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[SOLVED] another decay problem

karush

Well-known member
Jan 31, 2012
2,769
If 40 grams of radioactive substance decomposes to 20 grams in 2 years, then to the nearest gram the amount left after 3 years is

well i used the $N(t)=N_{0}e^{-kt}$

So $20=40e^{-k2}$ thus deriving k=.3466

Thus $N(3) = 40e^{-.3466(3)}$ resulting in: $N(3)= 14.1410$ or approx $14g$

just seeing if this correct....thnx much(Cool)
 

pickslides

Member
Feb 1, 2012
57
I agree with your method and solution.
 

CaptainBlack

Well-known member
Jan 26, 2012
890
If 40 grams of radioactive substance decomposes to 20 grams in 2 years, then to the nearest gram the amount left after 3 years is

well i used the $N(t)=N_{0}e^{-kt}$

So $20=40e^{-k2}$ thus deriving k=.3466

Thus $N(3) = 40e^{-.3466(3)}$ resulting in: $N(3)= 14.1410$ or approx $14g$

just seeing if this correct....thnx much(Cool)
As you are given a half-life of 2 years it makes more sense to express the decay by:

\[N(t)=N(0)\times 2^{-{t}/{t_{1/2}}}\]
where \(t_{1/2}\) is the half-life.

So for this problem:

\[N(3)=40 \times 2^{-3/2}\approx 14.1421\ {\rm{gram}}\]

CB