Silkroad Online Forums

Ok so today, I was late for class so I missed like half of the crap...

I need to know how to solve this cause its like, a test on this crap tomorrow V.V
Tritium, is the radiocative isotope of Hydrogen (H-3) . It has a half life of 12.4 years. If 56.2 mg of Tritum are released from a nuclear power plant during the course of an accident, how much will remain after 60 years of the accident.

Using this formula

Except that A is a K for Some reason..

logic says non but i know thats wrong because it halfs and halfs again not half gone then another half gone and leaving non, well interested to know the answer but seriously learn to do your homework your self, its what i do less i actually requires outside help ie surveys

[SD]Master_Wong wrote:logic says non but i know thats wrong because it halfs and halfs again not half gone then another half gone and leaving non, well interested to know the answer but seriously learn to do your homework your self, its what i do less i actually requires outside help ie surveys

I was there and took note through part of this.
I don't understand how to find Decay Constant..

After I find that out I can do rest,,

the best i can do for u is tell u what i think the answer might be using my own method
4.18mg will remain after the given time

60/12.4=4.84

56.2/2
Ans/2
Ans/2
Ans/2
Ans/2(1/.84)

lukefleming1 wrote:the best i can do for u is tell u what i think the answer might be using my own method
4.18mg will remain after the given time

60/12.4=4.84

56.2/2
Ans/2
Ans/2
Ans/2
Ans/2(1/.84)

but for some reason i don't think it is this simple
otherwise u wouldn't need natural logs to figure it out

lukefleming1 wrote:

lukefleming1 wrote:the best i can do for u is tell u what i think the answer might be using my own method
4.18mg will remain after the given time

60/12.4=4.84

56.2/2
Ans/2
Ans/2
Ans/2
Ans/2(1/.84)

but for some reason i don't think it is this simple
otherwise u wouldn't need natural logs to figure it out

No we did that yesterday, for this you do need natural logs.
Can't divide by 2 anymore =(

Squirt wrote:

lukefleming1 wrote:

lukefleming1 wrote:the best i can do for u is tell u what i think the answer might be using my own method
4.18mg will remain after the given time

60/12.4=4.84

56.2/2
Ans/2
Ans/2
Ans/2
Ans/2(1/.84)

but for some reason i don't think it is this simple
otherwise u wouldn't need natural logs to figure it out

No we did that yesterday, for this you do need natural logs.
Can't divide by 2 anymore =(

bah can't help u then my 10th grade math will only go so far xD

http://en.wikipedia.org/wiki/Exponential_decay
hate to refer u to wikipedia but honestly the equations halfway down the page seem to be related to finding ur decay constant
worth a shot anyways imo

I can somehow see that this is incredibly simple...
I haven't actually had this kind of stuff, but here's my go...

60/ 12.4 = 4.84
56.2/(2^4.84) = 1.36 gr

if you wanna check it:
you know it halved approx. 5x
56/2 = 28
28/2 = 14
14/2 = 7
7/2 = 3.5
3.5/2 = 1.75

That number is very close to it! There's your answer!

poehalcho wrote:I can somehow see that this is incredibly simple...
I haven't actually had this kind of stuff, but here's my go...

60/ 12.4 = 4.84
56.2/(2^4.84) = 1.36 gr

if you wanna check it:
you know it halved approx. 5x
56/2 = 28
28/2 = 14
14/2 = 7
7/2 = 3.5
3.5/2 = 1.75

That number is very close to it! There's your answer!

I think thats wrong, you didn't find the constant..

Squirt wrote:

poehalcho wrote:I can somehow see that this is incredibly simple...
I haven't actually had this kind of stuff, but here's my go...

60/ 12.4 = 4.84
56.2/(2^4.84) = 1.36 gr

if you wanna check it:
you know it halved approx. 5x
56/2 = 28
28/2 = 14
14/2 = 7
7/2 = 3.5
3.5/2 = 1.75

That number is very close to it! There's your answer!

I think thats wrong, you didn't find the constant..

Well you didn't ask for one -.-

poehalcho wrote:


Well you didn't ask for one -.-

I specifically remembered no more multiplying by 2.

It won't work for this..

You could set it up like this:

N = N(0)(1/2)^(t/h)
N = ?
N(0)= 56.2
t = 60
h = 12.4
N = 56.2(1/2)^(60/12.4)
N = 1.963991312 mg


Got that off yahoo answers.. don't know if it's correct. It's been two years since I've taken chemistry.

Doing a quick dirty calculation.. if you go through the half lifes until 10.6 years are left you get 1.75.. so not going the entire 12 years for the decay would cause you to have a little more.. 1.96 looks right.

You're solving for N(t) = the ammount left at time t

All you need to do is plug in the variables you have been given.

λ = the half-life (keep it in the same units as you use for t, that is in years) = 12.4 years

N(0) = the ammount at time = 0 = 56.2mg in your case

ln = natural logarithm, i'm assuming you'll have access to a calculator and can use this function.

so,
ln[N(t)/56.2] = -12.4*60

applying some maths, [ ln(a/b) = ln(a)-ln(b) ]
ln[N(t)] - ln(56.2) = -744
ln[N(t)] = -739.971
e^{ln[N(t)]} = e^-739.971
N(t) = e^-739.971

fuccn genoiss on the forums now

Kaigar wrote:You're solving for N(t) = the ammount left at time t

All you need to do is plug in the variables you have been given.

λ = the half-life (keep it in the same units as you use for t, that is in years) = 12.4 years

N(0) = the ammount at time = 0 = 56.2mg in your case

ln = natural logarithm, i'm assuming you'll have access to a calculator and can use this function.

so,
ln[N(t)/56.2] = -12.4*60

applying some maths, [ ln(a/b) = ln(a)-ln(b) ]
ln[N(t)] - ln(56.2) = -744
ln[N(t)] = -739.971
e^{ln[N(t)]} = e^-739.971
N(t) = e^-739.971

luls, lambda's the decay constant, not the half life. OP gonna fail.

SM-Count wrote:luls, lambda's the decay constant, not the half life. OP gonna fail.

Haha, yep. Then there are also different equations for 0, first, and second order reactions..

Find K with equation: T = (-1/k)ln(2)

Where T is halflife.. so (-ln(2))/12.4 = -0.055898966

K = -0.055898966

Plug that into the original equation and see what ya get.

ln(Nt/56.2)= -(-0.055898966(12.4))

Solve for Nt.




I think, lol.

Hmmm... so where's the radioactive decay constant come from then? Do you need to derive it from the half-life some how?

Anyway, sorry if bad advise was bad.

ok, so lambda = log2/(half-life)

too late but....FFFUUU--

it's official...
srf sucks at chemistry...

This is physics

Kaigar wrote:This is physics

No its Chemistry..

Semantics!

What level of study is this at? high school? university?

high school

Kaigar wrote:Semantics!

What level of study is this at? high school? university?

AP course, High school

should ask a classmate