Problem Solving Strategies

Abcd

5:26 AM

@JohnRennie Morning

John Rennie

@Abcd morning :-)

Abcd

Please explain

@JohnRennie ??

John Rennie

I'm not sure I'v ever got apparent depth really clear in my head, but basically if you look into a fluid like water from above it looks shallower than it is.

Abcd

I know

John Rennie

If the fluid has a depth $d$ then it appears to have a depth $d/\mu$ where $\mu$ is the refractive index of the fluid

And that's all this question is using.

Abcd

5:36 AM

Why adding them?

@JohnRennie oil is above water and if oil appears to be raised will you see a gap between oil and water?

John Rennie

The apparent thickness of the two fluids is the apparent thickness of the oil plus the apparent thickness of the water.

Abcd

How?

Could you show in a diagram

John Rennie

I don't think there's anything to show ...

If you have two layers stacked together then the aparent thickness of the total is the sum of the two apparent thicknesses ...

ayc

@JohnRennie Morning!.....and Please ping me when done with abcd

Abcd

@JohnRennie the problem is will there appear to be a gap between the oil layer and water layer

Because both won't be raised by same amount

John Rennie

5:49 AM

@Abcd Consider the optical path to the oil-water junction.

And consider the two optical paths to an infinitesimal distance upwards from the junction into the oil, and an infinitesimal distance downwards into the water.

Abcd

@JohnRennie didn't get

John Rennie

For a gap to appear those two paths would have to have be different by a non infinitesimal amount.

I guess I just can't understand why you think a gap should appear where the water and oil join ...

I need a few minutes to sort out something at work ...

harambe

@JohnRennie good morning

Needed help in a math question

John Rennie

6:07 AM

Something has come up at work. Back in a few minutes.

harambe

6:42 AM

@JohnRennie will you be active after 1-2 hr?

John Rennie

@harambe I'll be around for several hours. Actually I've just about finished the stuff I needed to do at work now.

harambe

@JohnRennie okay.

I needed some help in a math question

John Rennie

OK ... ?

ayc

@JohnRennie Are you free now?

harambe

@JohnRennie imgur.com/a/ZiqSi74

Q9

For intersecting at right angles, I have calculated the derivative of the curves and put their product equal to -1

But I think I am missing something more... Is there any other step to do here

Abcd

6:47 AM

Find the point of intersection too

And put that in the product of derivatives=-1 equation

John Rennie

My approach would be to find the point of intersection and calculate the derivatives at that point. I don't know if there is a more elegant solution ...

harambe

Thought so but point of intersection is going to be tough

John Rennie

Have you sketched the curves?

harambe

No. Just going by equation

ayc

@harambe it is y^3 or y^2 in first equation

harambe

6:59 AM

In my textbook, it's y^3.But I am starting to think it is misprint

John Rennie

Oh wait, I thought it was $x^2$ and $y^2$ in the first equation ...

Jasmine

7:51 AM

@JohnRennie while calculations we take everything in lowest significant digits?

Avnish Kabaj

Whoa whoa guys

Confocal conics are orthogonal

Ah that's a y^3

I apologise

John Rennie

@Jasmine I think you need to be a bit cautious about oversimplifying. It will depend on the calculation. Have you got a specific example you're looking at?

Jasmine

@JohnRennie I don't have it at the moment :(

ayc

@JohnRennie can we talk about the yesterday question?..regarding the E=mc^2

John Rennie

@ayc OK ... ?

Avnish Kabaj

8:05 AM

ayc

This paragraph was responsible for the confusion I had and still have:
"The nuclear force plays an essential role in storing energy that is used in nuclear power and nuclear weapons.. The mass of a nucleus is less than the sum total of the individual masses of the protons and neutrons. Work (energy) is required to bring charged protons together against their electric repulsion. This energy is stored when the protons and neutrons are bound together by the nuclear force to form a nucleusThe difference in masses is known as the mass defect, which can be expressed as an energy equivalent. Energ…

Avnish Kabaj

@harambe

John Rennie

@ayc I'm afraid that statement is mostly rubbish

ayc

@JohnRennie It was there on wikipedia

@JohnRennie en.wikipedia.org/wiki/Nuclear_force

John Rennie

Not everything on Wikipedia is correct

ayc

8:09 AM

@JohnRennie So, thats wrong?.......

