For the two-step decay stable

PHYSICS · Subjective · A Level · Physics · Nuclear physics · Nuclear physics

Problem

For the two-step decay

A\rightarrow B\rightarrow

stable, the number of nuclei of

B

at time

t

is known to be

N_B(t)=\frac{\lambda_A N_{A0}}{\lambda_B-\lambda_A}\bigl(e^{-\lambda_A t}-e^{-\lambda_B t}\bigr).

The half-life of

A

6\,\text{h}

and that of

B

1.5\,\text{h}

. Initially

N_{A0}=5.0\times10^{19}

. (a) Calculate the number of

B

nuclei present after

5.0\,\text{h}

. (b) Determine the time (in hours, correct to one decimal place) when the activities of

A

and

B

become equal and state the value of that common activity. (c) On the basis of your answer in (b), identify whether the system approaches secular equilibrium or transient equilibrium and justify your choice briefly.

Related theory

Radioactivity
Nuclear physics
Atomic Structure
Nuclei
Radioactive decay

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