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.

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.