Question

•A radioactive material A (decay constant λA) decays into a material B (decay constant λB) and then into material C (decay constant λC) which is also radioactive. Determine the amount of material C remaining after a time t?

Answer 1

Answer:

 The amount of C remaining after time t is

       $N_C__{R}} =N_D = (N_0 -N_0 e^{\lambda_A t}) - (N_0 -N_0 e^{-\lambda_A t})e^{-\lambda_B } [e^{-\lambda_C } ]$

Explanation:

We can represent the decay sequence as

      $A \to B \to C \to D$

The reason we added D is because we are told from the question that C is also radioactive so it has the  tendency to decay

Generally for every decay the remaining radioactive element can be obtained as

     $N = N_0 -N_0 e^{- \lambda t}$

Where N is the amount of the remaining radioactive material

            $N_0$ is the original amount amount of the radioactive material before decay

    and  $\lambda$ is the decay constant

Now for the decay from  $A \to B$  amount of radioactive element B formed from A after time t can be obtained as

          $N_b = N_0 -N_0 e^{- \lambda_A t}$

Where $\lambda _A$ is the decay constant of A

  Now for the decay from  $B \to C$  amount of radioactive element C formed from A after time t can be obtained as

       $N_c = (N_0 -N_0 e^{\lambda_A t}) - (N_0 -N_0 e^{\lambda_A t})e^{-\lambda_B t}$

Where $\lambda _B$ is the decay constant of B

  Now for the decay from  $C \to D$  amount of radioactive element D formed from A after time t can be obtained as

       $N_C__{R}} =N_D = (N_0 -N_0 e^{\lambda_A t}) - (N_0 -N_0 e^{-\lambda_A t})e^{-\lambda_B } [e^{-\lambda_C } ]$

So this amount of  D is the reaming amount of the radioactive material C