Problem Solving Strategies

Nobody recognizeable

00:10

@sammygerbil under development this drive.google.com/file/d/1wfsejemkevLHkjPlJDuphd6pu8YvLhPz/… is helpful.

@sammygerbil so is this drive.google.com/file/d/1Y8lsaLE0cNHalrFdNiP64VSu8OBix8gy/…

Nobody recognizeable

05:00

@JohnRennie are you here ?

John Rennie

@Nobodyrecognizeable morning :-)

Nobody recognizeable

chat.stackexchange.com/transcript/message/47316331#47316331

John Rennie

Just use conservation of angular momentum

Nobody recognizeable

@JohnRennie please solve it using angrangian mechanics.

John Rennie

Why on Earth would I use Lagrangian mechanics to solve that?

I'm not even sure you can as I'm not sure how to write down the action for that system.

Nobody recognizeable

05:06

@JohnRennie ok then lets go with conservation of momentum.

John Rennie

@Nobodyrecognizeable you don't need help with that, surely?

Nobody recognizeable

@JohnRennie yep. I was thinking that how to solve that using langrangian mechanics.

@JohnRennie are you somehow angry with me ?

John Rennie

I don't get angry with students for asking questions. I do think you should consider your questions more carefully. This chat room works well when members have a clearly stated question that they have already thought about.

If you are interested in learning Lagrangian mechanics then I think this is an excellent aim because it's an important skill if you want to work in physics or engineering. But the best way to do this is find some suitable study material and go off and study it.

Nobody recognizeable

@JohnRennie ok . I will make that note.

John Rennie

If you have specific questions about Lagrangian mechanics then I'm happy to try and help, though it's a long time since did those sorts of problem so my memory is likely to be rusty.

Nobody recognizeable

05:16

@JohnRennie ok. Im coming back with another question. Please wait.

@JohnRennie question number 16.

John Rennie

Have you already attempted this? How far did you get?

Nobody recognizeable

@JohnRennie if the angular velocity is $\omega$ then the velocity is higher in the farther points .

John Rennie

The way I would approach this is to write the equation for the angular acceleration. From this you can calculate the vertical component of the linear acceleration as a function of distance along the metre rule.

Nobody recognizeable

On the way i cant think what would happen if gravitational acceleration comes in the place also.

John Rennie

The angular acceleration is given by $T = I\alpha$, where $T$ is the torque $T = 0.5LMg\cos\theta$ and $\theta$ is the angle to the horizontal.

Nobody recognizeable

05:27

@JohnRennie why 0.5 factor comes here ?

John Rennie

$L$ is the length of the rod so the distance from the pivot to the centre of mass of the rod is $L/2$.

Nobody recognizeable

@JohnRennie fine.

John Rennie

Once you know the angular acceleration you can calculate the tangential (linear) acceleration, then multiply this by $\cos\theta$ to get the vertical component. The question is then how this compares with $g$.

user64829

@Nobodyrecognizeable Is the answer C?

Nobody recognizeable

@JohnRennie I =$mL^2 /12$

John Rennie

05:32

If the vertical component of the acceleration is greater than $g$ the coins will come off the rod, as in (A) to (C), while if is less than $g$ the coins will stay on the rod, as in (D).

@Nobodyrecognizeable the rod is rotating about its end. The value of $I$ you have given is for a rod rotating abut its centre.

For a rod rotating about one end $I = mL^2/3$

user64829

Ignore the message above, it must be B.

Nobody recognizeable

@JohnRennie right i integrated between L/2 TO -L/2 where i had to do it from 0 to L.

@JohnRennie acceleration comes out as 1.5gcos$\theta$

John Rennie

@Nobodyrecognizeable I got $1.5 (g/L) \cos\theta$

Nobody recognizeable

@JohnRennie that is the angular acceleration but if you multiply L to get linear acceleration you get the tangential acceleration. Although i had to specify it.

@JohnRennie if you multiply another cos$\theta$ then how to compare with g?

John Rennie

I need to work for five minutes ...

