Problem Solving Strategies

Sarabsri mt

5:32 AM

@JohnRennie hello

Are you free sir

John Rennie

@Sarabsrimt hi :-)

Yes I'm free.

Sarabsri mt

Planck was right
Is there any of his derivation which I can understand
Also , Q 2 is that energy is F * displacement .
So , whose F are we talking about. EM field

@JohnRennie These Q sir

From Q 2 , I mean energy as the energy of light

Are we talking about energy = F*Disp. So , force by EM FIELD or on EM FILED. Or it is photons

John Rennie

You mean Planck was right to say that in light the energy is quantised into photons with energy hf?

Sarabsri mt

Yes. Did he give any derivation

John Rennie

Planck was working on black body radiation. He didn't derive the quantisation of light. He assumed light was quantised and then showed that this gave the correct prediction for the spectrum of black body radiation.

Sarabsri mt

5:37 AM

Ohk. Sir

@JohnRennie q2 sir

John Rennie

The derivation of the black body spectrum is quite complicated but only involves regular calculus, so you could probably follow it. I must admit i don't remember the details.

@Sarabsrimt you mean this one?

13 hours ago, by Sarabsri mt

A question of my institute . There are two person(1 &2)falling down at a separation of 5 m from each other . The 1st person is falling with a velocity u or 0 m/s and 2nd person with velocity 2m/s . Both the persons fall into a deep of hole of 60 m .Find the amount of force exerted on each other . Taking into consideration that person 1 falls on person 2 . Take g as $10m/s^2$.

Sarabsri mt

Q 2 is that energy is F * displacement .
So , whose F are we talking about. EM field

This one sir

@JohnRennie this one I have solved . I will show you the solution.

Did you get the Q sir

John Rennie

What was Q2?

Sarabsri mt

So , we energy light imparts energy

formula of energy is F*Disp

So , whose force are we talking about . EM field or photons.

Also ,is it the force on it or it is the force by it.

I think it is on it

It ( EM FILED or photons )

John Rennie

A single photon carries an energy E = hf and a momentum p = h/λ

Sarabsri mt

5:53 AM

Yes

John Rennie

If a photon hits you an is absorbed then it transfers that moemntum to you, so your momentum changes by h/λ

Sarabsri mt

F * disp = H * f

Or f* Disp = p (lambda) * f

John Rennie

The complication here is that photons are often absorbed in inelastic collisions i.e. some of their energy is converted to heat.

Sarabsri mt

So , we are talking about the EM field carrying 1 photon here with energy = hf

John Rennie

So you cannot simply argue that hf = force x displacement.

Sarabsri mt

5:55 AM

Ohk.

John Rennie

When we calculate the force due to light we need to use the momentum.

Momentum is always conserved, so the calculation using momentum always works.

Sarabsri mt

Ohk. One thing

So , the force in the equation

Is it a force by light or it is force on light

If on light , does it mean on EM field

John Rennie

The force is exerted by the light on the object that it hits.

Sarabsri mt

Ohk. So Disp is of light as well.

John Rennie

The equation energy = force x displacement assumes that all the energy is converted to work, and this often isn't true for light.

Usually all the light is converted to heat not work.

Sarabsri mt

5:58 AM

Ohk.

John Rennie

So the equation doesn't apply.

Sarabsri mt

@JohnRennie ohk.

Then , how can we calculate F . We can write E = p * lambda * frequency

Where , p is mv and not ma

John Rennie

We calculate the force using the momentum.

Shall I explain the calculation?

Sarabsri mt

Force is directly proportional to momentum upon time

So , is it that we divide both eq by t

John Rennie

Yes: F = dp/dt

Sarabsri mt

6:02 AM

Oh. Great

John Rennie

Suppose I shine a light on you with intensity I

Sarabsri mt

Ok

John Rennie

Where the intensity means the energy hitting you per unit time is I joules per second.

Sarabsri mt

yes

John Rennie

Each photon has an energy E = hf, so the number of photons hitting you per unit time is N = I/hf.

