The radioactive element americium-241 has a half-life of 432 years. Suppose we start with a 20-g mass of americium-241.
How much will be left after 349 years? Compute the answer to three significant digits.

Question

The radioactive element americium-241 has a half-life of 432 years. Suppose we start with a 20-g mass of americium-241.
How much will be left after 349 years? Compute the answer to three significant digits.

anonymous · Answer

1/2= e^-(c*432)

c= ?

anonymous · Answer

http://openstudy.com/users/henpen#/updates/504250b2e4b060a4c5a2f27c

unklerhaukus · Answer

$$[\text{Am}]_t=[\text{Am}]_02^{{-t}/{t_{½}}}$$$$[\text{Am}]_0=20[\text{g}]\qquad t_{1/2}=432[\text{yr}]$$$$[\text{Am}]_t=20[\text{g}]2^{{-t}/{432[\text{yr}]}}$$$$t=349[\text{yr}]$$$$[\text{Am}]_t=20[\text{g}]2^{{-349[\text{yr}]}/{432}[\text{yr}]}$$