1. ## -HELP-

After the release of radioactive material into the atmosphere from a nuclear power plant at Chernobyl (Ukraine) in 1986, the hay in Austria was contaminated by iodine 131 (half-life 8 days). If it is alright to feed the hay to cows when 10% of the iodine 131 remains, how long do the farmers need to wait to use this hay?

2. by using

$\displaystyle N_t = N_0 (1/2)^{t/t_{1/2}}$

we get

$\displaystyle .1 = 1(1/2)^{t/8}$

now just solve for t, using logarithims

i think thats how you set it up, if not change the 1 and 0.1 to 100 and 10

3. Originally Posted by treetheta
by using

$\displaystyle N_t = N_0 (1/2)^{t/t_{1/2}}$

we get

$\displaystyle .1 = 1(1/2)^{t/8}$

now just solve for t, using logarithims

i think thats how you set it up, if not change the 1 and 0.1 to 100 and 10
Um...I don't get it.

4. No= Original
So we know that the orginal has to be 100% so we have to let it equal 1, or we can let it equal 100, it doesn't matter

Nt = Amount leftover
So we want 10% to be leftover so it is usable to feed the cows thats why we let it equal .1 or 10, it doesnt matter,

t1/2 = constant half-life decay
Since it tells us in the questions that radiocative decay in the hay is 8 days, we can let this be 8 then when we solve for t

We can find out how many days the hay hasn't to not be eaten for it to decay to 10% so it can be eatable for the cows

I hope that clears it up

you can always use the beloved,

Half-life - Wikipedia, the free encyclopedia

to clear things up

5. Originally Posted by treetheta
No= Original
So we know that the orginal has to be 100% so we have to let it equal 1, or we can let it equal 100, it doesn't matter

Nt = Amount leftover
So we want 10% to be leftover so it is usable to feed the cows thats why we let it equal .1 or 10, it doesnt matter,

t1/2 = constant half-life decay
Since it tells us in the questions that radiocative decay in the hay is 8 days, we can let this be 8 then when we solve for t

We can find out how many days the hay hasn't to not be eaten for it to decay to 10% so it can be eatable for the cows

I hope that clears it up

you can always use the beloved,

Half-life - Wikipedia, the free encyclopedia

to clear things up

o k... i got this so far.
ln .1 = ln (1/2)^(t/8)

6. you forgot the 1, and u can bring the exponent down, by properties of logarithms, then isolate for t,

if you need to know the logarithms:

Logarithm - Wikipedia, the free encyclopedia

7. Originally Posted by treetheta
you forgot the 1, and u can bring the exponent down, by properties of logarithms, then isolate for t,

if you need to know the logarithms:

Logarithm - Wikipedia, the free encyclopedia
o k...this is what I have now.
log .1 = log 1(1/2)^(t/8)

8. er...
did you look at the wikipedia site?
$\displaystyle .1 = 1(1/2)^{t/8}$

take ln of both sides

$\displaystyle ln.1 = ln(1/2)^{t/8}$
$\displaystyle ln.1 = t/8[ln0.5]$
$\displaystyle 8(ln.1)/(ln0.5) = t$