Question

Nuclear decay is a first-order kinetics process. What is the half-life of a radioactive isotope if it takes 233 minutes for the concentration of the isotope to drop from 0.500 M to 0.0125 M? Give your answer in minutes.

Answer 1

Answer: The half life of the given radioactive isotope is 43.86 minutes

Step-by-step explanation:

Rate law expression for first order kinetics is given by the equation:

`k=(2.303)/(t)\log([A_o])/([A])`

where,

k = rate constant = ?

t = time taken for decay process = 233 minutes

`[A_o]` = initial amount of the reactant = 0.500 M

[A] = amount left after decay process = 0.0125 M

Putting values in above equation, we get:

`k=(2.303)/(233)\log(0.500)/(0.0125)\\\\k=0.0158min^(-1)`

The equation used to calculate half life for first order kinetics:

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

where,

`t_(1/2)` = half-life of the reaction = ?

k = rate constant =
`0.0158min^(-1)`

Putting values in above equation, we get:

`t_(1/2)=(0.693)/(0.0158min^(-1))=43.86min`

Hence, the half life of the given radioactive isotope is 43.86 minutes