Question

A patient is given 0.055 μg of technetium-99m, a radioactive isotope with a half-life of about 6.0 hours.

How long does it take for the radioactive isotope to decay to 2.7×10⁻³ μg ? (Assume no excretion of the nuclide from the body.)

Express your answer using two significant figures.

Answer 1

We know that radioactive decay follows first order kinetics

Half life for a first order reaction=t_1/2=0.693/k

Where k=rate constant

So k=0.693/t_1/2=0.693/6 h=0.116 h^{​​​​​​-1}​

We know that for a first order reaction

Where [A]=Concentration of reactant A at time t=2.7x10^-3 x10^-6g=2.7 x 10^-9 g

[A]_o=Initial concentration of A=0.055 x10^-6 g

k=rate constant

t=time

Substituting the values in the integrated rate law

(2.7 x 10^-9 g)/(0.055 x 10^-6 g)=

49.1 x 10^-3 g=

Taking natural log on both sides we get

ln (49.1 x 10^-3 g)=(-0.116 h^{​​​​​​-1} x t)

-3.0 =-0.116 h ^-1 x t

t=-3.0/-0.116 h^-1​=25.9 h26 h

So time taken=26 h