When decay is 10 % complete, if N0 = 100 , then N = 100 – 10 = 90 and t = 20 minutes When decay is 19 % complete, N = 100 – 19 = 81
Substituting these values in formula we get,
For 10% decay completion, \(\lambda\) = \(\frac{2.303}{20}\) log10 \((\frac{100}{90})\).....(i)
For 19% decay completion, \(\lambda\) = \(\frac{2.303}{t}\) log10 \((\frac{100}{81})\).....(ii)
From equation (i) and (ii).
\(\frac{2.303}{20}\) log10 \(\frac{100}{90}\) = .\(\frac{2.303}{t}\) log10 \((\frac{100}{81})\)
∴ t log10 \((\frac{100}{90})\) = 20 log10 \((\frac{100}{81})\)
∴ t x 0.0458 = 20 x 0.0915
∴ t = \(\frac{20\times 0.0915}{0.0458}\) = 40 min