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

Explanation:

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

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

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:

$k=\frac{2.303}{233}\log\frac{0.500}{0.0125}\\\\k=0.0158min^{-1}$

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

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

where,

$t_{1/2}$ = half-life of the reaction = ?

k = rate constant = $0.0158min^{-1}$

Putting values in above equation, we get:

$t_{1/2}=\frac{0.693}{0.0158min^{-1}}=43.86min$

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