Question

Americium-241 is used in smoke detectors. It has a first-order rate constant for radioactive decay of k = 1.6x10^−3yr−1. By contrast, iodine-125, which is used to test for thyroid functioning, has a rate constant for radioactive decay of k = 0.011 day−1.

(a) What are the half-lives of these two isotopes?
(b) Which one decays at a faster rate?
(c) How much of a 1.00-mg sample of each isotope remains after three half-lives?
(d) How much of a 1.00-mg sample of each isotope remains after 4 days?

Answer 1

A. 433 years and 63 days

B. Iodine

C. 0.125mg

D. 1.00mg

Step-by-step explanation:

A. Americium

The formula of a radioactive decay constant half life is t = 0.693/k

Where k is the decay constant.

For americium, k = 0.0016

t = 0.693/0.0016 = 433.125 apprx 433 years

For iodine, k = 0.011

Half life = 0.693/0.011 = 63 days.

B. Iodine decays at a faster rate.

C. After three half lives

For both, first half life yields a mass of 0.5mg, next yields 0.25mg, next yield 0.125mg

D. 1.00mg still remains.

Half life is high for both so the decay after one day is insignificant.