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Suppose the daughter nucleus in a nuclear decay is itself radioactive - If \mathrm{\lambda_d} and \mathrm{\lambda m} denote the decay constant of daughter and mother nuclei and \mathrm{N_d} and \mathrm{N_m} the number of daughter and mother nuclei present at a time, then the number of daughter nuclei becomes constant then 

Option: 1

\mathrm{\lambda_m N_m=\lambda_d N_d}


Option: 2

\mathrm{\lambda_m N_d=\lambda_d N_m}


Option: 3

\mathrm{N_m-N_d=\lambda_m-\lambda_d}


Option: 4

\mathrm{N_m+N_d=\lambda_m+\lambda_d}


Answers (1)

best_answer

The number of mother nuclei decaying in a short time interval at is \mathrm{= Nm \lambda_m\, dt}.But death of a mother nucleus implies the birth of a daughter nucleus.
 
The number of daughter nuclei decaying in the same time interval is \mathrm{N_d \lambda_d \, d t}

the number of daughter nuclei will remain constant when \mathrm{\lambda_m N_m d t}
\mathrm{=\lambda d N_d d t}

\mathrm{\Rightarrow N_m \lambda_m=N_d \lambda_{d}}

Posted by

Anam Khan

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