Problem Solving Strategies - 2021-01-04

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user586228

03:26

user image

Aridhan

04:25

I got 7 upvotes on 1 post

I can upload now

Images

1 hour later…

John Rennie

05:28

@Aridhan cool :-)

Aridhan

user image

@JohnRennie check this image

Red colour is shape of electron or orbitals

Black colour is shells

(Blue)n is the size of shells

Is this correct ?

If you notice 1s orbital I have drawn

it is totally inside K shell

so , electron is that cloud inside

Isn’t it already present there . Why do we say that we need to find the probability of electron to be present there

Also , the green colour is where the orbital is not drawn

what is that region

John Rennie

The orbitals don't have precisely defined edges. Remember that the equation for the 1s orbital was $\psi(r) = A e^{-r/a}$ (where $A$ is a constant).

Aridhan

Yes

John Rennie

This is spherically symmetric, so we draw it as a sphere, but it isn't a sphere. The wavefunction decreases gradually with distance and has no sharp edge.

Aridhan

so is it like what you meant by shape of orbital is that , in that drawing of orbital , somewhere am electron is present

that electron is of Of course a cloud

John Rennie

05:37

@Aridhan No, the electron is spread out over the whole orbital. The electron is present everywhere in the orbital simultaneously.

Aridhan

Ok.@JohnRennie

John Rennie

The electron "cloud" and the orbital are the same thing.

I need a few minutes to make a coffee. Back soon.

Aridhan

Sure.

I’ll just type out my confusion

So , what have I understood.I’m gonna upload a pic

user image

user image

user image

By that green colour

I mean is , that whether do we find possibility of electron there

which region actually

yes , I also imagine it in 3D

I hope you got what I meant

John Rennie

05:53

The orbital is a function that describes the electron. The orbital isn't some object that the electron has to fit in, it's just a function that describes the shape of the electron cloud.

If we call this function $\psi$, then the density of the cloud is given by $|\psi|^2$. We use the modulus because $\psi$ is a complex number.

Suppose we look at some small volume $dV$, then the "amount" of the electron in that volume is the density times the volume i.e. it is is $|\psi|^2 dV$.

And when we say the "amount" of the electron in the volume $dV$ this actually means the probability of interacting with the electron in this volume $dV$.

1 hour later…

user586228

07:13

user image

Can someone help me out with this?

user586228

07:31

I am having a tough time with this.The earlier the response the better it is..4

Danny LeBeau

08:06

@JohnRennie Is there any experiment that verify speed of light is absolute constant?

user586228

09:04

@JohnRennie I would be very grateful to you if you help me solve this problem.

2 hours later…

user586228

11:04

Qhy is notbody responding to my query

pi-π

11:40

@JohnRennie Hello

3 hours later…

napstablook

14:29

@user586228 I think you will have better luck at h bar ( this room is for high school physics (i am assuming that because i don't get your qn at all and i am in 12th grade))

2 hours later…

John Rennie

16:14

@user586228 I don't know the answer ...

The activity, $a$, is given by $G = G_0 + RT \ln(a)$, but I don't see how you're going to get that from your equation.

$G_0$ is the free energy of the pure metal, Au or Ni, and $G$ is the free energy in the mixture.

user586228

user image

I have only one doubt...How did they calculate GAu xs partial

I mean curious mind told me that

But I am asking for a standardised mathematical approach for the same..

Rover

17:15

Hi @JohnRennie

John Rennie

@Rover hi :-)

Rover

Can you help me with kinetic energy of system of particles concept..

John Rennie

Yes, what's the question?

Rover

It says in frame attached to com k.E of system w.r.t com is a internal property and don't depend on frame of reference, this point didn't digest..

John Rennie

I suspect that what it means is we can define an internal energy for the system.

Rover

17:22

Okk

John Rennie

If a system is moving then it has a KE related to it's speed, and we can always eliminate this KE by changing to the rest frame of the system i.e. the frame in which it's velocity is zero.

Rover

K.E of system depends on frame of reference ..?

John Rennie

But when we have a system made up of lots of randomly moving particles then even in the rest frame, i.e. the COM frame, we still have some ebergy due to all this random motion.

Rover

Ok , so K.E don't depend on frame of reference, right?

John Rennie

An obvious example is a container of gas. Even when the container is stationary the gas molecules are all buzzing around so it has an internal energy due to this motion.

@Rover KE does depend on the frame of reference, but the internal energy doesn't depend on the frame of reference.

Take my example of a gas. For an ideal, monatomic gas the internal energy is $U = C_v T$. Yes?

Rover

17:26

Yes

John Rennie

And this internal energy is due to the KE of all the atoms in the gas. The average KE of an atom is 3/2kT. Yes?

Rover

Yes

John Rennie

So for one mole of the gas we multiply this by Avagadro's number to get U = 3.2 Na kT, and Na k = R. So we get U = 3/2RT.

And Cv = 3/2R for a monatomic ideal gas, so this is the same equation as U = Cv T

OK so far?

Rover

Ok

John Rennie

But nowhere in this have we said anything about a frame of reference.

Rover

17:30

Yes..

John Rennie

In fact we've assumed we are using the COM frame, but the point is that U is defined this way so U doesn't change as the frame of reference changes.

If U changed when we changed the frame that would mean the temperature must be frame dependent, because U = Cv T

So if U changes T must change.

Rover

How can we say it is from com frame..?

John Rennie

But it doesn't make sense to say the temperature of a gas changes just because its container is moving.

@Rover that assumption is buried in the working when we derive the expression for the internal energy.

The equation for the average KE of the molecules 3/2 kT is only true in the rest frame.

Rover

@JohnRennie Ohk I will refer once..

Sir , can you suggest anything for revision of rigid body dynamics, I always felt this topic overwhelmed ..

John Rennie

Do problems.

The best way to revise is doing JEE problems.

2

Rover

17:39

Ok , thanks for your time sir ,bye :)

John Rennie

Bye :-)

2 hours later…

Aridhan

19:36

user image

Can’t believe I got this badge

This is from my 1st question

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Jan '214

Problem Solving Strategies

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