Nobody recognizeable

05:47

@JohnRennie ok .

@JohnRennie when you come back ,ping me.

Nobody recognizeable

06:00

@JohnRennie are you back ?

John Rennie

@Nobodyrecognizeable sorry, that took longer than I expected.

We can get rid of the $\cos\theta$ term as follows ...

The condition we are looking for is that the downwards acceleration due to the angular acceleration of the rod is greater than $g$. The angular acceleration is greatest when $\cos\theta=1$ i.e. when the rod first starts rotating. So we look at the acceleration at this moment.

That means the angular acceleration at the moment the rod starts rotating is just $$ \alpha = \tfrac{3}{2} \frac{g}{L} $$ Yes?

Nobody recognizeable

@JohnRennie surely.

John Rennie

Consider the coin at a distance $x$ from the pivot, where $0 \le x \le L$

Nobody recognizeable

@JohnRennie x=2L/3

@JohnRennie @user64829 answer is b.

John Rennie

Yes, the value of $x$ for which the vertical acceleration exceeds $g$ is $x = 2L/3$. So for all values of $x$ greater than this the rod is accelerating downwards faster than the coin.

So, yes, the answer is (B).

user64829

06:09

Yeah

Nobody recognizeable

@JohnRennie really really thanks, professor.

user64829

@JohnRennie I have a doubt on same topic

Hold on, let me find that question

John Rennie

@user64829 Yes?

user64829

Question is 5-6 lines and I don't have my cell atm to take it's picture, do you have I.E Irodov nearby in any case?

It's from that book itself

If not, I have no problem typing it

John Rennie

Yes, I have a copy of Irodov. Which question?

user64829

06:11

mechanics

245th question

John Rennie

That one?

user64829

Yeah

I got it as $sqrt {6Fϕ/ml}$

I didn't understand how Sine function came there.

John Rennie

Looks like a conservation of energy problem to me.

The torque is $T = F\ell\cos\phi$ so the work done at an angle is:

$$ W = \int_0^\phi F\ell\cos\phi' d\phi' $$

And that work has to be equal to the kinetic energy $\tfrac{1}{2}I\omega^2$

user64829

@JohnRennie How's that? The force is always perpendicular to the rod right? Why did you take Cos phi?

John Rennie

> The force is always perpendicular to the rod right

user64829

06:20

Yes

John Rennie

Wrong! Read the question :-)

user64829

Oh Damn!

Sorryy,

Hold on sir, I'll try it again, misread the question

Nobody recognizeable

Lol i read the same.

John Rennie

I need to work now for about half an hour.

user64829

Okay, Np

Let me know when you're back

Nobody recognizeable

06:22

Sure.

user64829

@Nobodyrecognizeable Hehe

Nobody recognizeable

@user64829 $\omega =\frac { \sqrt {6Fsin\phi}}{\sqrt l}$

user64829

Wrong :-(

Nobody recognizeable

@user64829 whats the answer?

user64829

Your answer/ sqrt {m}

Nobody recognizeable

06:35

@user64829 didn't want to feel heavy so didnt give the $\sqrt m$ so anyway its my fault.

user64829

m was in denominator so it would've made it lighter ;p

Nobody recognizeable

Sure.

ostrichguy

anyone help me with it?

Option A is not true. But im confused with the other ones

Nobody recognizeable

@ostrichguy with this only johnrennie may be able to help you. He has to work for about 10 minutes more. Then you can ask. Though in the main site higher level questions arent regarded as check my work question. You can post it on the main site or wait a while.

John Rennie

@ostrichguy the probability distribution is just $\psi^*\psi$. You can write that down and show that it is not constant with time.

user64829

06:49

@JohnRennie I got that problem, Thank you!

John Rennie

@user64829 cool :-)

ostrichguy

@JohnRennie but the first two eigenstates are in UNEQUAL linear superposition

John Rennie

@ostrichguy at a guess the third point is true on the grounds that $\psi_2$ has a node there. I'd have to write down the wavefunction to be sure though.

