Question

The half-life of iodine-131 is 8.10 days. How long will it take for three-fourths of a sample of iodine-131 to decay?

can someone explain to me how to do half-life problems, please?

Answer 1

For a first order reaction, you can use: \(\Large t_{1/2}=\dfrac{ln2}{k}\), to find k, the decay constant.

After you can use the exponential growth/decay equation: \(\Large A_{t}=A_0*e^{-kt}\),

where \(A_{t}\) is the amount after t time elapsed, \(A_0\) is the initial amount. t is time, and k is the decay constant (you found above).

Answer 2

what is In2? im stilll confused?

Answer 3

can you show me how the problem is done?

Answer 4

"ln" is the natural logarithm

Answer 5

oh wowzers

Answer 6

lol okay, so plug in your half life into the first equation, it should lok ike this:

8.10 = \(\dfrac{ln2}{k}\)

solve for k

Answer 7

use a calculator, obvsiously.

Answer 8

2.44?

Answer 9

ln2=0.693

0.693/8.1=0.008557=k

Answer 10

is that just common knowledge that ln = 0.693?
lol

Answer 11

nope, you have to use your calculator ln(2) lol

Theres an easier method for this question, but it won't work for every question of this type.

Answer 12

ohh well then dont tell me the easier way

Answer 13

is the answer what k equals?

Answer 14

no, thats just the decay constant for the next equation.

you would plug in your values into this: \(\large A_{t}=A_0*e^{-kt}\)
where \(A_{t}\) is the amount after t time elapsed, \(A_0\) is the initial amount. t is time, and k is the decay constant (you found above).

can you try to figure out which values go where?

Answer 15

would At be 8.10 days and Ao would be 3/4?

Answer 16

no, 8.10 days is the half-life \((t_{1/2})\).
we used it in the first part and we don't need to use it again.

Ao is the initial amount, so 1
At is the amount you have after 3/4 of the amount decayed, so what is At?

Answer 17

im just going to take a stab in the dark and say 10.9? i did 181 divided by 3/4

Answer 18

where did you get 181 from :P lol

Answer 19

iodine -131? haha

Answer 20

im still confused as to what A equals

Answer 21

131 is just the mass of the isotope, thats not really relevant to the question here.

so Ao is the initial amount, that we set as 1 arbitrarily (we could've said it was any number, but we choose 1 to make it easy).

so At is the amount after "t" time elapsed, they said 3/4 of the sample decayed,

so At=1-3/4=1/4

cool?

Answer 22

got it!

Answer 23

so 1/4=1 times e ^ -0.0008557 ?

Answer 24

okay, so we plug everything in: \(\large \dfrac{1}{4}=(1)*e^{-0.008557\color{red}t}\)

now, we only need to solve for t. do you know how to solve logarithms?

Answer 25

ahh no :( sorry.. dont i just plug it in my calulator?

Answer 26

yeah, but you first need to rearrange it using logarithm rules. Unless you have a really good calculator.

Answer 27

how do i arrange it?

Answer 28

by the way i wrote too many zeroes for k, k=0.08557

\(\dfrac{1}{4}=(1)*e^{-0.08557t}\)

take "ln" of both sides

\(ln(\dfrac{1}{4})=ln(e^{-0.08557t})\)

using logarithm rules the exponent can come down

\(ln(\dfrac{1}{4})=-0.08557t*ln(e)\)

\(ln(e)=1\; because \;ln=log_e, \; so \;ln_e(e)=1\)

were left with:

\(ln(\dfrac{1}{4})=-0.08557t\)

from here just isolate t

\(\dfrac{ln(\dfrac{1}{4})}{-0.08557}=t\)

plug into your calculator

Answer 29

16.20?

Answer 30

yep. thats right.

if you notice the easier way would've been finding how many half-lives it takes to get to 1/4 of the original.

if you have 1 to start with
after one half life you have 1/2
then after other half life you have 1/4

2 * half-life=2*8.1 days = 16.2 days

Answer 31

hahah oh wow so much easier that way! but now i know for other problems(: thanks so much i appreciate your time!(:

Answer 32

haha yeah, it's a lot easier but it won't work if they give you random numbers like, find how long it takes to have 0.18491923 of the original. With the method i showed you you CAN solve that. lol no problem !