Question

A source contains two species of phosphorous nuclei, \(^{32}_{15}P\)(T_1/2 = 14.3 d) and \(^{32}_{15}P\) (T_1/2 = 25.3 d). At time t = 0, 90% of the decays are from \(^{32}_{15}P\). How much time has to elapse for only 15% of the decays to be from \(^{32}_{15}P\)? 

Answer 1

Data :  \(^{32}_{15}P\) : T_1/2 = 14.3 d

∴ \(\lambda_1\) = \(\cfrac{0.693}{14.3d}\) = 0.04846 d^-1

\(^{33}_{15}P\) : T_1/2 = 25.3d

∴ \(\lambda_2\) = \(\cfrac{0.693}{25.3d}\) = 0.02739d^-1

∴ (0.04846 – 0.02739) t = 2.303 (2.1847 – 0.4771)

∴ t = \(\cfrac{(2.303)(1.7076)}{0.02107}\) = 186.6 days 

This is the required time.