Question #240105

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?


1
Expert's answer
2021-09-22T00:27:02-0400

Radiactive decay law:

N=N0eλtN=N_0e^{-λt}

N0 = initial amount of radiactive substances\

N=amount of substance at time t

λ =decay constantly

Half life

T1/2=0.693λN0=13  gN=0.625  gT1/2=28.8  yearsT_{1/2}= \frac{0.693}{λ} \\ N_0=13 \;g \\ N= 0.625 \;g \\ T_{1/2}=28.8 \;years

Decay constant

λ=0.69328.8=0.024  year1N=N0eλt0.625=13×e0.024t0.04807=e0.024tln(0.04807)=0.024t3.035097=0.024tt=3.0350970.024=126.5  yearsλ = \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


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