OK so far?

Sarabsri mt

6:03 AM

Yes

John Rennie

Each photon has a momentum p = h/λ so the total momentum of all the photons hitting you per second is:

Δp = I/hf x h/λ

Sarabsri mt

Yes

John Rennie

So Δp = I/fλ, and fλ = c

So Δp = I/c

Sarabsri mt

Oh yes

John Rennie

What we've calculated is the momentum change per second i.e. dp/dt, and that is just equal to the force so the force is F = I/c.

Sarabsri mt

6:07 AM

Yeah. Since momentum is always conserved

John Rennie

Yes

Sarabsri mt

@JohnRennie here. Did we replace E with I

John Rennie

I is the intensity of the light beam i.e. the total energy per second of the light beam hitting you.

Sarabsri mt

Ohk

Thanks a lot sir.

Really helpful it was

@JohnRennie

John Rennie

:-)

Sarabsri mt

6:13 AM

I just like to understand things in deep

Just one thing. More is the energy of photon , more brighter is the light or photon right @JohnRennie

John Rennie

A light beam is made up from lots of photons, and each photon has a fixed energy E = hf.

Sarabsri mt

Yes

John Rennie

So when we say the light beam is brighter we mean it contains more photons i.e. increased brightness means more photons per second hitting you.

Sarabsri mt

But if it red light , it has a fixed hf and so does yellow light

but their hf is different

@JohnRennie ohk ohk.

John Rennie

The "brightness" of a light beam means the intensity i.e. the total energy per second.

Sarabsri mt

6:18 AM

Ohk

it is nothing but our amplitude as well.

I got it

John Rennie

And the energy per second is equal to the energy per photon times the number of photons per second.

Ohk ohk

@JohnRennie hi

@AshishAhuja hi :-)

nevermind I spotted my error yet again. It's kind of this weird thing where I think differently while I'm typing it out, sorry.

John Rennie

6:25 AM

I often find as soon as I start explaining a problem to someone else I immediately realise the solution.

satan 29

@JohnRennie very true!!!

John Rennie

I suspect the effort of ordering it in your mind so you can describe the problem helps you spot the solutin.

satan 29

I cant count how many times i have started to write a detailed question on stackex but realised the solution midway

John Rennie

:-)

Sid

@AshishAhuja I suspect you will have the same experience if you write a question down on a paper.

Ashish Ahuja

6:28 AM

Yeah possibly. But I get really lazy when writing stuff on paper and write it down in an abridged way so I guess I'll try typing out my whole question before posting it here or on SE.

Sarabsri mt

Yeah. It happens with me too. Even after lots of thinking alone , I am not able to find the solution . As I start to type it , I get the solution.

Sometimes . Not all the time.

Moreover , when John Rennie sir ask me if I can try it. I usually think in a way I could not think of I would ever do

@JohnRennie sir ,

I / hf = no of photons * h/ lambda (momentum of a photon )

So , in end we got p = I / c

Why do we use delta p.

John Rennie

Yes, where p is the total momentum change per second.

@Sarabsrimt typically we would use p to mean the momentum and Δp to mean the momentum change.

Sarabsri mt

Ohk. But we are finding momentum right,not momentum change there.

@JohnRennie This one sir

John Rennie

In this case the final momentum of the light is zero because all the light is absorbed by your clothes when it hits you, so the initial momentum and the momentum change are the same.

Sarabsri mt

6:43 AM

Ohhhh

inital momentum is just a micro second before touching my clothes ?

John Rennie

Yes

Sarabsri mt

Nice.

one more thing.

@JohnRennie Here

it should be I / c * dt right

$\delta$ p / dt = I / c * dt

John Rennie

I/c is the momentum change per second.

That because I is energy per second i.e. energy/time, so we are already dividing by time.

Sarabsri mt

Ohk. So , isn’t I = E

I is energy per time and E is energy of a photon

I is intensity

John Rennie

No, I is "energy per second" not just "energy"

Sarabsri mt

6:47 AM

Ohk.

