Question
$99 \%$ of a radioactive element will decay between
$N=N_{0}\left(\frac{1}{2}\right)^{n}$
$\frac{N}{N_{0}}=\left(\frac{1}{2}\right)^{n}$
$\frac{1}{100}=\left(\frac{1}{2}\right)^{n}$
$2^{n}=100$
The number of half lives is calculated as,
$2^{n}=100$
$n$ lies between the $6$ and $7 .$ So, the radioactive element will decay between $6$ and $7$ half lives.
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