John Rennie

That paragraph as a whole doesn't make sense

Consider a uranium nucleus

If you imagine starting with all the 238 protons and neutrons widely separated, then you bring them together to form a uranium nucleus then energy is released because the energy of a uranium nucleus is lower than the eneries of the separated nucleons.

@ayc That's our starting point, so you need to be comfortable with this idea.

Jasmine

Ok

John Rennie

@Jasmine yes

ayc

@JohnRennie In reality, how do we bring protons and neutrons together?

John Rennie

It happens naturally in stars

Stellar nucleosynthesis

Stellar nucleosynthesis is the theory explaining the creation (nucleosynthesis) of chemical elements by nuclear fusion reactions between atoms within stars. Stellar nucleosynthesis has occurred continuously since the original creation of hydrogen, helium and lithium during the Big Bang. It is a highly predictive theory that today yields excellent agreement between calculations based upon it and the observed abundances of the elements. It explains why the observed abundances of elements in the universe grow over time and why some elements and their isotopes are much more abundant than others. The...

Basically two protons collide, one converts to a neutron then the proton and neutron fuse to form deuterium. That then fuses with another proton to form helium 3, then reacts again in a way I've forgotten to form helium 4.

Then all the heavier elements are built up by combining protons and helium nuclei in various proportions.

ayc

8:25 AM

@JohnRennie Until iron?

John Rennie

Nuclei heavier than iron are formed in supernovae. The energy density in supernovae is so great that the heavier elements form even though it's energetically unfavourable.

See:

Neutron capture

Neutron capture is a nuclear reaction in which an atomic nucleus and one or more neutrons collide and merge to form a heavier nucleus. Since neutrons have no electric charge, they can enter a nucleus more easily than positively charged protons, which are repelled electrostatically.Neutron capture plays an important role in the cosmic nucleosynthesis of heavy elements. In stars it can proceed in two ways: as a rapid (r-process) or a slow process (s-process). Nuclei of masses greater than 56 cannot be formed by thermonuclear reactions (i.e. by nuclear fusion), but can be formed by neutron capture...

> Neutron capture plays an important role in the cosmic nucleosynthesis of heavy elements.

ayc

@JohnRennie Does a star instantaneously collapse when it runs out of fusion reactions?...I mean lets say that until t=t1 star is stable.And if next moment start cannot support its gavitational pressure then How long will take for the star to collapse into a so called blackhole

John Rennie

The kinetics of a supernova are exceedingly complicated and I'm not sure they are fully understood. The "bang" isn't instantaneous but it's pretty fast.

ayc

@JohnRennie How big is "fast"?..second? hours? days?

John Rennie

Seconds

You could Google for this ...

Type II supernova

A Type II supernova (plural: supernovae or supernovas) results from the rapid collapse and violent explosion of a massive star. A star must have at least 8 times, but no more than 40 to 50 times, the mass of the Sun (M☉) to undergo this type of explosion. Type II supernovae are distinguished from other types of supernovae by the presence of hydrogen in their spectra. They are usually observed in the spiral arms of galaxies and in H II regions, but not in elliptical galaxies. Stars generate energy by the nuclear fusion of elements. Unlike the Sun, massive stars possess the mass needed to fuse elements...

The core collapse takes only milliseconds. The overall explosion around ten seconds.

John Rennie

8:52 AM

@ayc Wikipedia is generally pretty good, but anything that really matters you should try to confirm from other sources. There must be a thousand web sites describing core collapse supernovae out there in Googlespace.

ayc

@JohnRennie I have plenty of questions to ask,but not time.I have to go.I'll see you in the evening or maybe tomorrow.Thank you for your time :) !