@ostrichguy so ... ? Just write down the wavefunction with $a$ and $b$ as free parameters

ostrichguy

yes you are right. I has a node at the centre

If there is a node at the center would that mean that the probability density is independent of time?

Abcd

Morning @JohnRennie ?

ostrichguy

06:53

at the center?

Abcd

@user64829 your gravatar changed again :P

John Rennie

@ostrichguy We write $\Psi = a\psi_1 + b\psi_2$ so the probability density is $(a\psi_1^* + b\psi_2^*)(a\psi_1 + b\psi_2)$. Yes?

@Abcd morning

user64829

Ik lol, it takes a random pattern everyday

Abcd

@JohnRennie busy with ostrich's question or can I ask?

John Rennie

@Abcd I think we're just about to finish Ostrich's question ...

ostrichguy

06:56

@JohnRennie yes

what what does it indicate when we are at UNEQUAL superposition of first 2 eigenstates?

John Rennie

@ostrichguy and at the centre of the box $\psi_2(t) = \psi_2^*(t) = 0$

ostrichguy

agreed

aahhhhh...

John Rennie

@ostrichguy BOOM! :-)

ostrichguy

WHabaaam\

probability densities at the nodes are independent of time

isn't thatt so

that*

John Rennie

If there is a node then the probability distribution is zero there.

Abcd

06:59

@user64829 since you have disclosed your real name already you can keep that or something related as username to get a fixed gravatar.

John Rennie

But note that the combined wavefunction $a\psi_1 + b\psi_2$ has no nodes.

ostrichguy

So I should go for option c ? or NONE OF THE ABOVE?

John Rennie

@ostrichguy I thought we'd answered this ...

5 mins ago, by John Rennie

@ostrichguy We write $\Psi = a\psi_1 + b\psi_2$ so the probability density is $(a\psi_1^* + b\psi_2^*)(a\psi_1 + b\psi_2)$. Yes?

ostrichguy

Yeah okay thanks @JohnRennie

John Rennie

If $\psi_2=0$ the probability distribution reduces to $\psi_1^*\psi_1$

That's constant because the time dependent term cancels when we multiply the wavefunction and its conjugate

I can do the sum in full if you want ...

2

ostrichguy

07:04

Nah, It's okay professor. Thanks a lot :-) I get it

2

John Rennie

@Abcd finished now

Abcd

Oct 19 at 20:30, by sammy gerbil

Oct 19 at 20:58, by sammy gerbil

@Abcd Returning to the last question, I have an idea : Suppose the neutron has momentum $p$ in the laboratory frame of reference. Then its KE is $K=p^2/2m$ where $m$ is its mass. In the centre-of-mass frame its momentum is approximately $p/2$ because it has approx the same mass as the hydrogen atom. Same for the H atom. So the initial KE is now $2\times \frac{(p/2)^2}{2m}=p^2/4m=K/2$.

John Rennie

So is it just (D) you're unsure about?

Abcd

@JohnRennie No (A)

@JohnRennie Sammy gerbil has the right idea of working in COM frame, but I think transition energies should also change in COM frame...

John Rennie

(A) is clearly false because the neutron could excite the hydrogen atom e.g. from $n=1$ to $n=2$. Then the collision would be inelastic and the energy required would be < 13.6eV.

Abcd

07:09

@JohnRennie Answer is A :P

Oct 19 at 21:09, by sammy gerbil

@Abcd In COM frame the collision can be totally inelastic. This happens when both particles come to rest. Initial KE is then converted in total to internal energy. In this frame initial KE is $K/2=6.8eV$. This is too small to excite H atom above ground state energy level. The next lowest level is $10.2eV$ above it. The gap is too high.

John Rennie

@Abcd then it's wrong

ostrichguy

Abcd

@JohnRennie can you please see SG's reasoning in the above two messages?

ostrichguy

thanks a lot professor :)

Abcd

Oct 19 at 20:58, by sammy gerbil

@Abcd Returning to the last question, I have an idea : Suppose the neutron has momentum $p$ in the laboratory frame of reference. Then its KE is $K=p^2/2m$ where $m$ is its mass. In the centre-of-mass frame its momentum is approximately $p/2$ because it has approx the same mass as the hydrogen atom. Same for the H atom. So the initial KE is now $2\times \frac{(p/2)^2}{2m}=p^2/4m=K/2$.

