In the mixture of P-32 and P-33 initially 10% decay came from P-33. Hence initially 90% of the mixture is P-32 and 10% of the mixture is P-33 .Let after time ‘t’ the mixture is left with 10% of P-32 and 90% of P-33. Half life of both P-32 and P-33 are given as 14.3 days and 25.3 days respectively. Let ‘x’ be total mass undecayed initially and ‘y’ be total mass undecayed finally. Let initial number of P-32 nuclides = 0.9 x Final number of P-32 nuclides = 0.1 y Similarly, initial number of P-33 nuclides = 0.1x Final number of P-33 nuclides = 0.9 y For isotope P-32 N = $N_0 2^{-\frac{t}{T_{\frac{1}{2}}}}$ or 0.1 y = 0.9 x $2^{-\frac{t}{14.3}}$ ... (i) For isotope P-33 N = $N_0 2^{-\frac{t}{T_{\frac{1}{2}}}}$ or 0.9 y = 0.1 x $2^{-\frac{t}{25.3}}$ ... (ii) On dividing, we get $\frac{1}{9}$ = 9 $\frac{2^{-\frac{t}{14.3}}}{2^{-\frac{t}{25.3}}}$ or $\frac{1}{81}$ = $2^{-\frac{t}{14.31}+\frac{t}{25.3}}$ $\frac{1}{81}$ = $2^{-t \cdot \frac{11}{14.3 \times 25.3}}$ or 81 = $2^{t \cdot \frac{11}{14.3 \times 25.3}}$ Taking log $\log_e$ 81 = $t \cdot \frac{11}{14.3 \times 25.3} \log_e$ 2 or 1.9082 = $\frac{11t}{25.3 \times 14.3}$ × 0.3010 t = 208.5 days = 209 days