John Rennie

@ayc Bye

harambe

10:35 AM

@JohnRennie Can I ask you a physical question though it is more astrological

John Rennie

@harambe yes, of course.

harambe

I was reading an article about the moon and the earth and it is said that the moon is going away from earth each day and it is affecting the rotation of earth

I don't understand how is the rotation of earth being affected?

John Rennie

Yes, that's correct.

harambe

Something about tides and centrifugal force but that didn't make sense

To me of course

John Rennie

15

Q: Why does the moon drift away from earth?

I once saw on TV that the moon is slowly drifting away from the earth, something like an inch a year. In relation to that the day on earth what also increase in time. I wonder why is that?

harambe

10:45 AM

Okay. That makes sense

@JohnRennie why does earth rotate and why do different planets have different rotation directions?

John Rennie

The Solar system formed from a large ball of dust, and that large ball of gas formed from an even bigger interstellar dust cloud that was turbulent i.e. within the interstellar dust cloud there were turbulent flows and eddies.

Because the flow was turbulent when any part of the cloud starts collapsing to form a ball it is very likely that the ball has some non-zero angular momentum.

harambe

Okay

John Rennie

And that's why the planets rotate around the Sun i.e. because the ball of dust that they formed from was already rotating.

harambe

Makes sense.

John Rennie

Now, the individual planets formed from smaller eddies in the ball of dust, so the same argument applies. Just by chance it's very likely that the planets ended up with some non-zero angular momentum.

harambe

10:59 AM

What are eddies?

John Rennie

"Eddies" are the small whirlpools you get in flowing water.

Eddy (fluid dynamics)

In fluid dynamics, an eddy is the swirling of a fluid and the reverse current created when the fluid is in a turbulent flow regime. The moving fluid creates a space devoid of downstream-flowing fluid on the downstream side of the object. Fluid behind the obstacle flows into the void creating a swirl of fluid on each edge of the obstacle, followed by a short reverse flow of fluid behind the obstacle flowing upstream, toward the back of the obstacle. This phenomenon is naturally observed behind large emergent rocks in swift-flowing rivers. == Swirl and eddies in engineering == The propensity of...

Or in this case in flowing interstellar dust.

harambe

Okay. That makes sense.

@JohnRennie do planets have shadow

Or anything like that in space

John Rennie

It's actually much more complicated than this because in the early stages of the Solar System planets were migrating inwards and outwards and colliding with each other and generally causing mayhem.

So the current planet rotation rates we see are a result of the initial dust cloud rotation plus all the later collisions

harambe

Wow. Are we taught these things in college

John Rennie

@harambe yes the planets cast shadows. The shadow of the Moon is what causes Lunar eclipses.

harambe

11:10 AM

@JohnRennie silly doubt. Literally forgot this

Esha Manideep

@JohnRennie I have a question, you free to answer ?

John Rennie

@EshaManideep yes?

Esha Manideep

Sun rays are focused with a lens of diameter d and focal length f to the black side of a thin plate. One side of the plate is perfectly black, and the other side is perfectly white. Angular diameter of the sun is 'a' and its intensity on the surface is I, stefan boltzmann constant is 'sigma'. Now find the temperature of the heated point of the plate and using thermodynamic arguments estimate the maximal diameter-to-focal length ratio of a lens

John Rennie

@EshaManideep The lens forms an image of the Sun on the plate. All the light collected by the lens is focused onto this image, so if you figure out how much energy from the Sun is being intercepted by the lens then you can work out how much energy is being concentrated into the image of the Sun on the plate. Yes?

Esha Manideep

11:26 AM

@JohnRennie wouldn't it be a point image ?

John Rennie

No. The light from the Sun is not parallel so its image is not a point. The question tells you the angular diameter of the Sun. You can calculate the angular magnification of the lens, so you can calculate the angular dimatere of the image and therefore the size of the image.

Esha Manideep

@JohnRennie Okay thanks.

Yes, we get the size of the image and then it will be related to f, so that's how f comes into picture

Wait, how does size matter ?

