method you like, provided it works. There are no problems in which you "have to" use a particular method - unless you are told to do so.

Dante

@sammygerbil I used 19th's method for 18, it didn't work, I thought there must be some reason.

sammy gerbil

@Dante Most likely you made a mistake in applying the method or in the calculations.

Mauricio Mendes

Hello everyone! Could anybody help me solve the "car and bicycle rider" from this problem set (ocw.mit.edu/courses/physics/…), please? Specifically, part b. I have integrated $a_c$ to get $v(t)$ and $x(t)$, but I have no idea how to manipulate these functions and determine $v_b$. Such functions are $$v(t)=-c(\frac{t^2}{2}-t_1t)+v_0-\frac{ct_1}{2}$$ and $$s(t)=-\frac{ct^3}{6}+\frac{ct_1t^3}{3}-\frac{ct_1t}{2}+v_0$$

Dante

23:43

@sammygerbil Did it work out for you?

sammy gerbil

@Dante I have not tried.

Dante

I don't think I made a calculation mistake, I got 3 equations, from 3 different paths, when I tried to solve them, I got LHS = RHS :/

All got cancelled and I Couldn't find any unknown

sammy gerbil

@Dante Do you want to show how you did the calculation?

Dante

It's kind of a rough work, is it ok?

sammy gerbil

:47350400 As long as I can read it.

Dante

23:50

@sammygerbil I don't think you can cuz I myself can't ;p Hold, on, let me solve clearly

sammy gerbil

@Dante There are no EMFs in the circuit for #19. If you want to apply the method of loop currents used in #18 you need to draw an extra circuit loop between a and b and including a source of EMF.

Dante

There's a mistake in this I suppose

@sammygerbil Oh ya, I was specifically talking about using 19th's method for 18th

Transcript for

Problem Solving Strategies