Oct 19 at 21:01, by sammy gerbil

So if $K<13.6eV$ in the lab frame then the energy available to excite the atom is only half that, ie $6.8eV$. This is too small to reach the 1st excited state, which requires $10.2eV$.

Oct 19 at 21:01, by sammy gerbil

So a neutron energy less than $13.6eV$ guarantees that the collision will be elastic. Option A.

John Rennie

07:11

Ah, OK, fair enough. I was forgetting that we have to conserve momentum as well. Yes, Sammy is correct.

Abcd

@JohnRennie My doubt is in the second (middle) message of sammy gerbil above.

John Rennie

OK, what's the problem with it?

Abcd

@JohnRennie Wont energies required for transition from 1 state to other be frame dependent?

John Rennie

No. The 1s to 2p transition is always 10.2eV. Why would it be frame dependent?

Abcd

@JohnRennie kinetic energy is not frame dependent?

$|E| = |KE|$ for nth orbit of Hydrogen atom.

John Rennie

07:14

The 1s to 2p is an electronic transition in the hydrogen atom. It has nothing to do with kinetic energy.

Abcd

14 secs ago, by Abcd

$|E| = |KE|$ for nth orbit of Hydrogen atom.

@JohnRennie ^^^

John Rennie

@Abcd the electron in an atom isn't a point particle like a planet orbiting a star. It doesn't have a kinetic energy - it just has a total energy. In the Bohr model it has a kinetic energy, but we all know the Bohr model is wrong.

Abcd

@JohnRennie in our modern physics syllabus we just have Bohr's model so I am discussing on its basis at the moment.

@JohnRennie ^^^

John Rennie

The Bohr model is wrong. If you're using the Bohr model to claim electronic transition energies are frame dependent you'll get the wrong answer.

user64829

@Abcd Done!

Abcd

07:17

@JohnRennie OK, but not even total energies? How can energy not be frame dependent?

John Rennie

That seems an odd question to me. Kinetic energies are frame dependent, but only kinetic energies. The energy of an excited hydrogen atom isn't a kinetic energy.

Abcd

8 mins ago, by Abcd

Oct 19 at 21:01, by sammy gerbil

So a neutron energy less than $13.6eV$ guarantees that the collision will be elastic. Option A.

@JohnRennie One doubt wrt this message^

@JohnRennie Just because a neutron isn't able to excite an electron, how does it imply collision will be elastic? What is the definition of elastic collision here?

John Rennie

Elastic collision means KE before = KE after

Abcd

oh then got it

John Rennie

The KE can only change if there is some mechanism for converting KE to some other form of energy e.g. electronic energy of the atom.

Abcd

07:23

@JohnRennie so is it possible that neturon goes to some other direction and hydrogen goes to some other direction with different velocities such that the sum of KEs comes out to be equal to the initial KE

John Rennie

@Abcd remember that you need to conserve momentum as well.

Abcd

oh

@JohnRennie so? what will be the result?

John Rennie

The neutron and proton can scatter at an angle, but (in the COM frame) their velocities must always be opposite so the momentum sums to zero.

Abcd

@JohnRennie so both will have velocities v/sqrt 2 in oppsoite directions where v is initial velocity of neutron?

opposite*

John Rennie

In the COM frame the neutron and H atom have initial velocities of $+v/2$ and $-v/2$ (approximately because the neutron and H masses are almost identical).

Abcd

07:29

oh i was talking from ground frame.

John Rennie

So after an elastic collision they must have velocities with a magnitude of $|v/2|$ in opposite directions, but not necessarily in the original (prescatter) direction.

Abcd

@JohnRennie please C ^^^

John Rennie

If you switch back to the ground frame by adding $\mathbf v/2$ to the velocities then the velocities of the two bits won't be opposite.