John Rennie

Because the plate will heat up until its black body radiation is equal to the radiation falling on it from the Sun.

Avnish Kabaj

For intensity

Esha Manideep

@JohnRennie I will take it from here, thanks !

John Rennie

11:34 AM

Cool :-)

Esha Manideep

Another one too

Estimate the time for a soap film to fall apart if the thickness of the soap film be h, the ring diameter D and the surface tension T. (I cut down a lot of history of the question, the soap film is on the circumference of a ring)

Avnish Kabaj

Where do you get such interesting questions from?

Esha Manideep

@AvnishKabaj Kalda ! :)

Avnish Kabaj

You're doing those

Wow

I printed but then chucked out the idea of giving the Olympians as there are only 6 seats from delhi

Esha Manideep

@AvnishKabaj For fun, today is a holiday !

Jasmine

11:46 AM

@JohnRennie are you available?

Avnish Kabaj

@EshaManideep I think you have to assume that the increase in Mass is uniform when the film is contracting

Esha Manideep

@Jasmine You mind if i have a look at your question ?

Avnish Kabaj

Density should cancel out

Esha Manideep

@AvnishKabaj Mass doesn't increase ?

Jasmine

@EshaManideep never mind

Avnish Kabaj

11:48 AM

@EshaManideep *density

John Rennie

@Jasmine hi Jasmine :-)

Avnish Kabaj

Wait

What I mean is

Mass per unit length

For a circular element

Density will remain the same

Esha Manideep

Okay, then ?

My idea was to use the decrease in surface energy as KE and then calculating time

But I don't know how to find velocity with out velocity distribution

Jasmine

In an electron capture process, a nucleus captures an electron from K shell of an atom,electron is having $ B_0$ followed by this capture of electron, several protons are emitted due to electronic transition. Find sum of energy of all photons.

Esha Manideep

B_0 is ?

Jasmine

11:55 AM

Binding energy of electron is B_0

Esha Manideep

Answer is B_0 ?

Jasmine

@EshaManideep yes..

How??

Esha Manideep

A top electron comes to it's place releasing B_0 - the BE of that electron, I guess you introduce another electron in it's place releasing BE of that electron energy, so total energy is B_0. Makes sense or am I being silly again ?

Jasmine

@EshaManideep I don't have proper understanding in this topic @JohnRennie can you help me please?

John Rennie

@Jasmine can you clarify what exactly the quantity $B_0$ is? It isn't clear from how you described the question.

Jasmine

12:06 PM

@JohnRennie B_0 is binding energy of electron In k shell

What I understand by binding energy is the energy required to move an electron from its orbit to infinity distance from nucleus

Esha Manideep

@JohnRennie please do look into my question too !

Avnish Kabaj

@EshaManideep

John Rennie

OK. It isn't a very good question since there is no simple answer. The total energy of the photons emitted will be approximately $B_0$, but that's only an approximate answer.

Avnish Kabaj

I'm getting a differential equations and which is proving to be quite difficult for me to solve

Esha Manideep

@AvnishKabaj They are ?

Jasmine

12:09 PM

@JohnRennie how will it be B_0

John Rennie

In an electron capture a proton captures the innermost electron and turns into a neutron. So the innermost electron basically just disappears leaving a hole for other electrons to fill.

Avnish Kabaj

$\rho$ is Mass density per unit area

John Rennie

If the binding energy of the electron was $B_0$ then energy released if we bring an electron in from infinity to fill that hole will be roughly $B_0$.

Avnish Kabaj

I have a paper tomorrow

This is all of my rough work

I hope some of it makes sense

John Rennie

In practice the electrons will all shuffle around in a series of transitions, but overall the total energy change will be approximately $B_0$.

