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The spent fuel of a nuclear reactor contains plutonium-239, which has a half-life of 24,000 years. If 1 barrel containing 10 kg of plutonium-239 is sealed, how many years must pass until only 10 g of plutonium-239 is left? Use y = y0 *e^ -(kt). (Round your answer to the nearest integer.

User Wes Winder
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Answer: This fuel will require 239230 years.

Step-by-step explanation:

All the radioactive reactions follow first order kinetics.

The equation used to calculate rate constant from given half life for first order kinetics:


t_(1/2)=(0.693)/(k)

We are given:


t_(1/2)=24000yrs

Putting values in above equation, we get:


k=(0.693)/(24000)=2.8875* 10^(-5)yr^(-1)

The equation used to calculate time period follows:


N=N_o* e^(-k* t)

where,


N_o = initial mass of isotope = 10 kg = 10000 g (Conversion factor: 1 kg = 1000 g)

N = mass of the parent isotope left after the time = 10 g

t = time = ? years

k = rate constant =
2.8875* 10^(-5)yr^(-1)

Putting values in above equation, we get:


N=N_o* e^{-(2.8875* 10^(-5)yr^(-1))* t}\\\\t=239230\text{ years}

Hence, this fuel will require 239230 years.

User Jesse Van Assen
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