# Thread: [SOLVED] help with half-life, formula with time and remaining time formula

1. ## [SOLVED] help with half-life, formula with time and remaining time formula

Okay I don't understand any of this & if I could get someones help I would appreciate it so much. We learned before the compounding formulas. I thin some how they relate but I don't know. Please help.

Problem:

At the beginning of an experiment, a scientist has 216 grams of radioactive goo. After 225 minutes, her sample has decayed to 27 grams.

What is the half-life of the goo in minutes?

Find a formula for , the amount of goo remaining at time .

How many grams of goo will remain after 8 minutes?

Hint: do part 2 first, i.e. you are just finding k from given pair. Then to find half-life, set outcome G(t) to be 1/2 of starting amount and solve for t.

2. Hello there,

From my general chemistry studies, I know that exponential growth and decay can be modelled by:

$G(t) = Ae^{kt}$, where A is the starting amount, k is just a constant, and t is the time.

I will show you #2 and then you will be able to do #1.

$G(t) = Ae^{kt}$

Your information indicates terminating mass as 27 g, A = 216 g, and t = 27 min:

$27 = 216 e^{k27}$

$ln\frac{1}{8} = k27$

$\Rightarrow k = 0.077016$.

I hope that this helps. Be sure to write back for more help.

3. Originally Posted by scherz0
Hello there,

From my general chemistry studies, I know that exponential growth and decay can be modelled by:

$G(t) = Ae^{kt}$, where A is the starting amount, k is just a constant, and t is the time.

I will show you #2 and then you will be able to do #1.

$G(t) = Ae^{kt}$

Your information indicates terminating mass as 27 g, A = 216 g, and t = 27 min:

$27 = 216 e^{k27}$

$ln\frac{1}{8} = k27$

k = -0.077016

I hope that this helps. Be sure to write back for more help.

Half life is given by:

$t_{\frac{1}{2}} = \frac{\ln(2)}{k}$

and is derived from the standard exponential decay equation:

$G(t) = G(0)e^{-kt}$

4. Thanksd scherz0 and e^(i*pi) but I am more lost than before. I am not saying it's both of your faults or what not. I am just not sure what is going on.

5. Originally Posted by scherz0
Hello there,

From my general chemistry studies, I know that exponential growth and decay can be modelled by:

$G(t) = Ae^{kt}$, where A is the starting amount, k is just a constant, and t is the time.

I will show you #2 and then you will be able to do #1.

$G(t) = Ae^{kt}$

Your information indicates terminating mass as 27 g, A = 216 g, and t = 27 min:

$27 = 216 e^{k27}$

$ln\frac{1}{8} = k27$

$\Rightarrow k = 0.077016$.

I hope that this helps. Be sure to write back for more help.

wouldn't the time be 225? why 27?

6. Originally Posted by lsnyder
Okay I don't understand any of this & if I could get someones help I would appreciate it so much. We learned before the compounding formulas. I thin some how they relate but I don't know. Please help.

Problem:

At the beginning of an experiment, a scientist has 216 grams of radioactive goo. After 225 minutes, her sample has decayed to 27 grams.

What is the half-life of the goo in minutes?

Find a formula for , the amount of goo remaining at time .

How many grams of goo will remain after 8 minutes?

Hint: do part 2 first, i.e. you are just finding k from given pair. Then to find half-life, set outcome G(t) to be 1/2 of starting amount and solve for t.
Since 216/2 = 108, 108/2 = 54 and 54/2 = 27 it should be clear that 225 minutes corresponds to 3 half lives. Therfore half life = 225/3 = 75 minutes.

7. Originally Posted by lsnyder
Okay I don't understand any of this & if I could get someones help I would appreciate it so much. We learned before the compounding formulas. I thin some how they relate but I don't know. Please help.

Problem:

At the beginning of an experiment, a scientist has 216 grams of radioactive goo. After 225 minutes, her sample has decayed to 27 grams.

What is the half-life of the goo in minutes?
This is an "exponential" problem but it isn't necessary to use "e" itself. If something has a half life of T minutes, then it is multiplied by 1/2 every T minutes. In t minutes there are t/T periods of T minutes so you multiply by $\left(\frac{1}{2}\right)^{t/T}$. If it starts at 216 grams then t minutes later you have $216\left(\frac{1}{2}\right)^{t/T}$. After 225 you hav 27 so $216\left(\frac{1}{2}\right)^{225/T}= 27$. Then $\left(\frac{1}{2}\right)^{225/T}= \frac{27}{216}= \frac{1}{8}= \left(\frac{1}{2}\right)^3 0.125$. So we must have 225/T= 3, T= 225/3= 75 minutes. That's very radioactive stuff!

Find a formula for , the amount of goo remaining at time .
The formula is $P(t)= 216\left(\frac{1}{2}\right)^{t/75}$.

[quote]How many grams of goo will remain after 8 minutes?
This is an ambiguous question. Does it refer to 8 minutes after the original time, when the mass was 216 grams or is it 8 minutes after the first 225 minutes? If the first, use $P(8)= 216\left(\frac{1}{2}\right)^{8/75}$. If the latter either set t= 225+ 8= 233 minutes so $P(233)= 216\left(\frac{1}{2}\right)^{233/75}$ or, equivalently, "restart" at 225 minutes: $27\left(\frac{1}{2}\right)^{8/75}$.

Hint: do part 2 first, i.e. you are just finding k from given pair. Then to find half-life, set outcome G(t) to be 1/2 of starting amount and solve for t.

8. wow, that made a lot more since.
Thank you Mr. Fantastic and Halls

9. HallsofIvy, it was the first one on the question about the 8 minutes. just incase you wondered which equalled to 200.6059846. Thanks again