Avnish Kabaj

12:11 PM

Start from rho'

Esha Manideep

@AvnishKabaj Not rightaway, let me look into it though

Avnish Kabaj

All the stuff above it is nonsense

Esha Manideep

@AvnishKabaj By the way, in the above pic, the way you assumed it to collapse is not as the one given in the question

John Rennie

@Jasmine the reason it's only approximate is because the electron capture changes the nuclear charge, and as a result all the energies of the orbitals will change too. So the binding energy of the innermost orbital will be less than $B_0$ after the capture.

Jasmine

@JohnRennie Great! I got it..

N+He>>O(8,17)+H

N mass-14.003u

He mass:4.003u

O mass -16.999u and H mass 1.008u

If oxygen nucleus has excitation energy of 1MeV . Find KEmin that He must have.

I am not getting correct answer

@JohnRennie help

John Rennie

12:25 PM

What do you get if you add up all the energies before and all the energies after?

Jasmine

@JohnRennie do you mean Qvalue?

John Rennie

Energy before is 4.003 + 14.003 = 18.006 u

Energy after is 16.999u + 1.008u + 1Mev = 18.007u + 1MeV

So the increase in energy is 0.001u + 1MeV. OK so far?

Esha Manideep

@Jasmine Don't forget the CM's KE after the collision !

Avnish Kabaj

@EshaManideep aw man

Not from the center?

Jasmine

@JohnRennie and multiplying it by 930/c^2

As you are adding in Mev

@EshaManideep did not

John Rennie

12:30 PM

1 amu = 931.5MeV so I get the difference to be 1.93MeV

Jasmine

@JohnRennie I forgot to mention that they have written 1u=930/c^2MeV

John Rennie

@Jasmine that doesn't significantly change my answer

Jasmine

@JohnRennie yes

John Rennie

What is the answer you're given?

Jasmine

2.20MeV

John Rennie

12:36 PM

Hmm, I guess we need to take into account the COM energy after the collision as Esha suggests.

Are you really only given the masses to 0.001u? That makes the calculation only very approximate.

Jasmine

@JohnRennie yes

I considered KE of com and got KE min is( M+m/M)change in energy

John Rennie

If the velocity of the He nucleus is $v$ (so it's energy is $\tfrac{1}{2}mv^2$) then the initial momentum is 4.003v. The final momentum has to be the same, so after the collision the velocity of the oxygen nucleus and proton will be $v 4.003/18.007$ (we assume the oxgen nucleus and the proton have roughly the same speed).

So the KE of the incoming He nucleus has to equal the energy required for the reaction (1.93MeV) plus the KE of the oxygen nucles and proton.

So we get ...

Jasmine

Now getting 2.9.. something

Not 2.20

@JohnRennie yes?

John Rennie

@Jasmine doing the calculation now ...

Esha Manideep

12:51 PM

About 2.48

John Rennie

Yes, I get 2.48 as well

Jasmine

1/2mu^2=(m+M)/M(1.93)

John Rennie

I have to go now. I'll be back tomorrow, but I'm going out tonight so it won't be as early as usual tomorrow!

Jasmine

@JohnRennie ok I got 2.48Mev

I took wrong mass

@JohnRennie have a great outing!

The answer is wrong I guess

Esha Manideep

1:29 PM

@sammygerbil Estimate the time for a soap film to fall apart if the thickness of the soap film be h, the ring diameter D and the surface tension T. (I cut down a lot of history of the question, the soap film is on the circumference of a ring) please answer this question

sammy gerbil

@EshaManideep I would try dimensional analysis.

Esha Manideep

The constant ?

Infact the question contained numbers for h,T etc... and the answer was 10 milliseconds !

sammy gerbil

The constant is not required for an order of magnitude calculation.

Esha Manideep

Alright, let me try. That's a nice idea :)

I am getting no where with dimensional analysis too

Avnish Kabaj

@EshaManideep if it's popping from the circumference then the calculations will be easier

What has been assumed?

Dimensional analysis doesn't tell you the value of the constant only the proportionality :/

Esha Manideep

1:43 PM

@AvnishKabaj True, but for estimation it's good enough

Avnish Kabaj

@EshaManideep what

No

K*(velocity)(force)

You won't be able to find out k

@EshaManideep can you send the entire question?

