- #1

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A substance decays from 20g to 15g in 7h.Determine the half-life of the substance.

I know that: [tex]M=M_{o}\left\frac{1}{2}\right^\frac{t}{h}[/tex]

I know that: [tex]M=M_{o}\left\frac{1}{2}\right^\frac{t}{h}[/tex]

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- Thread starter thomasrules
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- #1

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I know that: [tex]M=M_{o}\left\frac{1}{2}\right^\frac{t}{h}[/tex]

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- #2

Diane_

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M = M0(1/2)^(t/T)

ln(M) = ln(M0(1/2)^(t/T))

ln(M) = (t/T)ln(M0/2)

T = t ln(M0/2)/ln(M)

Since you have t in hours, this will give you the half-life in hours.

There are other approaches using more standard exponential decay formulas (decay constants, for instance), but all of them end up with logs eventually.

- #3

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ok thank you but by the way whats wrong with my latex:

"[tex]M=M_{o}\frac{1}{2}^\frac{t}{h}[/tex]"

"[tex]M=M_{o}\frac{1}{2}^\frac{t}{h}[/tex]"

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- #4

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you miss the "\" in front of the frac

- #5

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Ok I did that and got an answer of 5.95 but thats not the answer

SOMEONE HELP

SOMEONE HELP

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- #6

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this step is wrong!ln(M) = ln(M0(1/2)^(t/T))

ln(M) = (t/T)ln(M0/2)

here is the right one...

[tex]M=M_{o}\left\frac{1}{2}\right^\frac{t}{h}[/tex]

[tex]\frac{M}{M_0}=\left\frac{1}{2}\right^\frac{t}{h}[/tex]

[tex]log_2(\frac{M}{M_0})=log_2(\frac{1}{2}^\frac{t}{h})[/tex]

[tex]log_2(\frac{M}{M_0})=\frac{t}{h}log_2(\frac{1}{2})[/tex]

[tex]log_2(\frac{M}{M_0})=\frac{t}{h}log_2(2^{-1})[/tex]

[tex]log_2(\frac{M}{M_0})=-\frac{t}{h}log_2(2)[/tex]

[tex]log_2(\frac{M}{M_0})=-\frac{t}{h}[/tex]

[tex]h=-\frac{t}{log_2(\frac{M}{M_0})}[/tex]

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