Question

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.

Answer 1

Answer: This fuel will require 239230 years.

Explanation:

All the radioactive reactions follow first order kinetics.

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

$t_{1/2}=\frac{0.693}{k}$

We are given:

$t_{1/2}=24000yrs$

Putting values in above equation, we get:

$k=\frac{0.693}{24000}=2.8875\times 10^{-5}yr^{-1}$

The equation used to calculate time period follows:

$N=N_o\times e^{-k\times t}$

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 = $2.8875\times 10^{-5}yr^{-1}$

Putting values in above equation, we get:

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

Hence, this fuel will require 239230 years.