@JohnRennie so , here. N is no of photons . So , E / dt = hf / n * dt it should have been. So , n = I*dt / hf

John Rennie

N is number of photons per second hitting you

So it is also already divided by time.

Sarabsri mt

Ohk.

Got it sir

John Rennie

:-)

Sarabsri mt

So Δp = I/c and is F = I/c.

Last Q

how come these two are same sir ?

John Rennie

Δp is the momentum change per second, so it is dp/dt. OK so far?

Sarabsri mt

6:52 AM

Yes . I is also divided by time

got it sir

really thanks a lot

Got it now

John Rennie

:-)

Sarabsri mt

Whenever you’re free , The F = I/ c here means the total force of all the photons. And force of one photon = F / n = I/c

John Rennie

Yes F = I/c is the total force.

The number of photons per second is N = I/(hf). Yes?

Sarabsri mt

Yes.

Right

John Rennie

So I guess you could divide F by N to get hf/c

That's just equal to h/λ i.e. the momentum of one photon.

Sarabsri mt

7:02 AM

Force exerted per photon is F/n . Force exerted per photon in 1sec or unit time is F/N ? Is it correct

@JohnRennie ohk

John Rennie

One photon doesn't really exert a force. It exerts an impulse.

Sarabsri mt

Omg. Ohk

John Rennie

What we are doing is calculating the average force.

Sarabsri mt

@JohnRennie Here, F divided by I / hf I.e F * hf / I = I /c . So , doesn’t cut actually but gets squared up.

John Rennie

So we add up the total impulse from all the photons in one second then divide by 1 second to get impulse/time.

Sarabsri mt

7:06 AM

Ohk.

Ohk. That becomes force.

Thanks sir again

John Rennie

:-)

Robin Singh

7:18 AM

A steel ball initially at a pressure of 10^5 Pa is heated from 20ºC to 120ºC keeping its volume constant. What is the final pressure inside the ball

(Given the coefficient of linear expansion of steel is 1.1×10−5 0C−1 and bulk modulus of steel is 1.6×1011 N/m2)

@JohnRennie Hello sir, this is my question

John Rennie

@RobinSingh hi :-)

The way you do this is to use the expansion coefficient to find how much the ball would expend when heated.

If the linear expansion coefficient is α then the volume expansion coefficient is 3α. Yes?

Robin Singh

Yes sir

John Rennie

So the amount the volume increases is given by ΔV = V 3α ΔT. Yes?

Robin Singh

Yes sir

John Rennie

We are told the volume of the ball stays fixed, so to return it to its original volume V we need to compress it by the same volume ΔV.

And the pressure needed to compress an object by a volume ΔV is given by P = B ΔV/V, where B is the bulk modulus. Yes?

Robin Singh

7:24 AM

Yes sir

John Rennie

So now you can substitute for ΔV in the equation for the pressure and calculate the pressure.

Robin Singh

But sir isn't it already under a pressure of 10^5 Pa? Wouldn't we add that to the bulk modulus equation?

John Rennie

Ah, OK, I should have written the equation as ΔP = B ΔV/V i.e. this gives the pressure change needed to compress by ΔV.

So the final total pressure would be that ΔP plus the initial pressure.

That initial pressure is just one atmosphere i.e. 101325 pascals, isn't it?

Robin Singh

Yes sir

John Rennie

So I guess they are just saying the ball is initially in air at one atmosphere.

So the final pressure is P_f = ΔP + 10⁵

Robin Singh

7:29 AM

Yes sir, thank you :-)

John Rennie

:-)

Sarabsri mt

Energy of a photon = K.E + P.E

So , if this is right.

Then , hv - $hv_0$ = Kinetic energy final + potential energy initial - kinetic energy final - potential energy initial .

So , how do we calculate this then as K.E of photo electron .

John Rennie

When the photoelectron is removed from the metal the potential energy change is the work function and is usually written Φ.

