1. ## Half Life

This one is half-life I know but I cannot figure out how to go through it, it isn't finding the amount so none of the examples my teacher gave me can show how to move things around...any help is appreciated.

The half-life of sulfur-138 is 4 days. After 12 days,a sample of this radioactive material has been reduced to a mass of 0.64g.

a) What is the initial mass?
b) What is the mass after 2 days?

Also if it wouldn't hurt could someone could walk through it, if not then just show all the steps to solve it.

2. Originally Posted by risteek
This one is half-life I know but I cannot figure out how to go through it, it isn't finding the amount so none of the examples my teacher gave me can show how to move things around...any help is appreciated.

The half-life of sulfur-138 is 4 days. After 12 days,a sample of this radioactive material has been reduced to a mass of 0.64g.

a) What is the initial mass?
b) What is the mass after 2 days?

Also if it wouldn't hurt could someone could walk through it, if not then just show all the steps to solve it.
$y_0$ = amount of $S_{138}$ at time $t = 0$ days

$y$ = amount of $S_{138}$ at any time, $t$ , in days

$y = y_0 \left(\frac{1}{2}\right)^{\frac{t}{4}}
$

$0.64 = y_0 \left(\frac{1}{2}\right)^{\frac{12}{4}}
$

$y_0 = 5.12$ g

you do part (b)

3. Originally Posted by risteek
This one is half-life I know but I cannot figure out how to go through it, it isn't finding the amount so none of the examples my teacher gave me can show how to move things around...any help is appreciated.

The half-life of sulfur-138 is 4 days. After 12 days,a sample of this radioactive material has been reduced to a mass of 0.64g.

a) What is the initial mass?
b) What is the mass after 2 days?

Also if it wouldn't hurt could someone could walk through it, if not then just show all the steps to solve it.

Formulæ for half-life in exponential decay
Main article: Exponential decay
An exponential decay process can be described by any of the following three equivalent formulae:

Nt = N0et / τ Nt = N0e − λt where
• N0 is the initial quantity of the thing that will decay (this quantity may be measured in grams, moles, number of atoms, etc.),
• Nt is the quantity that still remains and has not yet decayed after a time t,
• t1 / 2 is the half-life of the decaying quantity,
• τ is a positive number called the mean lifetime of the decaying quantity,
• λ is a positive number called the decay constant of the decaying quantity.
You've got $N_t=0.64$, and $t_{1/2}=4$
and
$t=12$

so

$0.64=N_0(\frac{1}{2})^{12/4}$

What's eight times 0.64?

4. Thank you so much! But where does the eight come from?

5. Originally Posted by risteek
Thank you so much! But where does the eight come from?
\begin{aligned}
0.64 &= N_0(\frac{1}{2})^{12/4} \\
0.64 &= N_0(\frac{1}{2})^3 \\
0.64 &= N_0(\frac{1}{8}) \\
8(0.64) &= N_0
\end{aligned}

That's where the eight came from.

01

6. Originally Posted by risteek
Thank you so much! But where does the eight come from?
$\left(\frac{1}{2}\right)^3 = \frac{1}{8}$

7. Wooooow. I feel dumb I can't even use exponents right. >.< Thanks!

8. $y = y_0 \left(\frac{1}{2}\right)^{\frac{t}{h}}
$

$y = 5.12 \left(\frac{1}{2}\right)^{\frac{2}{4}}
$

$y = 2.56$ g

Is that right?

9. (1/2)^3=8

10. Originally Posted by risteek
$y = y_0 \left(\frac{1}{2}\right)^{\frac{t}{h}}
$

$y = 5.12 \left(\frac{1}{2}\right)^{\frac{2}{4}}
$

$y = 2.56$ g

Is that right?
Nope...
\begin{aligned}
y &= 5.12 \left(\frac{1}{2}\right)^{\frac{2}{4}} \\
y &= 5.12 \left(\frac{1}{2}\right)^{\frac{1}{2}} \\
y &= 5.12 \left(\frac{1}{\sqrt{2}}\right) \\
y &\approx 3.62
\end{aligned}

01

11. Originally Posted by VonNemo19
(1/2)^3=8
You mean 1/8.

01

12. Originally Posted by yeongil
Nope...
\begin{aligned}
y &= 5.12 \left(\frac{1}{2}\right)^{\frac{2}{4}} \\
y &= 5.12 \left(\frac{1}{2}\right)^{\frac{1}{2}} \\
y &= 5.12 \left(\frac{1}{\sqrt{2}}\right) \\
y &\approx 3.62
\end{aligned}
My teacher never taught us to square route anything with in the equation... is this question an exception?

13. Originally Posted by risteek
$y = y_0 \left(\frac{1}{2}\right)^{\frac{t}{h}}
$

$y = 5.12 \left(\frac{1}{2}\right)^{\frac{2}{4}}
$

$y = 2.56$ g

Is that right?
nope ...

$y = 5.12 \left(\frac{1}{2}\right)^{\frac{1}{2}} = 3.62$ g

14. Oh I get it, i forgot to put the exponent into the lowest possible form. Thanks guys

15. Originally Posted by risteek
Oh I get it, i forgot to put the exponent into the lowest possible form. Thanks guys
come back anytime