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?

Answer 1

Is the decay exponential, is it linear ?
It must be stated

Answer 2

linear

Answer 3

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

Answer 4

Thanks Reiny