Question

1. 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

1. 31.1 hr

Answer 1

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