What is the half life of -0.02m?

Question

solomonzelman · Answer

idk

wolf1728 · Answer

psam24
Don't you have a similar question posted on this site?

anonymous · Answer

yea is that wrong?

wolf1728 · Answer

No - I just figured it is easier to work in one place.

wolf1728 · Answer

For example in the other question you state the time is in days but nothing is stated here.

wolf1728 · Answer

When you say half-life is -.02m does that mean after one day, .02 of the sample has decayed?

anonymous · Answer

yes

anonymous · Answer

other instructions were more descriptive

wolf1728 · Answer

See attached graphic:
(I'm not exactly sure if I understand the problem correctly, but here's my explanation).
Based on the data that after 1 day, .02 of the original sample has decayed, then we can say original amount is 100, present amount = 98 and time is one day.
half-life = (time)*log(2)/log(orig amt / present amt)
half-life = 1 day * 0.3010299957 / log (100/98)
half-life = 1 day * 0.3010299957 / log (100/98)
half-life = 1 day * 0.3010299957 / log (1.0204081633)
half-life = 0.3010299957 / 0.0087739243
half-life = 34.31 days
(Base 10 logs were used)