Americium-240 has a half-life of 2.12 days. (Can you see why americium-241 is used in smoke detectors rather than americium-240? See exercise 4.) How many hours does it take for 34.5% of a sample of americium-240 to decay?
answer should be
31.1 hr
Given:
Half life = 2.12 days = 2.12*24 hr = 50.88 hr
use relation between rate constant and half life of 1st order reaction
k = (ln 2) / k
= 0.693/(half life)
= 0.693/(50.88)
= 1.362*10^-2 hr-1
we have:
[A]o = 100 (Let initial amount be 100)
[A] = 65.5 (34.5 % has decayed. So, remaining is 65.5 %)
k = 1.362*10^-2 hr-1
use integrated rate law for 1st order reaction
ln[A] = ln[A]o - k*t
ln(65.5) = ln(1*10^2) - 1.362*10^-2*t
4.182 = 4.605 - 1.362*10^-2*t
1.362*10^-2*t = 0.4231
t = 31.1 hr
Answer: 31.1 hr
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