@JohnRennie Hello :-)

@Jasmine hi :-)

I had a doubt in a nuclear physics problem

The question-

In a fusion reaction 1kg Deauteriam undergoes complete fusion. Find amount of total energy released.

H(2,1)+H(2,1) => He(3,2) + n(1,0)

Have you had a go at this? How far have you got with it?

H(2,1) : 2.015amu He(3,2) :3.017 amu n(1,0):1.005amu

So the mass of the two deuterium atoms in the reaction is 2 x 2.015 = 2.030 amu.

@JohnRennie found the Qvalue for one reaction ( where 2 atoms of Deauteriam undergoes reaction ) as 3.726Mev

The mass of the He-3 atom and the neutron is 3.017 + 1.009 = 3.026 amu.

@JohnRennie its 1.009 sorry

So every two deuterium atoms that react produce 0.004 amu worth of energy.

Are you only given 3 significant figures in the weights? That's going to make the calculation rather inaccurate.

@JohnRennie yes

I get 0.00351 amu using more accurate weights from Wikipedia.

@JohnRennie So 2D can give 3.726 Mev energy

@JohnRennie I guess we have to consider what the question setter has provided or maybe it is the reason why I am getting a wrong answer

@Jasmine Yes.

So now you just have to figure out how many D atoms there are in 1 kg.

@JohnRennie is it correct to say 2×2.015amu gives 3.726 Mev energy

@Jasmine yes

@JohnRennie so then we can calculate for 1 amu and then convert 1kg to t amu and the answer would be t × energy for 1 amu

Yes, though that seems like doing it a slightly roundabout way. 1 amu is 1.66054e-27 kg, so the mass of a D atom is 2.105 x 1.66054e-27 = 3.344E-27 kg.

So the number of D atoms in a kg is N = 1/3.344E-27 = 2.990E+26

@JohnRennie I actually got the correct answer

Cool :-)

I was making some calculation error seems like

It's easy to lose track when you're doing a multistep calculation and make a mistake.

@JohnRennie particularly in exam

I don't know why we are expected to do such calculations without calculator

@JohnRennie Sir I know your lunch is over and now you're making your mind for rest :-), but may I ask about a definition in which I'm getting some problems?

@Jasmine Gosh, you did that without a calculator?

@JohnRennie yes

I couldn't do that. I used Microsoft Excel :-)

@JohnRennie Its not allowed in JEE :-)

@JohnRennie I had one more question from diffraction

@Jasmine yes?

@Knight hi :-)

@Knight if you would like to ask first

@JohnRennie Hi Hi !

@Jasmine So nice of you, may I?

@Jasmine you ask first. Knight's question will be complicated. They always are :-)

HAHAHHAHHA

LOL

@JohnRennie Ok :p

Its from YDSE not diffraction

just ask it

Sir knows everything

The question says at O (origin) 3rd order maxima is found to be formed

We need to calculate the distance from O where the 4 th maxima will be formed

Are you given any other dimensions? Slit spacing? Wavelength of light? Screen distance?

And thin air film has a thicknedd 't' not given

@JohnRennie yes

D/d=1000 wavelength of light =0.78×10^-6m

The top ray has a longer optical path length than the bottom ray. The difference is $t(n-1)$. Yes?

@JohnRennie yes

I could find t as 7.8×10^-6m

So the difference in the path lengths is $\Delta s = d \sin\theta - t(n-1)$ where I'm taking $\theta$ to be positive if it's anticlockwise from the horizontal axis.

@JohnRennie Ok, there are 2 answers given, so ± could be better I think

And the positions of the maxima are when $n\lambda = d \sin\theta - t(n-1)$

@JohnRennie yes

@Jasmine well there is a 4th maxima either side, so I guess the two answers are those two distances. In that case one answer will be eight times the other.

@JohnRennie yes indeed one is -0.6micrometre and other 4.2micrometre

Hmm, seven times. Did I mean seven times or eight. Let me recount.

Ah, I miscounted. I did mean one is seven times the other.

It would probably be easiest if I drew a diagram. Shall I do that?

@JohnRennie yup

@JohnRennie yes :-)

@Jasmine OK, give me a moment.

Ok

Is the wavelength of light given in the medium

Or its the wavelength in air

I've drawn the maxima up to $n=4$ either side of the horizontal.

We are told that the $n=3$ fringe is at the centre, so the two distances to the two $n=4$ fringes are marked by the vertical arrows on the right.

The way I would do this is find the position of the (shifted) central fringe, then divide it by three and that gives you the spacing between the fringes.

@JohnRennie Ok

Once the have this spacing, let's call it L, then the distances from the centre to the fourth fringes at -L and 7L. Yes?

@JohnRennie yes

That's how I knew one distance was seven times the other.

@JohnRennie Ok

The zeroth fringe is where the paths have equal lengths, i.e. $d\sin\theta = t(n-1)$. Yes?

@JohnRennie yes

Let me scroll up and check what all the dimensions are ...

$$ \sin\theta = \frac{t(n-1)}{d} $$

and the istance along the screen is (approximately) $D\sin\theta$, so the distance to the zero fringe is:

$$ 3L = D\sin\theta = t(n-1)\frac{D}{d} $$

@JohnRennie please tell if the wavelength of light is the wavelength in the given medium

@Jasmine I don't know. What does the question say?

@JohnRennie says nothing

But I guess you have considered that

In that case I would guess it is the wavelength in air

So will that mean we have to multiply $dsin{\theta}$ by 1.3

I would just divide λ by 1.3 instead

@JohnRennie in that case the other term will also get multiplied by 1.3

Here is where I generally get confused

Why is this wrong then ^

I get $t = 10λ'$, where $λ' = 780/1.3 = 600 nm$ is the wavelength in the medium not in air.

So $t = 6$ microns

@JohnRennie thats wrong, I mean as there was other part of the question asking to calculate t which I got as 7.8 microns indicating wavelength is in medium

And the answer key confirms t = 7.8 microns?

@JohnRennie yes

OK, then the wavelength given must be the wavelength in the medium.

No, wait, that's going to give the wrong answer ...

I'm going to have to look at this tomorrow. I'm tired and getting mixed up.

@JohnRennie Sure

This was a bit complicated

Back tomorrow :-)

I'm just getting mixed up with which wavelength is which. It's unusual having the light travel through a medium with a thin air layer. It's usually the other way around.

@JohnRennie Thats what I was also having problem with

Its fine we can discuss it tomorrow