A sample of 35S complete its 10% decay in 20 min, then calculate the t

1.	A sample of 35S complete its 10% decay in 20 min, then calculate the time required to complete decay by 19%.
Answer» 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

A sample of 35S complete its 10% decay in 20 min, then calculate the time required to complete decay by 19%.

Answer»

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