2

Abcd

@JohnRennie oh okay!

John Rennie

@Abcd I can't believe they are asking questions about the Bohr model. Good grief. Are they supposed to be teaching you physics?

Nobody recognizeable

07:32

John Rennie

Anyhow there's a sneaky shortcut you can use. You can take the proton mass into account by replacing the electron mass with the reduced mass:

Abcd

@JohnRennie its important to learn it, isn't it? This is the last part of JEE physics syllabus under "Modern physics".

John Rennie

$$ \mu_e = \frac{m_e m_p}{m_e + m_p} $$

Nobody recognizeable

John Rennie

@Abcd no it isn't. It is of historical interest only. I guess if you're going to be asked questions about it then you have to learn it. But no university physics course will teach the Bohr model.

Nobody recognizeable

07:35

Abcd

@Nobodyrecognizeable dude answer is A despite that formula

@Nobodyrecognizeable I am aware of these formulas. Dont post.

Nobody recognizeable

@Abcd fine,sorry.

Abcd

@JohnRennie pleasetell how to do this question

John Rennie

@Abcd when you're calculating orbits and energies in the Bohr model you use the electron mass. To take the proton mass into account replace the electron mass by the reduced mass.

Nobody recognizeable

@Abcd A is always true as you have Z in denominator for r.

Abcd

07:38

@Nobodyrecognizeable Z of deuterium is same as Z of hydrogen.

John Rennie

@Nobodyrecognizeable hydrogen and deuterium have the same value for Z

Abcd

@JohnRennie didnt get

John Rennie

@Abcd which bit didn't you get? What the reduced mass is?

Nobody recognizeable

@JohnRennie yep. Sorry.

John Rennie

@Nobodyrecognizeable if you feel the urge to keep a dignified silence don't let us stop you

Abcd

07:39

we were taught that after subsituting the value of constants the radius of nth orbit of hydrogen like atoms is:
$0.529 \dfrac{n^2}{Z} \pu{Angstrom}$

Loop Back

@Nobodyrecognizeable you there

Abcd

So how can (A) be true?

Nobody recognizeable

@LoopBack yep.

Loop Back

@sammygerbil 's solution is incorrect

John Rennie

That value 0.529 is the Bohr radius in Angstroms. If we write the equation for the Bohr radius we get:

Loop Back

07:42

@Nobodyrecognizeable He had done a mistake. And if you correct that mistake you will get answer as B

John Rennie

$$ a_0 = \frac{4 \pi\epsilon \hbar^2}{m_e e^2} $$

That's derived by equating the electrostatic and centripetal forces for the $n=1$ orbit.

Abcd

@JohnRennie OK?

@JohnRennie yes I am aware of the derivation.

John Rennie

So the key point is that $a_0 \propto 1/m_e$

To take the nuclear mass into account we need to replace $m_e$ by the reduced mass, so we get $a_0 \propto 1/\mu$

Nobody recognizeable

4 mins ago, by John Rennie

@Nobodyrecognizeable if you feel the urge to keep a dignified silence don't let us stop you

@JohnRennie what do you mean by that ?

Abcd

@JohnRennie Not getting why we are doing this.

John Rennie

07:45

@Abcd why we replace the electron mass by the reduced mass?

Abcd

@JohnRennie yes

John Rennie

I don't think there is an easy way to prove this. Unless you want to grind through the calculations you'll just have to accept it.

Abcd

@JohnRennie and when do we do this?

Nobody recognizeable

Lengthening Pendulum Problem #19 : I think I have solved it. Angular momentum is conserved according to eq 3 on p 268 of the following article : audiophile.tam.cornell.edu/…. So at the lowest point we can write Lv0=(L+δ)v1Lv0=(L+δ)v1 or v_0^2=(1+x)^2v_1^2wherewherev_0, v_1arethevelocitiestoleftandrightofthelowestpointandarethevelocitiestoleftandrightofthelowestpointandx=\frac{\delta}{L}$.

