# Half Life

• June 10th 2009, 12:51 PM
risteek
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.
• June 10th 2009, 01:05 PM
skeeter
Quote:

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)
• June 10th 2009, 01:14 PM
VonNemo19
Quote:

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:

http://upload.wikimedia.org/math/a/1...773909d256.png 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?
• June 10th 2009, 01:29 PM
risteek
Thank you so much! But where does the eight come from?
• June 10th 2009, 01:51 PM
yeongil
Quote:

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
• June 10th 2009, 01:51 PM
skeeter
Quote:

Originally Posted by risteek
Thank you so much! But where does the eight come from?

$\left(\frac{1}{2}\right)^3 = \frac{1}{8}$
• June 10th 2009, 01:53 PM
risteek
Wooooow. I feel dumb I can't even use exponents right. >.< Thanks!
• June 10th 2009, 02:44 PM
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?
• June 10th 2009, 02:53 PM
VonNemo19
(1/2)^3=8
• June 10th 2009, 03:11 PM
yeongil
Quote:

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
• June 10th 2009, 03:13 PM
yeongil
Quote:

Originally Posted by VonNemo19
(1/2)^3=8

You mean 1/8. (Doh)

01
• June 10th 2009, 03:26 PM
risteek
Quote:

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?
• June 10th 2009, 03:32 PM
skeeter
Quote:

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
• June 10th 2009, 03:39 PM
risteek
Oh I get it, i forgot to put the exponent into the lowest possible form. Thanks guys :D
• June 10th 2009, 04:21 PM
VonNemo19
Quote:

Originally Posted by risteek
Oh I get it, i forgot to put the exponent into the lowest possible form. Thanks guys :D

come back anytime(Wink)