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In a decay process \mathrm{A} decays to \mathrm{B}
     \mathrm{A \rightarrow B}
Two graphs of number of nuclei of \mathrm{A } and \mathrm{B } versus time is given then, which of the following option is incorrect.

 

Option: 1

\mathrm{t_{2}-t_{1}= 4}


Option: 2

\mathrm{t_{2}-t_{1}= 2}


Option: 3

\mathrm{t_1=2\, \log _2 5}


Option: 4

\mathrm{t_2=\log _2 100}


Answers (1)

\mathrm{N=\frac{N_0}{5}=N_0 e^{-\lambda t_1}}

\mathrm{\Rightarrow \frac{1}{5}=e^{-\lambda t_1} \Rightarrow \lambda t_1=\ln 5-(1)}

\mathrm{\text { For A }\: 3 t_{1 / 2}=6 \Rightarrow t_{1 / 2}=2\, \mathrm{sec} .}
\mathrm{\Rightarrow \lambda=\frac{\ln 2}{t_{1 / 2}}=\frac{\ln ^2}{2} . }

\mathrm{\Rightarrow t_1=\frac{2 \ln 5}{\ln ^2}=2 \log _2 5 \: -(2)}
\mathrm{Al so \, \quad N_0-\frac{9 N_0}{10}=N_0 e^{-\lambda t_2} }
\mathrm{\Rightarrow t_2=\frac{2 \ln 10}{\ln^{2}}=2 \log _2 10=\log _2 100 -(3) }

equation (2) - equation (3)

\mathrm{\Rightarrow t_2-t_1=\log _2 100-\log _2 25=\log _2 4=2 . }
\mathrm{ \Rightarrow t_2-t_1=2 . }

In correct option is (a)

Posted by

Kshitij

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