Question

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 1

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.

Answer 2

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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.

HOPE SO IT WILL HELP.....