Math problem regarding the Chernobyl accident.

[BlueTimberwolf]

How can 2E998 atoms equal 14 moles when 6E23 (Avo) * 14 moles = 8.4E+24 atoms

 

[Narthlotep]

Simple, I haven't slept in about 24 hours, that's why. It looked about right to me, though I got most of the math wrong. I can either redo it in this post, or edit the above one, if you prefer. I also had one semester of physics and one semester of chemistry, neither of which dealt with mole conversions, mostly it was watching tangential movies in physics, and listening to the chemistry teacher go on and on about how he did analytical work on some of the Challenger wreckage.

 

[Rolling Thunder]

I think I'll stick to what I'm good at- negotiations, arguing, violence and writing, rather than a series of rather pointless sums.

 

[Sayvara]

Spoiler: I think I might have done somthing wrong but

I got the diameter to 1 * 10^64 know universes...

That is what I got too.

i got:

5.3575E+300 number of atoms

9.00E+276 number of moles

1.18E+279 grams

1.18E+273 tons

5.78E+273 volume m3

4.57143E+32 volume of universe light years

4.57143E+47 volume of uni m3

1.26E+226 number of universes in size

Thats alot of I 131 i might have made a mistake in my math.

Nope... that seems perfectly reasonable.


Conclusion: the probablility that even one atom of I-131 that was released from Chernobyl still exists is a probability so small that it is less than 1 divided by the numbers of atoms in the entire universe... many times over.


Of course regarding Chernobyl there is still the problem of Caesium-137 and Strontium-90 which does not have quite so timid half-lives...

Still, I recommend this video: Chernobyl 2006 [http://www.youtube.com/watch?v=101OEaksU0s] - australian Carl Montgomery visting Pripyat, the worker town just outside the powerplant.

/S

 

[Lukeje]

So, there has to be extraneous factors. I-131 is chemically equivalent to I-127, so will be absorbed into carbon life forms and water, where it may bond into molecules, thus altering it's half-life.

Wait a minute... you can't change something's half-life by forming a chemical bond!
Radioactive decay is strictly a first-order process. d/dt = k , whether the I is in a molecule or monatomic. { means 'concentration of I' and 'k' is a proportionality constant.}

 

[Anarchemitis]

And no, this is not homework or an assignment. This is for fun. Yes, for fun. I am a nerd!! I think things like this are fun. Suck it up, average-boy!

HaHA!

Let's see....
Decay would be equal to 1 atom=X(number of atoms (/6.02x10[sup]23[/sup]))(0.5)[sup]8000/8[/sup]
Therefore
1=X(.05)[sup]1000[/sup]
Therefore
(((Logarithm 1)/(Logarithm 0.5[sup]1000[/sup]))/6.02*10[sup]23[/sup])*131) is equal to the amount of Iodine-131 in metric grams, rounded off to the whole gram.
(I think I may have made a mistake regarding the Logarithm part.)
And for those without a scientific calculator present, this is the answer, according to me:

• 42

Whoops, that's not even possible because 1 Logarithms can't be performed on exponents and
.............................................................................2 Logarithm of 1 is 0.
Let's retry with Mols, then go from there.
Actually, if the question is supposed to factor in light-speed and the estimated volume of the universe, te question is nigh possible with the given information.

 

[Lukeje]

ANOTHER maths thread?! *sigh*

Yes, but this one uses Science.

 

[The_root_of_all_evil]

So, there has to be extraneous factors. I-131 is chemically equivalent to I-127, so will be absorbed into carbon life forms and water, where it may bond into molecules, thus altering it's half-life.

Wait a minute... you can't change something's half-life by forming a chemical bond!
Radioactive decay is strictly a first-order process. d/dt = k , whether the I is in a molecule or monatomic. { means 'concentration of I' and 'k' is a proportionality constant.}

My mistake, I meant to say that it would be taken out of the measurable bio-sphere as it would no longer be strict I-131.

 

[corporate_gamer]

Do you need all of that data. whats the speed of light got to do with it.

Its just 1x2^1000 isnt it?


Number of half-life 8000/8 =1000

There is 1 now. 1000 half life since the accident. So you just double 1 a thousand times

... I think?

 

[Lukeje]

HaHA!

Let's see....
Decay would be equal to 1 atom=X(number of atoms (/6.02x10[sup]23[/sup]))(0.5)[sup]8000/8[/sup]
Therefore
1=X(.05)[sup]1000[/sup]
Therefore
(((Logarithm 1)/(Logarithm 0.5[sup]1000[/sup]))/6.02*10[sup]23[/sup])*131) is equal to the amount of Iodine-131 in metric grams, rounded off to the whole gram.
(I think I may have made a mistake regarding the Logarithm part.)
And for those without a scientific calculator present, this is the answer, according to me:

• 42

Whoops, that's not even possible because 1 Logarithms can't be performed on exponents and
.............................................................................2 Logarithm of 1 is 0.
Let's retry with Mols, then go from there.
Actually, if the question is supposed to factor in light-speed and the estimated volume of the universe, te question is nigh possible with the given information.

I think that the eq'n you are searching for is
1 = N(2[sup]1000[/sup]),
log1 = log(N) + log2[sup]1000[/sup]
log1 = log(N) + 1000
where all the logs are base 2.

 

[The_root_of_all_evil]

Do you need all of that data. whats the speed of light got to do with it.

Its just 1x2^1000 isnt it?


Number of half-life 8000/8 =1000

There is 1 now. 1000 half life since the accident. So you just double 1 a thousand times

... I think?

No...try the chess board problem.
Put 1 grain of rice on the first square, 2 on the second...etc.

You've got the right idea, but the maths part is changing it from atoms to volume.

2^1000 is 107150861 with 292 0's after it. Roughly

 

[Anarchemitis]

[img_inline caption="This goes to show how particularly accurate and relevant your source data is." width="500"]http://upload.wikimedia.org/wikipedia/commons/2/20/Airdosechernobyl2.png[/img_inline]

 

[jim_doki]

oh for the love of.....

I'm gonna say 10 million

 

[Lukeje]

[img_inline caption="This goes to show how particularly accurate and relevant your source data is." width="500"]http://upload.wikimedia.org/wikipedia/commons/2/20/Airdosechernobyl2.png[/img_inline]

What's the percentage scale measuring?

 

[Anarchemitis]

Amount of radioactive isotope fallout (percentage) over time since the explosion of Chernobyl Reactor 4.

 

[Lukeje]

Amount of radioactive isotope fallout (percentage) over time since the explosion of Chernobyl Reactor 4.

The percentage make-up off the fallout? Because I don't see how that contradicts the data we've been given. Without knowing the half-lives of the other isotopes, the graph doesn't really help us except in the fact that we can tell that there is very little [sup]131[/sup]I at t = 8000 days.

 

[Sayvara]

Sayvara, this is another attempt to prove conclusively to the forum members that nuclear power is nowhere near as unsafe as people think, and that it's in fact one of the safest sources of energy out there, isn't it?

Not really no. While my sentiment is that nuclear power is a safe power source, that was not the point of this thread. It was just a fun fact I found out while I did some numbers on it.

[img_inline caption="This goes to show how particularly accurate and relevant your source data is." width="500"]http://upload.wikimedia.org/wikipedia/commons/2/20/Airdosechernobyl2.png[/img_inline]

I'm sorry but I have a hard time knowing if you mean that in earnest or ironically. I'd say you mean it in earnest because the graph does show that within 100 days (i.e.) > 10 half-lives, the contribution from I-131 to the radiation is negligable.

/S