Math problem regarding the Chernobyl accident.

[Sayvara]

On april 26 1986, due fundamentally crappy design, lack of operator training and non-adherence to safety procedures, reactor number 4 at the V.I. Lenin Memorial Chernobyl Nuclear Power Station quite literary went sky-high. A number of substances were released into the atmosphere in this accident, many of them harmful due to them being toxic and radioactive.

When an atom releases radioactivity it changes into another substance. Alternatively it changes into a state where it won't release radioactivity again. This means that radioactivity diminishes over time as the material in question is consumed. One important quality of radioactive substances is their half-life. What half-life means is that when the time specified by the half-life has passed, half of the substance remains. Conversely, this means that one half-life ago, twice as much of the substance existed as does now.

One of the major worries with nuclear fallout is the isotope Iodine-131. I-131 exists in a relatively large amount in a reactor and it radiates quite a bit. However, as it decays it is eventually transformed to chemically inert and non-radioactive Xenon-131 which is harmless.

So here is the question:

Assuming that there is one atom of I-131 left in the biosphere from Chernobyl accident, how much I-131 would have had to be released at the time of the accident?

And I would like to have the answer expressed in diameters of the known universe. Do your numbers right and it will become clear to you why.

Some figures to help you. They have been rounded, but mostly not more than some percent, to make things easier.

Time since the accident: 8000 days
Half-life of I-131: 8 days
1 mole of atoms = 6 * 10^23 atoms
Atomic weight of I-131: 131 g / mole
Density of Iodine: 4.9 g/cm^3 (or kg/dm^3 or ton/m^3)
One light-year: 9.4 * 10^15 m
Size of the known universe: 93 billion lightyears

And to help a bit more: the forumla for calculating the volume of a sphere: 4 * PI * radius^3 / 3

The calculator in Windows will be able to handle the numbers. Just switch it to "Scientific" mode.

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!

/S

 

[Cheesus333]

Wuh? Buh?

Why?

 

[Fraught]

*something explodes*

Was that my head?!

 

[Sayvara]

Wuh? Buh?

Why?

See bottom of OP.

/S

 

[Lukeje]

You can't actually answer that; theoretically one atom could last for the entire span of the universe without decaying. It is a random process. Hence, the alternative could be true. One atom could be left alot sooner than you would predict. One atom is not enough. One MOLE however...

 

[Sayvara]

You can't actually answer that; theoretically one atom could last for the entire span of the universe without decaying. It is a random process. Hence, the alternative could be true. One atom could be left alot sooner than you would predict. One atom is not enough. One MOLE however...

Of course I generalize here, I am perfectly aware of the randomness of radioactive decay. However as soon as we get over a few hundred atoms the process is predictable enough to be approximated by the half-life model since the deviations cancel eachother out. I could have said: "assuming 1024 atoms of I-131 remains" and removed 10 half-lives from the 1000 in the OP, and we would have gotten the same answer while now being statistically in the clear.

/S

 

[zirnitra]

Time since the accident: 8000 days
Half-life of I-131: 8 days
1 mole of atoms = 6 * 10^23 atoms
Atomic weight of I-131: 131 g / mole
Density of Iodine: 4.9 g/cm^3 (or kg/dm^3 or ton/m^3)
One light-year: 9.4 * 10^15 m
Size of the known universe: 93 billion lightyears

/S

I think my audible groan when I saw this largely explained why I was kicked out of a level chemistry so thankyou.

 

[Sir_Substance]

Time since the accident: 8000 days
Half-life of I-131: 8 days
1 mole of atoms = 6 * 10^23 atoms
Atomic weight of I-131: 131 g / mole
Density of Iodine: 4.9 g/cm^3 (or kg/dm^3 or ton/m^3)
One light-year: 9.4 * 10^15 m
Size of the known universe: 93 billion lightyears

/S

I think my audible groan when I saw this largely explained why I was kicked out of a level chemistry so thankyou.

actually, its more likely you were kicked out for not knowing the difference between physics and chemistry

there is a certain amount of overlap, but few people would call this a chemistry problem

 

[Lukeje]

Time since the accident: 8000 days
Half-life of I-131: 8 days
1 mole of atoms = 6 * 10^23 atoms
Atomic weight of I-131: 131 g / mole
Density of Iodine: 4.9 g/cm^3 (or kg/dm^3 or ton/m^3)
One light-year: 9.4 * 10^15 m
Size of the known universe: 93 billion lightyears

/S

I think my audible groan when I saw this largely explained why I was kicked out of a level chemistry so thankyou.

actually, its more likely you were kicked out for not knowing the difference between physics and chemistry

there is a certain amount of overlap, but few people would call this a chemistry problem

Actually, its a physical chemistry problem.

 

[Syntax Error]

If there was one atom today, 8 days ago, there were two atoms, then 8 days prior to that, there were four.

