Asked by Ernie
If 18g of a radioactive substance are present initially and 5 yr later only 9g remains, how much of the substance will be present after 11yr?
Answers
Answered by
Reiny
Is the decay exponential, is it linear ?
It must be stated
It must be stated
Answered by
Ernie
linear
Answered by
Reiny
not linear in real life, but ....
so you have two points (0,18) and (5,9)
slope = (9-18)/(5-0) = -9/5
so y-18 = (-9/5)(x - 0)
y = (-9/5)x + 18
so when x = 11
y = (-9/5)(11) + 18 = -1.8
Now do you see how silly it is to label this linear ?
(after 10 years there would be nothing left)
So it has to be exponential, (and all the physics tutors now have stopped cringing) .
m = a e^(kt) , where m is the mass, a is the initial mass and t is the time in years
when t = 0 ---> 18 = a e^0 , a = 18
when t = 5 ---> 9 = 18 e^5k
.5... = e^ 5k
ln .5... = 5k lne
k = ln .5.../5
so when t = 11
m = 18 e^(11 ln .5../5) = appr 3.9 g
so you have two points (0,18) and (5,9)
slope = (9-18)/(5-0) = -9/5
so y-18 = (-9/5)(x - 0)
y = (-9/5)x + 18
so when x = 11
y = (-9/5)(11) + 18 = -1.8
Now do you see how silly it is to label this linear ?
(after 10 years there would be nothing left)
So it has to be exponential, (and all the physics tutors now have stopped cringing) .
m = a e^(kt) , where m is the mass, a is the initial mass and t is the time in years
when t = 0 ---> 18 = a e^0 , a = 18
when t = 5 ---> 9 = 18 e^5k
.5... = e^ 5k
ln .5... = 5k lne
k = ln .5.../5
so when t = 11
m = 18 e^(11 ln .5../5) = appr 3.9 g
Answered by
Ernie
Thanks Reiny
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