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

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 ) ]`

Step-by-step 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