Question 1150245: Iodine 131, which is produced by nuclear explosions, has a half-life of 8 days. In the spring of 1986, a nuclear explosion occurred at a nuclear power plant in Chernobyl, near Kiev in the Soviet Union. Suppose that the hay from nearby farms was contaminated by the fallout and could not be fed to dairy cows, since the iodine 131 would contaminate their milk. The hay was found to contain six times the allowable level of iodine 131 shortly after the explosion. How long must the hay be stored to be certain that the amount of iodine 131 is reduced to an acceptable level?
Use A = A0 ekt
Found 3 solutions by Boreal, ikleyn, Theo:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! A=Aoe^(kt)
find k with half-life of 8 days, so A=0.5 Ao
0.5Ao=Aoe^8k
0.5=e^8k
ln both sides
-0.693=8k
k=-0.0866
Ao=6*A
so this becomes
ln(1/6)=e^(-0.0866t)
-1.7918=-0.0866t
t=-20.68 days if I round at the very end
This is about 2.5 half-lives, and 1/2.5^2 is 1/6.25 or about 1/6 as a rough check.
Answer by ikleyn(52864)  (Show Source):
You can put this solution on YOUR website! .
The Chernobyl accident was not a nuclear explosion, in the strict meaning of this term.
It was a steam explosion, which destroyed the reactor.
The Chernobyl accident in 1986 was the result of a flawed reactor design that was operated with inadequately trained personnel.
The resulting steam explosion and fires released at least 5% of the radioactive reactor core into the atmosphere and downwind
https://www.world-nuclear.org/information-library/safety-and-security/safety-of-plants/chernobyl-accident.aspx
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! i think the formula you want is:
A = A0 * e ^ (kt)
A is the future value.
A0 is the present value.
k is the rate per time period.
t is the number of time periods.
you are given that iodine 131 has a half life of 8 days.
set A = 1
set Ao = 2
set t = 8 days
solve for k
the formula becomes 1 = 2 * e ^ (k * 8)
divide both sides of the equation by 2 to get:
1/2 = e ^ (k * 8)
take the natural log of both sides of this equation to get:
ln(1/2) = ln(e ^ (k * 8))
since ln(e ^ x) = x * ln(e) and since ln(e) = 1, then:
ln(1/2) = ln(e ^ (k * 8)) becomes:
ln(1/2) = k * 8 * ln(e) which becomes:
ln(1/2) = k * 8
solve for k to get:
k = ln(1/2) / 8 = -.0866433976.
replace k in the original equation with that to get:
1/2 = e ^ (-.0866433976 * 8) which becomes:
1/2 = .5 = 1/2.
this confirms the value of k is correct.
the hay was found to contain 6 times the allowable amount of iodine 131.
to be acceptable, it would have to contain 1 times the allowable amount of iodine 131.
using the value of k calculated above, the formula becomes:
1 = 6 * e ^ (-.0866433976 * t)
divide both sides of this formula by 6 to get:
1/6 = e ^ (-.0866433976 * t)
take the natural log of both sides of this equation to get:
ln(1/6) = ln(e ^ (-.0866433976 * t)).
by the same log rules given above, this becomes:
ln(1/6) = -.0866433976 * t
divide both sides of this equation by -.0866433976 to get:
ln(1/6) / -.0866433976 = t
solve for t to get:
t = 20.67970001 days.
it will take that many days for the level of iodine 131 to be acceptable.
you get:
1 = 6 * e ^ (-.0866433976 * 20.67970001).
thie becomes 1 = 1, confirming the solution is correct.
this equation can be graphed.
the equation for graphing is y = 6 * e ^ (-.0866433976 * x)
the graph is shown below.
the points on the graph are:
(0,6)
(8,3)
(20.68,1)
the points are in (x,y) format.
x represents the number of days since the nuclear explosion.
y represents the number of times the level of radiation is above the acceptable level.
at (0,6), there is 6 times the acceptable level of radiation at day 0, when the explosion occurs.
at (8,3), this level has been reduced in half to 3 times the acceptable level in 8 days.
at (20.68,1) this level has been reduced to 1 times the acceptable level of radiation in 20.68 days.
the hay must be stored a minimum of 20.68 days to get the radiation down to an acceptable level.
that's your solution.
|
|
|
