Determine half-life of this substance

  • #1
243
0
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]
 
Last edited:

Answers and Replies

  • #2
Diane_
Homework Helper
390
1
I'm insufficiently skilled with Latex (and more than sufficiently lazy not to want to figure it out right now) to post this properly, but there is a small problem with the equation you've posted. The power on the (1/2) should be (t/T), where T = the half-life in units of whatever t is. Given that, you can solve your equation for T, thus:

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
243
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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]"
 
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  • #4
609
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you miss the "\" in front of the frac
 
  • #5
243
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Ok I did that and got an answer of 5.95 but thats not the answer

SOMEONE HELP
 
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  • #6
609
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ln(M) = ln(M0(1/2)^(t/T))
ln(M) = (t/T)ln(M0/2)
this step is wrong!

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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