On the LHS of the swing the mass falls a distance h0=L(1−cosθ0)h0=L(1−cos⁡θ0) so that v20=2gh0v02=2gh0. On the RHS of the swing the mass rises through a height of h1=(L+δ)(1−cosθ1)h1=(L+δ)(1−cos⁡θ1) when …

Loop Back

@Nobodyrecognizeable sorry. @sammygerbil's answer is correct

Nobody recognizeable

07:48

@LoopBack ok fine then.

John Rennie

@Abcd actually you should always use the reduced mass. It's just that because $m_p \gg m_e$ the reduced mass is very close to the electron mass so it's a good approximation to just use the electron mass. The point of this question is that it specifically takes the nuclear mass into account.

Abcd

@JohnRennie does m_p include the mass of both proton and neutron?

Loop Back

@Nobodyrecognizeable but have something to ask

John Rennie

@Abcd Yes. I guess I should have written $m_N$ for the mass of the nucleus

Abcd

@JohnRennie OK, what's the next step?

John Rennie

07:52

If we calculate the reduced masses we'll find $\mu_H < \mu_D < m_e$. It should be clear this is the case if you write out the equations for the two reduced masses.

And remember that $a_0 \propto 1/\mu$

Nobody recognizeable

Lengthening Pendulum Problem #19 : I think I have solved it. Angular momentum is conserved according to eq 3 on p 268 of the following article : audiophile.tam.cornell.edu/…. So at the lowest point we can write Lv0=(L+δ)v1Lv0=(L+δ)v1 or $v_0^2=(1+x)^2v_1^2$ wherewhere $v_0, v_1$ are the velocities to left and right of the lowest point and are the velocities to left and right of the lowest point and $x=\frac{\delta}{L}$.

On the LHS of the swing the mass falls a distance h0=L(1−cosθ0)h0=L(1−cos⁡θ0) so that $v_0^2=2gh_0v0^2=2gh_0$. On the RHS of the swing the mass rises through a height of h1…

Abcd

$\mu = \dfrac{m_e m_N}{m_e + m_N} = \dfrac{m_e}{1+ \dfrac{m_e}{m_N}}$

@JohnRennie right.

John Rennie

Yes. If $m_N \to \infty$ then we get $\mu = m_e$, but for finite $m_N$ we always get $\mu < m_e$.

Abcd

@JohnRennie what about part B C and D?

John Rennie

Offhand I can't remember the equation for the speed in a Bohr orbit, but I assume it is a function of mass. I'd guess it's proportional to $\sqrt{m}$.

Abcd

07:58

@JohnRennie wait i have it

Hmm that one^

Nobody recognizeable

@JohnRennie ^^^

John Rennie

That's the frequency not the velocity

Abcd

lol

posting mine then"

$v_o = \dfrac{nh}{2 \pi m r}$

And we know the the value of r

$v_o = \dfrac{nh }{2\pi m \dfrac{n^2h^2}{\pi me^2 Z}}$

@JohnRennie its independent of m

John Rennie

That nice Mr Google tells me $$v = \sqrt{\frac{ke^2}{\mu r}} $$

Abcd

@JohnRennie why is my equation wrong?

@JohnRennie remember that r is inversely proportional to $\mu$

John Rennie

08:03

Ah, yes, and $r \propto 1/\mu$

So the velocity is indeed independent of the mass

Abcd

ok so B is false

John Rennie

@Abcd yes

Abcd

option C?

Energy $\propto \mu$

so energy difference $\propto \mu$

John Rennie

Yes

Abcd

so option C is false

option D:

no

$L = \dfrac{nh}{2\pi}$

from bohr's quantisation rule.

L independent of $\mu$

Rigth @JohnRennie ?

so D false.

John Rennie

08:07

Yes

Abcd

@JohnRennie please tell me why he has written "consider motion of nucleus" in the question.

John Rennie

@Abcd the reduced mass arises because we are taking the motion of the nucleus into account, so including the motion of the nucleus necessarily means you need to use the reduced mass.

Maybe the question expected to to calculate the orbital motion for the two bodies. You can do that by working in the COM frame. It's a bit messy but not too bad.

