What is the rate of decay of awesome-10 which has a half-life of 650 days?

- anonymous

What is the rate of decay of awesome-10 which has a half-life of 650 days?

- schrodinger

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- anonymous

I can't figure out how to solve these rate of decay questions, so I just made up my own as a way to help me learn this. How would I solve this?

- anonymous

if it has a half life of 650 time units it means every 650 time units 1/2 of the samle dissapears.

- anonymous

Yep. :) How would I find the rate of decay, though?

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## More answers

- anonymous

halflife= -ln2/k
that is the half life equation

- anonymous

I have a formula \[N=N _{0}e ^{kt}\]
How/where would I plug everything in?

- anonymous

I'm not sure I understand @Spectrum

- anonymous

I know spectrum

- anonymous

i gave him all right answers

- anonymous

for his teset
he did good to

- anonymous

I think they are asking for k
you got N=1/2No
t=650
so you plug all numbers in and find k

- anonymous

I mean, I don't understand what he meant. I'm having a lot of trouble with logarithms, so I'm not quick on my feet. :P

- anonymous

\[N=\frac{ 1 }{ 2 }N _{0}^{650k}\]
Okay, so what do I do next?

- anonymous

Would I put some test number in for No? Like 100 or something?

- anonymous

no,|dw:1354966748637:dw| you plug 1/2No for Nthen you can simplify for No

- anonymous

|dw:1354967053153:dw|

- anonymous

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- anonymous

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- anonymous

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- anonymous

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- anonymous

I think now it makes sense. Doesn't it?

- anonymous

KIS means keep it simple.

- anonymous

Derp still need help understanding =/

- anonymous

The easiest way I believe is to derive it from calculus (if you've taken Diff EQ , you should know how to). Doing this you get an equation much like yours only N_0 is denoted c
\[A=ce^{kt}\]

- anonymous

now we're given two values
to find what c is you need to know that
\[N(0)=N_0\]
\[N_0=ce^0=c\]
\[N=N_0e^{kt}\]

- anonymous

same as your equation now we know that after 650 days , we will have half of the original amount soo
\[N(650 days)=.5N_0\]
\[.5N_0=N_0e^{650k}\]
solve for k, divide by \[N_0\]

- anonymous

\[\frac{.5N_0}{N_0}=e^{650k}\]
the initial N's cancel so you get
\[.5=e^{650k}\]
to get rid of e take the ln of both sides
\[ln(.5)=650k\]
\[\frac{ln(.5)}{650}=k\]

- anonymous

your rate of decay is \[\frac{ln(.5)}{650}=k\]
your equation will be
\[N=N_0e^{\frac{ln(.5)t}{650}}\]

- anonymous

@Outkast3r09 : same thing I did. That is why I was confused with @mahmit2012 had done

- anonymous

\[\frac{dP}{dt}=kP\]
\[\frac{dP}{dt}-kP=0\]
\[e^{\int{kdt}}=e^{kt+c}=ce^{kt}\]

- anonymous

you integrate the left side you'll just get that and the right side will be P

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