Not the copy pasted one

Esha Manideep

No, I can't upload images

Avnish Kabaj

In which booklet is it

Esha Manideep

Thermodynamics, problem 54

Avnish Kabaj

Questions number

Thanks

Jasmine

2:26 PM

Help help

Esha Manideep

@Jasmine please make it clear

Jasmine

@EshaManideep what can you not understand?

Esha Manideep

How does the current exit ? through those two wires ?

Jasmine

@EshaManideep yes

Esha Manideep

resistances between ab and ac are not given?

Jasmine

2:30 PM

@EshaManideep except for AC and that 2R/5 all are having resistance as R/5

sammy gerbil

@EshaManideep Suppose the circular film pops from the centre and spreads outwards. The force on the film is proportional to the circumference of the hole, which increases from $0$ to $2\pi r$ so the average circumference is $\pi r$ and the average force is $F=\gamma \pi r$ (using $\gamma$ for surface tension). The mass of the film is $m=\rho \pi r^2 h$. The average acceleration is $a=F/m$. The distance the edge of the soap film moves is $r=\frac12at^2$ where $t$ is time taken.

Using $r=6cm, h=1\mu m, \rho=10^3 kg/m^3, \gamma=72mN/m^2$ I get $t=1ms$.

Jasmine

@sammygerbil can you see the question I just posted

Esha Manideep

the question says h is one micro meter and diameter is ten centimeters and surface tension is 25 milli newton per meter

sammy gerbil

@Jasmine Please can you explain the diagram. Is the question asking for resistance of a network?

Jasmine

@sammygerbil the question asks for resistance in AC such that the current through AC is twice as in AB. Its given that all are having resistance as R/5 except for one that is 2R/5 which I have marked

sammy gerbil

2:40 PM

@Jasmine Do all the edges parallel to the one marked $2R/5$ also have the same resistance $2R/5$?

Esha Manideep

2R/15 ? I don't see how what i did makes sense though

Jasmine

@sammygerbil no

sammy gerbil

@EshaManideep Can you explain how you did it?

Esha Manideep

Oh no, I misunderstood, wait.. sorry :P

Jasmine

@EshaManideep options- 9/10R 8/9R 7/9R 3/20R

sammy gerbil

2:49 PM

@Jasmine You can use symmetry and conservation of current to find currents in other edges. eg current in BC = current in AB. Let the 3 nodes at the front, opposite ABC, be DEF. Then current in DA and CF is 3i and current in BE is 0.

Jasmine

@sammygerbil can I say the ratio of equivalent resistances across AC and AB=1/2

sammy gerbil

@Jasmine I don't think so. That would be too easy!

Jasmine

@sammygerbil which is E?

@sammygerbil i said equivalent

Not resistance

@sammygerbil how can I apply symmetry if one of the resistance is unequal

sammy gerbil

@Jasmine E is opposite and connected to B.

Jasmine

@sammygerbil oh

sammy gerbil

2:52 PM

@Jasmine There is a vertical plane of symmetry down the middle of the prism.

Perpendicular to face ABC.

Jasmine

But I have to find equivalent through say FG

sammy gerbil

@Jasmine Oh I see. No I don't see how that can be assumed.

Esha Manideep

Resistance between AB is not given right ?

sammy gerbil

Oh I think I see it now. PD across AC equals PD across ABC.

Esha Manideep

AC

Jasmine

2:58 PM

@EshaManideep it is R/5

Esha Manideep

Can you please post the complete question, from whichever book you got it from ? Something doesn't feel correct

Jasmine

Its from AIITS

sammy gerbil

Oh sorry there is no plane of symmetry because resistance of DA is not same as CF.

Esha Manideep

Complete question ?

Sid

@sammygerbil I am not sure I see a plane of symmetry there...

The resistances are different, no?