Abcd

@JohnRennie my teacher was solving this question using Binary star system. No idea what he was doing because I have totally forgotten binary stars.

John Rennie

But you'll just end up with the result that you can simply replace the electron mass by the reduced mass.

@Abcd it's not a hard calculation. Just a bit tedious. Just work in the COM frame and equate the centripetal force to the electrostatic/gravitational force as usual.

Abcd

@JohnRennie can you show me how its done

John Rennie

08:12

OK. I'll draw a diagram. But I'm going to make a coffee first.

Abcd

Okay.

Abcd

08:26

@JohnRennie can we do the derivation tomorrow instead?

John Rennie

@Abcd have a go at it yourself.

Abcd

@JohnRennie i have no idea what binary star is ... and I dont recall doing many problems on it.

so I have few questions wrt it.

John Rennie

The force is $ke^2/(r_e + r_p)^2$ while the centripetal acceleration is $v_e^2/r_e$

Just equate the two, then use $m_e r_e = m_p r_p$

The answer drops out fairly easily.

Abcd

@JohnRennie tomorrow please.

John Rennie

@Abcd OK

Abcd

08:31

@JohnRennie please tell why he has subtracted energy of electrons in $U_i$

John Rennie

Because the calculation starts with the energy of an isolated nucleus i.e. the mass of the atom minus the mass of the electrons in the atom (times $c^2$. Then it ends with the new isolated nucleus plus the newly created electron.

It's done that way because the mass of atoms is easy to measure while the mass of isolated nuclei isn't easy to measure.

Abcd

ok got it

Abcd

08:50

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

John Rennie

@Abcd working at the moment ...

Abcd

09:06

@JohnRennie please tell when its done

John Rennie

@Abcd I'm waiting for some data to restore off a backup, so you can ask now ...

Abcd

16 mins ago, by Abcd

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

John Rennie

Intensity of all radiation or just the intensity of the distinct lines?

Abcd

They have just written intensity

@JohnRennie i think all the radiations.

John Rennie

As a general rule the overall x-ray intensity increases with electron energy. However I think the intensity of a specific line is maximised when the electron energy matches the line energy (though I wouldn't swear to that).

Abcd

09:10

@JohnRennie why does overall intensity increase?

John Rennie

I don't know the reason ...

But obviously if you take the energy down to zero the x-ray intensity goes to zero.

So if you increase the energy from zero the x-ray intensity must increase.

Whether it peaks at some energy I don't know.

@Abcd circular?

Abcd

@JohnRennie I mean I dont like that reason.

John Rennie

myscope.training/legacy/analysis/eds/xrayintensity

That's the intensity of a distinct line not the continuous background.

harambe

09:37

@JohnRennie good morning

John Rennie

@harambe busy atm, sorry

harambe

No problem

Dante

10:57

Hi, anyone?

$f(x) =maximum of g(x),h(x)$

Does it mean $f'(x)=maximum of g'(x),h'(x)$

Dante

11:13

Discuss the differentiability of $f(x)=maximum of (2sinx,1-cosx)$ in the domain (0,pi)

Is there any other method to do this problem than tracing the graph?

John Rennie

@Dante consider the points where $g(x) = h(x)$. If $g'(x)$ is not equal to $h'(x)$ at these points it means the gradient at that point is undefined because it could have two values.

Dante

12:20

Yeah, got it!

Abcd

14:54

@sammygerbil please let me know when you are free

Dante

16:39

@JohnRennie Hello

John Rennie

@Dante I'm finishing for the day now I'm afraid. You need to catch me in the morning.

Dante

Okay

UK's morning?

Nobody recognizeable

@JohnRennie what the meaning of the sentence on the right said by you . I need it to rectify my faults.

sammy gerbil

18:01

@Abcd I am back. Available for the next 6 hours.

sammy gerbil

18:31

@Nobodyrecognizeable No I can only edit my own current posts before they become locked, the same as you. Room Owners can move posts to trash, or ban users from the chatroom for a while. That is all, I think.

Transcript for

Problem Solving Strategies