1. Growth/Decay Question

This problem uses radioactive isotopes. I need to find the amount of grams after 4000 years. I'm given:
Isotope:14-C
Half-life (years): 5715
Initial Quantity: 15 g
Amount After 4000 Years: ?? g

Also unsure as to starting this one.

2. Originally Posted by BeSweeet
This problem uses radioactive isotopes. I need to find the amount of grams after 4000 years. I'm given:
[i]Isotope:[/]14-C
Half-life (years): 5715
Initial Quantity: 15 g
Amount After 4000 Years: ?? g

Also unsure as to starting this one.
Use the following two formulae:

$\displaystyle {\lambda} = \frac{ln2}{t_{1/2}}$

$\displaystyle A(t) = A_0e^{-\lambda t}$

Where:

• $\displaystyle \lambda$ = Decay Constant
• $\displaystyle t_{1/2}$ = half life
• $\displaystyle A(t)$ = Amount left at time t
• $\displaystyle A_0$ = Amount at t=0
• $\displaystyle t$ = time.

3. Originally Posted by e^(i*pi)
Use the following two formulae:

$\displaystyle {\lambda} = \frac{ln2}{t_{1/2}}$

$\displaystyle A(t) = A_0e^{-\lambda t}$

Where:

• $\displaystyle \lambda$ = Decay Constant
• $\displaystyle t_{1/2}$ = half life
• $\displaystyle A(t)$ = Amount left at time t
• $\displaystyle A_0$ = Amount at t=0
• $\displaystyle t$ = time.
[tex]\lambda[/MATH - 15?
$\displaystyle t_{1/2}$ - 5715
$\displaystyle A(t)$ - ?
$\displaystyle A_0$ - 15
$\displaystyle t$ - ?

Help?

4. Originally Posted by BeSweeet
This problem uses radioactive isotopes. I need to find the amount of grams after 4000 years. I'm given:
[i]Isotope:[/]14-C
Half-life (years): 5715
Initial Quantity: 15 g
Amount After 4000 Years: ?? g

Also unsure as to starting this one.
Let t denote the time measured in years, A_0 the initial amount and A(t) the amount after t years. Then A(t) is given by:

$\displaystyle A(t)=A_0 \cdot e^{k\cdot t}$ where k is a constant which you have to calculate first.

You already know: If the initial value is A_0 then $\displaystyle A(5715) = \frac12 A_0$. That means:

$\displaystyle \frac12 A_0 = A_0 \cdot e^{k \cdot 5715}$ . Solve for k. I've got $\displaystyle k \approx -0.0001212856$

Now calculate A(4000) with $\displaystyle A_0 = 15\ g$

5. Originally Posted by BeSweeet
[tex]\lambda[/MATH - 15?
$\displaystyle t_{1/2}$ - 5715
$\displaystyle A(t)$ - ?
$\displaystyle A_0$ - 15
$\displaystyle t$ - ?

Help?
$\displaystyle \lambda = \frac{ln2}{5715}$ (which gives the same value as k as indicated by earboth)

A(t) is what you're trying to find.

t = 4000

6. I am so freaking confused right now.

This may be asking a lot, but could someone find the answer, and show the steps? The best way for me, personally, to figure out how to do something, is to analyze steps that get an answer.