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Mathematics 24 Online
OpenStudy (anonymous):

who is good at half-life problems?

OpenStudy (anonymous):

Shoot...

OpenStudy (anonymous):

Thirty percent of a radioactive substance decays in four years. Assuming the decay is exponential, find half life of the substance.

OpenStudy (anonymous):

I am stumped. Sorry. Need to go study this one!

OpenStudy (zarkon):

pretty easy...where are you stuck

OpenStudy (anonymous):

the whole problem

OpenStudy (anonymous):

no worry's GT

OpenStudy (zarkon):

if you start with one unit of substance , how much would you have after 4 years

OpenStudy (zarkon):

can you answer that

OpenStudy (anonymous):

no

OpenStudy (anonymous):

0.7 units

OpenStudy (zarkon):

yes..now plug that into the exponential decay model and solve for k \[A=A_0e^{kt}\]

OpenStudy (anonymous):

You are solving for e^kt = 1/2?

OpenStudy (zarkon):

not yet...we have to find k first

OpenStudy (anonymous):

\[\ln .70/4=k\]

OpenStudy (zarkon):

good

OpenStudy (anonymous):

-0.089

OpenStudy (zarkon):

yes

OpenStudy (zarkon):

now use your k and solve this for t \[\frac{1}{2}=e^{kt}\]

OpenStudy (anonymous):

@zarkon - thanks! I learned (rather refreshed my memories!) of it.

OpenStudy (zarkon):

no problem

OpenStudy (anonymous):

ln(.5)/(?)=t

OpenStudy (anonymous):

would the -0.089 replace the ?

OpenStudy (anonymous):

It has to.

OpenStudy (zarkon):

yes...that is where k goes

OpenStudy (anonymous):

7.788=t rite

OpenStudy (zarkon):

I get 7.7734328395

OpenStudy (zarkon):

I let the calculator keep all the digits

OpenStudy (anonymous):

ohh okay i see now

OpenStudy (anonymous):

The radioactive element polonium has a half-life of 140 days. If 15 mg of this substance decays exponentially, find: A.) How much of the substance would remain after 60 days. B.) how long would it take before it becomes 4 mg.

OpenStudy (zarkon):

solve \[\frac{1}{2}=e^{k\cdot140}\] for k

OpenStudy (zarkon):

after finding k A) evaluate \[15e^{k\cdot 60}\] B) solve \[4=15e^{k\cdot t}\] for t

OpenStudy (anonymous):

okay let me try it

OpenStudy (anonymous):

-0.004951051=k

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