The work function is a property of the metal because it is related to how tightly the electrons are bound into the metal.

So the photon energy is related to the KE of the photoelectron by:

hf = KE + Φ

Sarabsri mt

Ohk.

Sir , is energy of photon = K.E + P.E only right ?

John Rennie

So the KE is just KE = hf - Φ

@Sarabsrimt yes

Sarabsri mt

7:37 AM

Ohk. Where p.e is work function

John Rennie

Yes

Sarabsri mt

P.Eis energy posed by body at rest. So,total amount of energy it absorb before it can ejected

so , it means same as work function

Is it his correct

John Rennie

That's correct.

Sarabsri mt

Let us say that I sent total energy of 8 W to satan29(let us say he is a Na metal)

work function is 5W

John Rennie

8W isn't an energy, it is a power i.e. one watt is one joule per second.

Sarabsri mt

7:40 AM

8W = K.E + 5 W

oh sorry

take it as J/S

John Rennie

Suppose you shine light with wavelength λ on sodium.

Sarabsri mt

Ohk

John Rennie

Then the energy of one photon is E = hc/λ. Yes?

Sarabsri mt

Yes

John Rennie

The work function of sodium is Φ, so the energy left over after removing the photoelectron from the sodium is hc/λ - Φ

And this left over energy goes into the KE of the electron.

Se we get: KE = hc/λ - Φ

Sarabsri mt

7:44 AM

So , that remaining energy gets converted

John Rennie

Yes

Sarabsri mt

It passes on its left over energy to electron

to give it speed

John Rennie

Correct :-)

Sarabsri mt

Kind of like that

Great. Thanks a lot sir again 🙏

John Rennie

I need to go and do some work now. I'll be back in about an hour.

Sarabsri mt

7:46 AM

Sure. Sir , what do you do btw now if I may ask. Just for interest if it is related to science

Ashish Ahuja

8:26 AM

If we have a conical pendulum, and we are talking about angular momentum and torque about the point of suspension, their magnitudes are constant but their directions are continuously changing. If $\vec{L}$ be the angular momentum then

$$ \frac{\vec{dL}}{dt} = \vec{\tau}$$

Taking the magnitude on both sides, and integrating (since magnitude of torque $\tau$ is constant), we get

$$|\triangle L| = |\tau| \times \triangle t$$

Is this a valid result? I'm getting the wrong answer in a problem and I have a feeling that this is where I went wrong.

I think we can't integrate $|\vec{dL}|$ like what I've done above, is that the issue?

satan 29

How did you end with a scalar result after a vector integration?

Ashish Ahuja

I took the magnitudes of LHS and RHS on both sides before integration, since writing torque as a vector would become an incredibly complicated function of time. (since direction is changing)

That's the incorrect part? I'm not sure what the integration would be like.

satan 29

Could you post the exact problem? the problem with vector integration is that the unit vectors may themselvves change with time, so we cant just move em out of the integral sign. If only I could get an idea of what the directions of $\tau$ and $L$ are..

Ashish Ahuja

Irodov 1.188

I can post a picture if you'd like

I was able to solve it in a different way, but this was my original method and I could never figure out what exactly went wrong, fairly sure the issue is in what I posted above.

satan 29

post the picture please

Ashish Ahuja

8:35 AM

ok give me a min...

brb

satan 29

ok, so what was youur approach?

@AshishAhuja aight

Ashish Ahuja

yeah I'm back

@satan29 I was able to solve it, I found that the magnitude of angular momentum and torque would remain constant, just their directions change. I was able to find both their magnitudes as well. It's much easier to just subtract the horizontal components of the initial and final angular momentums which gave me the correct answer, but I first tried finding the change...

... in angular momentum using the dL/dt = torque relation and integrating, which failed. I'm not entirely sure why it did, and I've posted above the relation I got.

