A Math Paradox!

[insane_alien]

how can the particles travel faster than c? even an 11 year old knows things can't go faster than c.

 

 

as a side note. does this socalled weapons class you attend go along the lines of "Heh Heh big fire! Heh Heh!" or does it actually have something scientific in it?

[Alpha-137]

Yes, you are right that there is NO physical meaning to these calculations and there was

 

never meant to be any.

 

This was in reply to a question of, if we ever did make a bomb to equal the total

 

energy of E=mc².

 

He was very big on everyone asking questions for there was NO dum questions when

 

it comes to safety with anything to do with nuclear power. This was just one way to put

 

the fear of God in all of us.

 

And yes there would be documentation of this because any time that he even sat in on

 

a class more or less one that he spoke at one he was filmed and I am sure that it is in the

 

archives.

 

Yes, insane_alien; No particle can move at c².

 

Alpha-137:)

[insane_alien]

So ehh why did you say that particles were travelling at c^2.

and i quote,

To make it as simple as I can we will look at just one particle’s action at first.

 

This particle will travel at c² = 8.987551787^16 meters per sec.

 

Your in a science forum, please obey the laws of physics.

[Alpha-137]

Yes, again: insane_alien; No particle can move at c².

 

Can’t you understand that this posting is all hypocritical,

 

Because man will never be able to totally use all of the energy

 

in mass at E=mc² and it all started back in the mid-60’s with a dum question of,

 

if we could. And this answer was the to put the fear of God into some very green men

 

that was about to go out in a nuclear sub for their first time.

 

PS: insane_alien the men was not really the color green.

 

Alphy-137:)

[timo]

Yes, again: insane_alien; No particle can move at c².

I think particles with a diameter of 5 Ampere actually can

[[Tycho?]]

What kind of crazy definition for hypothetical do you have Alpha? You're just making stuff up. It isn't hypothetical, its incoherent ramblings.

[Alpha-137]

Tycho?????????????

 

Hypothetical

 

According to Microsoft Word

 

It is just what I wanted to say.

 

Theortical; Imaginary; Supposed;

 

Antonym ( real)

 

Alpha-137

[insane_alien]

In science a hypothetical situation still has to obey the laws of physics. which you seem to see as inconsequential.

[ydoaPs]

According to Microsoft Word

i'm not even going to bother with a witty remark.

[Janus]

Yes' date=' again: insane_alien; No particle can move at c².

 

Can’t you understand that this posting is all hypocritical,

 

[/quote'] Yes I quess it is all hypocritical, in the fact that you are feigning to have an understanding on a subject that you obviously know next to nothing about.

[5614]

Mate, you clearly have no idea how to use E = mc²... I'll show you how to using an example of 1g of matter being converted into energy.

 

1g matter = 0.001kg

c² = 300,000,000 * 300,000,000

 

E = mc² = 0.001 * 300,000,000 * 300,000,000 = 9 * 10^13 J or 90TJ

 

This is about twice the power of the Hiroshima bomb.

 

Quite what the hell you were doing I have no idea.

 

Also here is some real nuclear bomb calculations: http://chemcases.com/nuclear/nc-09.htm

[the tree]

5641, So the Hiroshima bomb converted roughly half a gram of matter?

 

Alpha, when are you going to accept that these guys understand more* than you?

 

*or at least are able to apply (very) common knowlege better

[5614]

5641, So the Hiroshima bomb converted roughly half a gram of matter?

Well, no.

 

The example I gave using E = mc² was showing the conversion of 1g of matter into energy. Like a particle-antiparticle annihilation.

 

A nuclear bomb is different because you are not converting the matter to energy, nuclear fission is occurring.

 

The energy released in this type of reaction is calculated using binding energies, see here for an example (at the bottom of the page): http://www.einstein-online.info/en/spotlights/binding_energy/index.html#Rechenbeispiel2

[insane_alien]

but it did convert half a gram of mass. possibly what the the tree was meaning, only he can tell us for sure though.

[5614]

Well if we get 166MeV (value from here, as posted previously) from each fission and we have half a gram of U-235 that's

[math]moles = \frac{0.5}{235} = 0.002mol \to 1.28 \times 10^{21} atoms[/math]

[using Avogadro's number]

so that many lots of 166MeV gives us [math]166 \times 1.28 \times 10^{21} MeV = 2.123 \times 10^{23} MeV[/math]

now converting that to joules: http://www.google.co.uk/search?hl=en&safe=off&q=convert+terajoules+to+2.126989126+*+10%5E23+megaelectron+volts&meta=

gives us 0.03TJ when we should be getting about 90TJ!

