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?
Is the decay exponential, is it linear ?
It must be stated
linear
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
Thanks Reiny
To determine how much of the radioactive substance will be present after 11 years, we need to determine the decay rate of the substance.
The decay rate of a radioactive substance is commonly measured by its half-life. A half-life is the amount of time it takes for half of the initial quantity of the substance to decay.
In this case, we know that after 5 years, only 9g of the substance remains. This means that half of the substance has decayed in those 5 years.
Using this information, we can calculate the half-life of the substance. If half of the initial quantity decays in 5 years, we can assume that the half-life is also 5 years.
Now, let's determine how many half-lives have passed between the initial amount of the substance and the time in question (11 years). To do this, we divide the total time (11 years) by the half-life (5 years):
11 years / 5 years = 2.2 half-lives
Since 2.2 half-lives have passed, we can calculate how much of the substance remains by multiplying the initial quantity (18g) by (1/2)^2.2:
Remaining quantity = Initial quantity * (1/2)^(number of half-lives)
Remaining quantity = 18g * (1/2)^2.2
Using a calculator, we can find that (1/2)^2.2 is approximately 0.334.
Therefore, the amount of the substance that will be present after 11 years is:
Remaining quantity = 18g * 0.334
Remaining quantity ≈ 6.01g
So, approximately 6.01 grams of the substance will be present after 11 years.