Question

What is the half-life of a radioactive substance if it takes 5 years for one-third of the material to decay?

Answer 1

If I use the formula P(t) = Ce^rt, I will just plug in to get

1/3 = e^5r and then solve for r.

I got r = ln |1/3| /5

and then i plug back into the equation to get half life and solve for time??

1/2 = e^(ln |1/3|/5)t

and solving for t i get
5 ln |1/2|/ ln |1/3| = t

and then i am done???

Answer 2

is this correct?

Answer 3

\[\frac{1}{3}=e ^{-k5}\]
Solve to find the decay constant k.
The plug in the value of k into the following and solve for t.\[\frac{1}{2}=e ^{-kt}\]

Answer 4

y did you use -k? is that part of the formula, i thought it was just a regular constant.

Answer 5

so k = -5/ln |1/3|

Answer 6

and t = (ln |1/2| ln |1/3| )/ 5

can you simplify the top

Answer 7

\[k=\frac{\ln \frac{1}{3}}{-5}\]

Answer 8

k=0.219722457

Answer 9

dang it. okay i see my mess up on k

Answer 10

In the formula for radioactive decay the exponent of e is negative.

Answer 11

so t will be

5(ln |1/2|)/ln |1/3|

Answer 12

can we simplify that?

Answer 13

\[Half\ life=\frac{\ln \frac{1}{2}}{-0.219722457}\]

Answer 14

No need to simplify. Just use a calculator.

Answer 15

so t = 3.15 years?

Answer 16

Correct.

Answer 17

but if it takes 5 years for 1/3, then how does that make sense???

Answer 18

Actually I misread the question. The amount of material left after 5 years is 2/3 of the original amount.
So recalculating:
\[\ln \frac{2}{3}=-5k\]
\[k=\frac{\ln \frac{2}{3}}{-5}=0.081\]
\[half\ life=\frac{\ln \frac{1}{2}}{-0.081}\]

Answer 19

Okay, so you use the amout left and not the amount used. makes much more sense now... thank you.

Answer 20

You're welcome :)