 

From that we can conclude that more than half a gram of U-235 underwent nuclear fission in the Hiroshima nuclear bomb explosion. In fact about 2640 times as much, that is 1320g or 1.32kg... err, the actual value should be about 700g (it is a known figure I looked up)... I can't see what went wrong. I can only assume there is a problem with using the 166MeV. Or maybe the isotopes which the U-235 turns into also decay, so they too release energy, more energy, it is still a lot to make up for though, I'm not sure.

[insane_alien]

no i wasn't talking about the amount of matter. just the mass, well binding energy mass.

[swansont]

"half a gram converted to energy" is a different calculation than "half a gram of U-235 undergoing fission." For the latter you still have most of the mass present as the fission fragments.

[5614]

no i wasn't talking about the amount of matter. just the mass, well binding energy mass.

Err, you lost me.

 

Half a gram of binding energy mass? wtf?

 

"half a gram converted to energy" is a different calculation than "half a gram of U-235 undergoing fission." For the latter you still have most of the mass present as the fission fragments.

Yeah I know, hence I used the binding energies to work out the total energy released when one atom of U-235 undergoes nuclear fission, (I found similar values on other sites as well)... do you know where my calculations went wrong?

[insane_alien]

sorry, 5614, even i don't know what the hell i was saying then, thats what alcohol and sleep deprivation do to you.

 

What i meant to say was, the binding energy has mass, when you split the atom the energy is released and so is the mass of the energy(obviously) so half a gram of energy was released but no matter was destroyed (number of protons, neutrons etc. in = number of protons, neutrons etc. out)

 

sorry again for being incomprehensible.

[5614]

E = mc² = 0.001 * 300' date='000,000 * 300,000,000 = 9 * 10^13 J or 90TJ

 

This is about twice the power of the Hiroshima bomb.[/quote']The only fault I could find is that this is not twice the power of the Hiroshima bomb, that bomb was a ~90TJ explosion.

 

A second site said the fission of U-235 was 200MeV, which makes no significant difference to the calculations in which I used 166MeV.

 

Changing estimated mass of U-235 required makes no difference to the end calculation because it is taken back into account at the end.

 

I can't find another side (other than wikipedia) to confirm that 700g of U-235 underwent nuclear fission.

 

One website suggested that the Hiroshima explosion was only 55TJ, but even then according to my calculations 1kg of U-235 was required. But I'd rather go with the more commonly used values of ~90TJ.

 

So I really can't see what was wrong with my post #40.

 

[edit]

 

in fact if you work backwards saying, right, you need 90TJ of energy and get 166MeV from each atom you can work out how many atoms you need and thus how many moles and thus how much, ie. the mass of U-235 required... if you use this method you still get just over 1000g when it should be 700g.

 

As many methods all show the same value of ~1kg of U-235 required I would assume that either the 700g as cited on wikipedia is incorrect or the value 166MeV for the fission of 1 atom of U-235 is incorrect.

[incadink]

I think there is something you are missing regarding yield calculations. The amount of fissile material used in an atom bomb is not the same as the critical mass of the element. The Hiroshima bomb was a gun-type weapon. It traded off effiency for simplicity and reliability. The bottom line is not all the U-235 was converted to energy. A good estimate is about 40% of the fissile material. That has to be factored into your equations.

[swansont]

I think there is something you are missing regarding yield calculations. The amount of fissile material used in an atom bomb is not the same as the critical mass of the element. The Hiroshima bomb was a gun-type weapon. It traded off effiency for simplicity and reliability. The bottom line is not all the U-235 was converted to energy. A good estimate is about 40% of the fissile material. That has to be factored into your equations.

 

Note (or note again, since I said it a few posts back) that "mass undergoing fission" is not the same thing as "mass converting to energy"

[5614]

Yep.

 

Because in mass-energy conversion all of the mass is converted into energy.

 

Whereas in fission only part of the binding energy (BE) of the U-235 is converted into energy (the other part of the BE used in the bi-products, what the U-235 decays into).