Question

Krypton-91 is a radioactive substance that decays very quickly. The function Q(t)=Qoe^-kt models radioactive decay of krypton-91. Q represents the quantity remaining after t seconds and the decay constant k is approximately 0.07. How long will it take a quantity of krypton-91 to decay t o10% of its origional amount? round your answer to the nearest second.

Answer 1

Cant tell if this is chemistry or math

Answer 2

Its math! precalc:)

Answer 3

Well can yah help with this?

Answer 4

i giv medal fan and testimony

Answer 5

dude I cant I dont have time sorry :(

Answer 6

awrh

Answer 7

Lets suppose the original amount Qo is 100 grams of the radioactive substance

Answer 8

Okay

Answer 9

Lets suppose the original amount Qo is 100 grams of the radioactive substance.

Therefore the amount of substance left at time t is modeled by the equation:
Q(t) = 100*e^(-.07*t)

We want to find how long will it take to decay to 10% of its original amount. so what is 10% of 100 grams? that is 10 grams
So we need to solve this equation:

10 = 100 * e^(-.07 * t )

Answer 10

alright so first thing I would do is divide both sides by 100:)

Answer 11

right

Answer 12

I get -1.43 seconds after dividing by 100 on both sides I get
.1=e^-0.07t
then divide by -0.07 on both sides and
so .1/-0.07=-1.43 when rounded up.
am I right??

Answer 13

@perl

Answer 14

I agree with your work up to here
.1=e^-0.07t

Answer 15

now you can't divide both sides by -.07, since -.07 is an exponent

Answer 16

oh then I have to take ln(.1)/-0.07

Answer 17

right
$$ \Large .1 = e^ {-.07t}
\\~\\

$$

Answer 18

t = ln (.1) / -.07

Answer 19

Okay so after doing that I get t= -32.89

Answer 20

t should be positive

Answer 21

but we cant have negative seconds

Answer 22

ln(.1) / (-.07) = 32.89

Answer 23

ah I redid it and got that! thanks:)) I forgot to do -.07