## anonymous 5 years ago Suppose the amount of ozone in the atmosphere is decaying exponentially, losing 0.25% each year. How many years does it take for the ozone level to drop 24%? A) 78 years B) 86 years C) 94 years D) 102 years E) 110 years

1. anonymous

Is this a first order reaction?

2. Yuki

these problems are easier than it looks like

3. Yuki

let's say N is the current amount and N_0 be the original amount

4. Yuki

in one year, N-0 becomes 99.75% of N_0 so $N_1 = .9975*N_0$

5. Yuki

next year, N_1 becomes 99.75% of N_1 so $N_2 = .9975*N_1$

6. Yuki

if you substitute N_1 with N_0, $N_2 = (.9975)*(.9975N_0)$

7. Yuki

as you can see, the number of years will just tell you how many times the initial amount decreased by .25%, so the general eqn after t years is $N_t = N_0(.9975)^t$

8. Yuki

I f we used this eqn, we can say that N_t = .24N_0, because it is 24% of the original state. or in other words, $.24 = (.9975)^t$

9. Yuki

all you have to do is to solve for t :)

10. Yuki

oops, sorry not .24 it is actually .76 because the wording says "drops 24%" not "is 24%" so you are solving for $.76 = (.9925)^t$

11. anonymous

12. Yuki

since$t = {\ln(.76) \over \ln(.9975)}$ which is approximately 109.6 I'd use 110

13. Yuki

yep

14. anonymous

thanks!

15. Yuki

np :) most exponential decays and growths work the same way, so let me know if you need more help ok ?