anonymous
  • anonymous
If you have 10.0 g of a substance that decays with a half-life of 14 days, then how much will you have after 42 days? A. 0.10 g B. 0.31 g C. 1.25 g D. 2.50 g
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Michele_Laino
  • Michele_Laino
here we have to apply this formula: \[\Large m\left( t \right) = {m_0}{e^{ - t/\tau }}\] where m_0 is the initial mass of our sample, m(t) is the mass of our sample at time t, and \tau is the half life
anonymous
  • anonymous
ok! what do we plug in?
Michele_Laino
  • Michele_Laino
here is the next step: \[\Large m\left( {42} \right) = 10 \times {e^{ - 42/14}} = \frac{{10}}{{{e^3}}} = ...grams\]

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anonymous
  • anonymous
ok! what is e?
anonymous
  • anonymous
oh wait sorry haha i totally blanked out :P
anonymous
  • anonymous
we get 0.497870684?
Michele_Laino
  • Michele_Laino
e is the Neperus constant, namely: \[e = 2.71828\]
anonymous
  • anonymous
yes:)
Michele_Laino
  • Michele_Laino
yes! I got 0.49
Michele_Laino
  • Michele_Laino
my formula is the correct one!
anonymous
  • anonymous
yay!! what would the solution be though? 0.49 is not a choice :/
Michele_Laino
  • Michele_Laino
please wait, I'm checking my computation
anonymous
  • anonymous
ok!
Michele_Laino
  • Michele_Laino
I think that \[\tau \] is the average life, and it is given by the subsequent formula: \[\tau = \frac{{14}}{{0.693}} = 20.2\;days\]
anonymous
  • anonymous
ohh okay! i am confused, though... which choice would be the accurate solution? :/
Michele_Laino
  • Michele_Laino
so the right answer is given by the subsequent computation: \[\Large m\left( {42} \right) = 10 \times {e^{ - 42/20.2}} = \frac{{10}}{{{e^{2.079}}}} = ...grams\]
anonymous
  • anonymous
oh! ok! so we get 1.25! so choice C is the solution?
Michele_Laino
  • Michele_Laino
yes! that's right!
anonymous
  • anonymous
yay! thank you!!:) okay! onto the next:)
Michele_Laino
  • Michele_Laino
:) ok!

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