\(N=N_{0}\left(\frac{1}{2}\right)^{n}\)
\(\frac{N}{N_{0}}=\left(\frac{1}{2}\right)^{n}\)
\(\frac{1}{100}=\left(\frac{1}{2}\right)^{n}\)
\(2^{n}=100\)
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.