Everything else I plugged into the relation, the torque, L etc. was the same for the other approach as well so I'm guessing that's not wrong.

satan 29

the torque about point o is mgr in magnitude, and its direction is changing all the time, right?

Ashish Ahuja

yes

satan 29

its pointing "outward"

Ashish Ahuja

8:48 AM

yes

satan 29

lets call that unit vector $\hat{n}$

then your equation becomes $d\vec{L}= mgr \hat{n} dt$

Ashish Ahuja

yup

satan 29

when you integrate both sides, the left side becomes $\vec{L_{final}} -\vec{ L_{initial}}$

which is what the question required. SO all we need is to figure out the integral on the right side.

Ashish Ahuja

yes

satan 29

which becomes $mgr \int{\hat{n} dt} $

the only thing to realise here is that $\hat{n}$ is a function of time, so you cant move it outside the integral sign.

consider a circle in the x-y plane.

Ashish Ahuja

8:54 AM

ok

satan 29

take a point on the circumference, draw the vector i horizontally, vector j vertically, draw the vectpr n outwards, and draw a vector "tangential" at that point, name it t. (such that n x t = k) . Let the angle between n and i be u.

then n = icosu + j sinu

=icos(wt) + jsin(wt)

Ashish Ahuja

"outwards" as in radially outwards, but still in the x-y plane?

satan 29

outwards is really what matters. we can translate a vector anywhere we want

Ashish Ahuja

uhh I'm confused.

satan 29

you can picture a circle centered at point o if you want, the argument wont change...

okay, consider a circle centered at point o.

Ashish Ahuja

8:59 AM

Nevermind I got it

I got confused with n

satan 29

oh

ok, so now you have n as a function of time. so you can carry out ndt easily.

Ashish Ahuja

okay I'll try it all out. I need to wrap my head around what you wrote above, and I'll let ping you if it works/does not work, thanks.

satan 29

the limits of t will be 0 to pi

@AshishAhuja alright. The only point here is that the unit vector itseld is a function of time, (evident ffrom the fact that the direction is changing all the time) . so all you need to do is to find How n changes with time, and carry out the simple integration.

Ashish Ahuja

got it.

satan 29

whats the answer?

Ashish Ahuja

9:04 AM

Oh by "got it" I meant I understood your message. I'm still doing it, this is probably the first time I'm doing a vector integration of this sort.

satan 29

no as in , whats the answer in irodov

Ashish Ahuja

$$2\sqrt{1 - (\frac{g}{\omega ^2 l})^2} \frac{mgl}{\omega}$$

the brackets are over the whole g/w^2l term, mathjax doesn't render it properly.

satan 29

oh. I got 2mgr/w. I suspect they did not give you the radius. In terms of the length, youll get 2mglcos(theta)/ w. so i guess you will also need to find the angle of the string. As i remember, this is fairly standard in these conical pendulum problems..

Ashish Ahuja

yeah I'll do that.

satan 29

alright. More importantly, do let me know if you now know what was wrong with your earlier approach

Ashish Ahuja

9:17 AM

@satan29 yeah my approach was messed mainly due to mathematical reasons. I did know that the vector would be changing direction, but had never seen such an integration so I tried weird stuff like taking magnitudes :-|

@satan29 you mean 0 to $\frac{\pi}{\omega}$ right?

satan 29

@AshishAhuja oops yes

Ashish Ahuja

@satan29 I'm getting the correct answer, but what's the point of defining the vector t? You haven't used it anywhere, right? The $t$ in the cos, sin terms is time, not that vector?

satan 29

11:03 AM

@AshishAhuja Yeah, it served no purpose for this particular problem. However, in general, whenever we have a problem involving rotating vectors, the vector t (tangential) comes handy in a lot of situations. Particulary because $d/dt(\hat{n})= \omega \hat{t}$

Ashish Ahuja

11:26 AM

@satan29 ah okay, thank you.

satan 29

@AshishAhuja you got 2mgr/w too?

Ashish Ahuja

yes. And since r = l sin theta by substituting sin theta the answer matches with the one in irodov.

satan 29

ok, cool.

