Answer:
The amount of C remaining after time t is
[tex]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 } ][/tex]
Explanation:
We can represent the decay sequence as
[tex]A \to B \to C \to D[/tex]
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
[tex]N = N_0 -N_0 e^{- \lambda t}[/tex]
Where N is the amount of the remaining radioactive material
[tex]N_0[/tex] is the original amount amount of the radioactive material before decay
and [tex]\lambda[/tex] is the decay constant
Now for the decay from [tex]A \to B[/tex] amount of radioactive element B formed from A after time t can be obtained as
[tex]N_b = N_0 -N_0 e^{- \lambda_A t}[/tex]
Where [tex]\lambda _A[/tex] is the decay constant of A
Now for the decay from [tex]B \to C[/tex] amount of radioactive element C formed from A after time t can be obtained as
[tex]N_c = (N_0 -N_0 e^{\lambda_A t}) - (N_0 -N_0 e^{\lambda_A t})e^{-\lambda_B t}[/tex]
Where [tex]\lambda _B[/tex] is the decay constant of B
Now for the decay from [tex]C \to D[/tex] amount of radioactive element D formed from A after time t can be obtained as
[tex]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 } ][/tex]
So this amount of D is the reaming amount of the radioactive material C
