Answer to Question #240105 in General Chemistry for cess

Question

a 13 g sample of an isotope of strontium-90 was formed in 1960 explosion of an atomic bomb at Johnson Island in the pacific test site. in wha year will only 0.625 grams of the strontium-90 remain?

Answer 1

Radiactive decay law:

$N=N_0e^{-λt}$

N₀ = initial amount of radiactive substances\

N=amount of substance at time t

λ =decay constantly

Half life

$T_{1/2}= \frac{0.693}{λ} \\ N_0=13 \;g \\ N= 0.625 \;g \\ T_{1/2}=28.8 \;years$

Decay constant

$λ = \frac{0.693}{28.8}= 0.024 \;year^{-1} \\ N=N_0e^{-λt} \\ 0.625 = 13 \times e^{-0.024t} \\ 0.04807 = e^{-0.024t} \\ ln(0.04807) = -0.024t \\ -3.035097 = -0.024t \\ t = \frac{-3.035097}{-0.024} = 126.5 \;years$

1960+126.5 = 2086.5 - year when only 0.625 grams of the strontium-90 will remain