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Remaining quantity (in\%) of radioactive element after 5 half lives is
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$3.125 \%$
Remaining quantity
$N=N_0\left(\frac{1}{2}\right)^n=N_0\left(\frac{1}{2}\right)^5=\frac{N_0}{32}$
In percentage, $N=\frac{N_0}{30 \times N_0} \times 100=3.125 \%$
$N=N_0\left(\frac{1}{2}\right)^n=N_0\left(\frac{1}{2}\right)^5=\frac{N_0}{32}$
In percentage, $N=\frac{N_0}{30 \times N_0} \times 100=3.125 \%$
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