Question

The half-life of Sr 90 is 28 years

A. how many grams of Sr 90 in a 8.00 g sample of this nuclide will
have decayed after 84 years?

B. how many hours are required to reduce the amount of Sr 90 present in 80.0g sample to 10.0mg?

Answer #1

all radio active decays are first order reactions.

for first order

K = 0.693 / t_{1/2} = 0.693 / 28

K = 0.02475 y^{-1}

for first order reactions

K = (2.303 / t) log [A_{0}] / [A]

A) 0.02475 = (2.303 / 84) log [8.0] / [A]

0.9027 = 0.903 - log [A]

log [A] = 0.0003

[A] = 1.00 g

after 84 years 8.00 - 1.00 = 7.00 g Sr decayed

**7.00 g Sr decayed.**

B) [A_{0}] = 80.0 g

[A] = 10.0 mg = 0.01 g

0.02475 = (2.303 / t) log [80] / [0.01]

0.02475 = (2.303 / t) 3.90

2.303 / t = 0.006346

t = 362.9 years = 3179004 hours

**answer = 3179004 hours**

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