Sarabsri mt

11:59 AM

@JohnRennie hi sir

Having a doubt sir

2

Q: What if kinetic energy of ejected electron = 0

So , I got to know that when hf≥work function. Then, electron still come out. So, if I say kinetic energy of ejected electron = 0. It should still come out. Then , how does the electron even move out or gets ejected if it’s v =0.

4

Q: What happens when work function = hf

What happens when the photon which hits a metal surface has energy equal to the work function of that surface? $$\phi = hf$$ I realise the emitted electron will have no kinetic energy after escape, but then what does it do? Hover above the surface? Or does it have momentum from escaping? I also d...

quantum-mechanics photons electrons photoelectric-effect

John Rennie

@Sarabsrimt hi :-)

Sarabsri mt

Are both the Q related

@JohnRennie Hello sir

John Rennie

@Sarabsrimt yes, they are basically the same question.

Sarabsri mt

Ohk. So , is my Q that I asked correct

If K.E Of photo electron = 0 , then v = 0

Or it doesn’t uncertainty principle.

John Rennie

There is an extra complication, because when a photon hits the metal surface it does not always eject an electron.

Sarabsri mt

12:03 PM

Yes. If it less than work function

John Rennie

There is a probability, P, that the photon will eject an electron, and in fact that probability is very low.

Sarabsri mt

Also , equal to

@JohnRennie So , are my conditions also wrong

hf > W

John Rennie

Typically only one photon in about 100,000 will eject an electron. The other 99,999 photons are just absorbed and their energy is turned to heat.

Sarabsri mt

Absorbed by other electrons or same electron ?

sir

John Rennie

I've written an answer on this on the main site. Let me see if I can find it ...

Sarabsri mt

12:05 PM

Sure sir. It would be a great help

John Rennie

9

A: Cause behind photoelectric effect

A light ray has an oscillating electric field associated with it, and this oscillating electric field will make electrons oscillate when the light ray passes through them. Note that I'm talking about light rays not photons - we'll get to photons in a bit. When the oscillating field of the light ...

Also:

4

A: Kinetic energy of photoelectrons

Photoemission in the usual experiment is a two step process. First the incident photon creates a photoelectron in the bulk of the metal: The quantum efficiency for this is almost $100$% i.e. almost every photon that hits the metal creates a primary photoelectron. However this photoelectron is ...

Sarabsri mt

Ohk. I will read both sir.Thank you very much for it

John Rennie

12:23 PM

I need to go know. I will be around tomorrow as usual.

Sarabsri mt

1:07 PM

@RishiNandhaVanchi if I shower rain drops on a hemi sphere
Then , amount of area covered by the rain drops equal pi * r^2 . I am not understanding why is it like this and not 4 * pi *$ r^2$ / 2.

John Rennie

3:54 PM

@Sarabsrimt If you are calculating the number of drops hitting the sphere per unit time then you need the cross sectional area of the sphere. The cross section of a sphere of radius r is just a circle of radius r so it's area is πr².

Sarabsri mt

Omg sir. Finally you’re here.

Sir , but if increase the radius of sphere

So , will the area of hemi sphere increase

then , how is it that we need to find pi * r^2

It’s like , if i have a hemi sphere,

Then total light on the hemi sphere should be 4 pi r^2 / 2 right

@JohnRennie. Sir , where am I wrong here

-1

Q: What is the connection of neils Bohr model with photo electric effect

When light falls on an atom , the electron side it gets excited and moves to excited orbit . Whereas in case of photo electric effect , when an electron (which is inside the atom ) gets excited , it leaves the shell i.e photo electric effect happens there. Also , why does the electron move to exc...

physical-chemistry photochemistry

John Rennie

4:16 PM

I'm starting my lunch now I'm afraid. We'll have to discuss this tomorrow.

Sarabsri mt

@JohnRennie For sure sir.

Transcript for

Problem Solving Strategies