Question

The half-life of carbon-14 is approximately 6000 years. Determine how much of 100 kg of this substance will remain after 12,000 years.

Answer 1

@terenzreignz

Answer 2

@amistre64 @SolomonZelman

Answer 3

Dang I forgot how to do these. -:(

I am replying to get a notification and to later view this question. (You guys are probably thinking that I am typing the greatest tip/solution in the world right now... well , not really)

Answer 4

I was... going to ask you if you were up to the task, Zelman XD

Okay, listen up...
What do you know about half life, @hello1213 ?

Answer 5

I know the formula for half-life decay... it is\[N=N _{0}(\frac{ 1 }{ 2 })^\frac{ t }{ h }\]

Answer 6

Well, It's just a matter of plugging in :D
t here is the time that passed
and h is the half life
\(N_0\) is the amount of the substance at the start.

Plug in and be done ^_^

Answer 7

Well I think that's where I am having trouble not plugging in, but simply solving the equation.

Answer 8

Okay... here we go...
\[\Large N = N_0 \left(\frac12\right)^{\frac t h}\]
Let's plug in, one at a time.
What is \(N_0 = \color{red}?\)

Answer 9

100?

Answer 10

Right.
\[\Large N =\color{blue}{100} \left(\frac12\right)^{\frac t h}\]
What is \(t =\color{red}?\)
Or, in other words... how much time has passed?

Answer 11

12,000

Answer 12

Correct again :D
\[\Large N =100\left(\frac12\right)^{\frac {\color{blue} {12000}} h}\]
What about h, the half life?

Answer 13

so h would be 6000.

Answer 14

Okay, we finally have N in terms of numbers only... lol
\[\Large N = 100 \left(\frac12\right)^{\frac {12000}{6000}}\]

But... first things first, what is \(\Large \frac{12000}{6000}\) equal to?

Answer 15

2

Answer 16

Right...
\[\Large N = 100 \left(\frac12\right)^2\]
Can you take it from here?

Answer 17

Ohhh, I didn't realize 2 was a power before, so I was writing it wrong, and multiplying 2 by 1/2, when 2 was a power... :/... Yes so the answer is N=25!

Answer 18

That is correct.
Well done ^_^

Answer 19

Thanks again... You always make these seemingly difficult problems easy!

Answer 20

I do try ^_^