So, taking into account that the accident happened 8000 days ago, divide that by 8, then you get 1000 (the number of half-lifes that occured since that day). The number of I-131 atoms in the atomosphere on the day of the accident would be 2^999.

I'm not too sure about this, so can you guys confirm if I'm on the right track?

 

[Cpt. Red]

Spoiler: I think I might have done somthing wrong but

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

 

[DeadpanLunatic]

I have removed my words from this site.

 

[brazuca]

So here is the question:

Assuming that there is one atom of I-131 left in the biosphere from Chernobyl accident, how much I-131 would have had to be released at the time of the accident?

And I would like to have the answer expressed in diameters of the known universe. Do your numbers right and it will become clear to you why.

Some figures to help you. They have been rounded, but mostly not more than some percent, to make things easier.

Time since the accident: 8000 days
Half-life of I-131: 8 days
1 mole of atoms = 6 * 10^23 atoms
Atomic weight of I-131: 131 g / mole
Density of Iodine: 4.9 g/cm^3 (or kg/dm^3 or ton/m^3)
One light-year: 9.4 * 10^15 m
Size of the known universe: 93 billion lightyears

And to help a bit more: the forumla for calculating the volume of a sphere: 4 * PI * radius^3 / 3

/S

I dont need all those figures to calculate it (using off course High School knowledge). So u gona need the fowlloing formulas:

Mi=Mo/2° where: Mi stands for inicial mass; Mo final mass and ° stands for numbers of HL

C= 1/35 X day^-1 So u can know that C is the Constant reaction rate constant.

The rest is Stoichiometry. As soon as I get home (I am in a lan house) I will see what I can do.

 

[Syntax Error]

Well, this thread is probably the one reason that makes me hate people: they need to complicate everything!

Joking aside, I'm liking what I'm seeing so far. And let's just say that I'm too tired to think right now. I'll probably come up with a better solution next time (if ever I can do so before this thread dies).

 

[Lukeje]

I get 2e^1000 atoms...
Edit: that should be e^(1000 ln2) = e^ln(2^1000) = 2^1000 atoms

 

[MajukaRinye]

I get 2e^1000 atoms...

So you need to get your answer in diameters of the known universe.

 

[BlueTimberwolf]

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.

 

[The_root_of_all_evil]

Similar to the chess problem.

If we're assuming all things are constant, then 2^(8000/8) atoms, or roughly (1/6)(2^977) moles.
Which would be (1/783)(2^977)g...Aproximately (1/4000)(2^977) g/cm3 (4096 is the closest power of 2)...(2^965) cubic centimetres.

Which is roughly 3.11850048 × 10^290

Centimetres are 15 powers smaller than Kilometres.

3.11850048x10^275 cubic kilometre cloud.

Now, given the Earth is Volume 1.0832073×10^12 km³; one can assume it's not a bilinear reduction.

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. There is also the factor that the gas will be kept at an unnatural pressure due to contact with the air. And the randomness of decay already mentioned.

The problem arises when the only information comes from the Soviets, I-131 was stated as being the main problem where Strontium-90 and Cesium-137 were also involved. Especially as Cesium-137 is indistinguishable from Potassium, which our bodies absorb(Half-Life 30 years...)

Interestingly, we can get it back to a purely maths problem by this

The pro-nuclear Time magazine reported in 1989 that perhaps "one billion or more" curies were released, rather than the 50 to 80 million estimated by Russian authorities.[5] One curie is the amount of radiation equal to the disintegration of 37 billion atoms -- 37 billion becquerels -- per second. It is a very large amount of radiation.

So on release, 37 billion billion (10^18 using the short scale) atoms decayed per second. As the half life is 8 minutes (And I-131 is the only one already likely to decay because of it's short half-life), that means there is likely to be 8*2*60 more atoms than that.

So, final answer Chris, is aproximately 3.5x10^21 atoms or 1/200th of a mole.

If we're using the long scale though, UK based, that's 5000 moles.

*sweeps up all the exploded heads*

 

[Narthlotep]

Funny, "Assuming that there is one atom of I-131 left in the biosphere from Chernobyl accident, how much I-131 would have had to be released at the time of the accident?" is the first question that the OP asked for. Assuming that radioactive decay follows a simple exponential decay (I never got too far into physics or chemistry, outside of metallurgy) function, then the inverse would be 2*10^998 atoms, or aprox. 14moles of I-131 released the day Chernobyl went glowy. Forgive me if I'm wrong, but that seems fairly logical to me.

 

[The_root_of_all_evil]

then the inverse would be 2*10^998 atoms, or aprox. 14moles of I-131 released the day Chernobyl went glowy. Forgive me if I'm wrong, but that seems fairly logical to me.

Uh...how is 2*10^998 atoms = 14 moles??????