Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Consider the decay scheme-   with <img alt="\lambda_A<\lambda_B" src="/l

Consider the decay scheme-
A \rightarrow B \rightarrow C  with \lambda_A<\lambda_B. After transient equilibrium is established between A and B, The interval of time \Delta t, such that activity of A at (t-\Delta t) is equal to activity of B at ' t ' is
 

Option: 1

\Delta \mathrm{t}=\frac{1}{\lambda_B} \ln \left[\frac{\lambda_B}{\lambda_B-\lambda_A}\right]


Option: 2

\Delta \mathrm{t}=\frac{1}{\lambda_A} \ln \left[\frac{\lambda_B}{\lambda_B-\lambda_A}\right]


Option: 3

\Delta \mathrm{t}=\frac{1}{\lambda_A} \ln \left[\frac{\lambda_A}{\lambda_B-\lambda_A}\right]


Option: 4

\Delta \mathrm{t}=\frac{1}{\lambda_B} \ln \left[\frac{\lambda_A}{\lambda_B-\lambda_A}\right]


Answers (1)

best_answer

Let \mathrm{N}_1 be the amount of parent atom of \mathrm{A} at any instant (\mathrm{t}-\Delta \mathrm{t})
Therefore, N_1=N_1^0 \quad e^{-\lambda_A(t-\Delta t)}

where N_1^0 is the number of parent atoms of A at t=0.

The number of daughter atoms at any time t is given by

N_2=\frac{\lambda_A N_1^0}{\lambda_B-\lambda_A}\left[e^{-\lambda_A t}-e^{-\lambda_B t}\right]

For transient atom \lambda_A<\lambda_B, hence e^{-\lambda_0} t term can be neglected in (2)

\therefore N_2=\frac{\lambda_A N_1^0}{\lambda_B-\lambda_A} e^{-\lambda_A t}

Given that activity of A at (t-\Delta t)= activity of B at t.

\begin{aligned} & \Rightarrow \lambda_A N_1^0 e^{-\lambda_A(t-\Delta t)}=\lambda_B \frac{\lambda_A N_1^0}{\lambda_B-\lambda_A} e^{-\lambda_A t} \\ & \Rightarrow e^{\lambda_A \Delta t}=\frac{\lambda_B}{\lambda_B-\lambda_A} \\ & \Rightarrow \Delta t=\frac{1}{\lambda_A} \ln \left[\frac{\lambda_B}{\lambda_B-\lambda_A}\right] \end{aligned}

Posted by

shivangi.bhatnagar

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Latest Question