## anonymous one year ago 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

1. 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

2. anonymous

ok! what do we plug in?

3. Michele_Laino

here is the next step: $\Large m\left( {42} \right) = 10 \times {e^{ - 42/14}} = \frac{{10}}{{{e^3}}} = ...grams$

4. anonymous

ok! what is e?

5. anonymous

oh wait sorry haha i totally blanked out :P

6. anonymous

we get 0.497870684?

7. Michele_Laino

e is the Neperus constant, namely: $e = 2.71828$

8. anonymous

yes:)

9. Michele_Laino

yes! I got 0.49

10. Michele_Laino

my formula is the correct one!

11. anonymous

yay!! what would the solution be though? 0.49 is not a choice :/

12. Michele_Laino

please wait, I'm checking my computation

13. anonymous

ok!

14. Michele_Laino

I think that $\tau$ is the average life, and it is given by the subsequent formula: $\tau = \frac{{14}}{{0.693}} = 20.2\;days$

15. anonymous

ohh okay! i am confused, though... which choice would be the accurate solution? :/

16. Michele_Laino

so the right answer is given by the subsequent computation: $\Large m\left( {42} \right) = 10 \times {e^{ - 42/20.2}} = \frac{{10}}{{{e^{2.079}}}} = ...grams$

17. anonymous

oh! ok! so we get 1.25! so choice C is the solution?

18. Michele_Laino

yes! that's right!

19. anonymous

yay! thank you!!:) okay! onto the next:)

20. Michele_Laino

:) ok!