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I'LL MARK AS BRAINLIEST ...if t is the time required for a radioactive substance to reduce to one third of its initial amount then what amount will be left over after 0.5t ... Plz answer ... NO SPAMMING...

Answer»

if t is the time required for a radioactive SUBSTANCE to REDUCE to ONE THIRD of its initial amount,

N_t=N_oe^{-t/ \tau}

\Rightarrow \frac{N_t}{N_o}= e^{-t/ \tau} \\ \\ \Rightarrow \frac{1}{3} = e^{-t/ \tau}


If t' = 0.5t

N_{t'}=N_oe^{-t'/ \tau}

\Rightarrow \frac{N_{t'}}{N_o}= e^{-t'/ \tau} \\ \\ \Rightarrow \frac{N_{t'}}{N_o}= e^{-0.5t/ \tau}

\\ \\ \Rightarrow \frac{N_{t'}}{N_o}= e^{0.5} \times e^{-t/ \tau}

\\ \\ \Rightarrow \frac{N_{t'}}{N_o}= 1.6487 \times \frac{1}{3}

\\ \\ \Rightarrow \frac{N_{t'}}{N_o}= 0.5496 \approx 0.55 =\frac{11}{20}


Thus, after 0.5t, 11/20 of the ORIGINAL amount will be left.


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