Jasmine

3:01 PM

Esha Manideep

Yeah that's what was troubling me, you calling equivalent resistance. Now, give me 2 min :D

Jasmine

@sammygerbil yes

Esha Manideep

Answer is ? I am getting none of them as the answer

Dante

@sammygerbil Hi!

sammy gerbil

@Dante hello

Dante

3:12 PM

How can Packing Fraction of a nucleus be negative when mass of nucleus is always less than it's constituent nucleons together?

Esha Manideep

@Dante emedicalprep.com/study-material/chemistry/nuclear-chemistry/…

Jasmine

@EshaManideep 3R/20

Dante

@EshaManideep What does negative packing fraction indicate? Is the mass of nucleus lower than it's cons or is it the other way round?

Jasmine

I dont want to solve it more

Dante

I'm asking because they've given different formula for packing fraction in my book

sammy gerbil

3:16 PM

@Jasmine ok well done.

Esha Manideep

@Jasmine Means ?

I don't think that should be the answer, I feel it's wrong, anyways ask somebody else to verify whether 13R/50 is the answer or not. :)

Dante

@sammygerbil

Is it true that atomic mass is always smaller than sum of mass of constituent particles?

sammy gerbil

@Dante yes

Dante

Ok, then how's that packing fraction is negative and positive in different cases?

sammy gerbil

@Dante What formula is your book using?

Dante

3:22 PM

mass defect per nucleon

sammy gerbil

Using formula in Esha's link PF will always be -ve.

@Dante That is the same, isn't it?

Dante

Mass defect is always positive (right?), so formula must be -mass defect per nucleon

And I don't understand how can it be positive also as given in the link sent by Esha.

sammy gerbil

@Dante I don't understand either. There is an explanation in the video youtube.com/watch?v=btj1b2gI9Bg&feature=youtu.be but I am not following because it is in Hindi.

Dante

Hmm

sammy gerbil

See 18:30 onwards.

Dante

3:41 PM

She said experimental mass of deuterium is more than expected :/

How's that possible?

sammy gerbil

@Dante Did you get an explanation from the video? I am not sure but it looks like the anomaly arises because of the definition of the atomic mass unit amu, which is based on the mass of the carbon-12 nucleus. If this is the explanation then it is rather silly, because the definition using carbon-12 is arbitrary. We could use any nucleus for the definition of amu, there is nothing special about carbon-12.

If we used a different definition of amu we would get different values for packing fraction.

Dante

In either case the fact that mass of nucleus>mass of the constituent particles will remain same ...right?

sammy gerbil

@Dante Yes that is always true.

Professor says so at start of the video.

@Dante Sorry - other way round : mass of nucleus < mass of constituent particles.

Dante

Hmm, so this contradicts what's given in my book

Then what does she mean when she says experimental mass of deuterium is more than theoretical mass?

sammy gerbil

@Dante Unless I am missing something, PF concept is not useful.

@Dante I think she means expected mass of deuterium is 2/12 of mass of Carbon 12 nucleus.

Dante

3:49 PM

Actually, you are right, I was just looking to get my concept clarified. Anyways, I think I must move on for now

@sammygerbil She clearly said mass of deuterium nucleus of greater than constituent particles I guess....

sammy gerbil

Carbon-12 nucleus has less mass than 6 free protons and 6 free neutrons.

So 2/12 of carbon-12 nucleus is less than mass of 1 proton + 1 neutron.

Ice Inkberry

@EshaManideep Oh, do the star ones from Gnadig :P Edit: I mean they are good too.

sammy gerbil

If you use the masses of free protons and neutrons to calculate expected mass of nucleus then PF is always -ve.

Dante

@sammygerbil Hmm maybe your are right. But she expressed theoretical and experimental mass in same units. If she was referring to 1/12 of carbon-12 nucleus as 1 a.m.u, in any case, mass of nucleus shouldn't have been greater.

@sammygerbil Hmmm I see.

Ice Inkberry

@AvnishKabaj Soap films fall apart from a circular end. Right?

Avnish Kabaj

3:56 PM

@IceInkberry yeah that's how it's shown

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