Question

Where N(t) = N(0)e^kt
k = -0.1386 what percentage of the isotope do I have after 15 days?

Answer 1

We have: \[
N(t) = N(0)e^{kt}
\]If we divide we get: \[
\frac{N(t)}{N(0)} = e^{kt}
\]Now \(N(t)/N(0)\) will be the remaining percentage of the isotope.

Answer 2

So we want to find \(N(15)/N(0)\), which equals: \[
e^{kt} = e^{-0.1386t}= e^{-0.1386(15)}
\]

Answer 3

half life = .693/k

k = -.1386 therefore, half life =
.693/-.1386

half life = -5

Answer 4

therefore 12.5%

Answer 5

I don't know why half life comes out negative but if it is 5 days, then every 5 days, we lose one half of the original sample.

So we start with 100% at day zero.

At day 5, we have lost half of the sample and so 50% remains

At day 10, we have lost half of THAT sample and so 25% remains

At day 15, we have lost half of THAT sample and so 12.5% remains

You are correct Kerry!!

Answer 6

Thank you so much everyone xoxo

Answer 7

u r welcome Kerriy

Answer 8

When will 95% remain?

Answer 9

Am I solving for t?

Answer 10

Using the formula from this web page:
http://www.1728.org/halflife.htm

Ending amt = beginnig amount / 2^n
where n is the number of half lives

Answer 11

I gave the wrong formula

ending amount = 95
beginning amount = 100
half life = 5

time = half-life * (log(beg'ng amt/ending amt))/log 2

Answer 12

time = 5 * log(100/95) / log (2)

time = 0.0222763947 / 0.3010299957

time = 0.0740005814 days

Answer 13

Forgot to multiply by 5 ARRGGGHHHH !!!!!!!!!!!!!!!!!!!!!!!!
5 * .0740005814=
0.3700029072 days

Answer 14

using formula N(t)=N(0)e^kt

Answer 15

By the way there is a calculator on that page
http://www.1728.org/halflife.htm
and it does calculate .37 days
Now for using that formula huh?

Answer 16

lol

Answer 17

We KNOW the answer is .37 days so that is a good start.

Answer 18

ok got it so does that mean 37 days?

Answer 19

no it means in .37 days if you started with 100% of a substance in .37 days you would have 95% of that

Answer 20

roughly 8.9 hours

Answer 21

I think I might be able to re write that whacky "e" formula.

Answer 22

time = half-life * (natural log(beg'ng amt/ending amt))/natural log (2)
time = 5 * natural log (100/95) / natural log (2)
time = 5 * 0.051293294343 / 0.69314718056

and that equals ................
time = 0.3700029069 days !!!

Answer 23

cooly dooly - thanks again :) So time is on x axis and percentages on y?

Answer 24

gee I'm not too sure how to graph it but the formula you need for that time calculation is
time = half-life * (natural log(beg'ng amt/ending amt))/natural log (2)
anyway, I have to go but it was fun helping you out !!!!!!!!!!!!!!!!