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 = N02−t∕T1∕2 or 0.1 y = 0.9 x 2−t∕14.3 ... (i)
For isotope P-33
N = N02−t∕T1∕2 or 0.9 y = 0.1 x 2−t∕25.3 ... (ii)
On dividing, we get

1

9

= 9

2−t∕14.3

2−t∕25.3

or

1

81

= 2(−

t

14.31

+

t

25.3

)

1

81

= 2−t[

11

14.3×25.3

] or 81 = 2t[

11

14.3×25.3

]
Taking log
loge 81 = t(

11

14.3×25.3

)loge 2 or 1.9082 =

11t

25.3×14.3

× 0.3010
t = 208.5 days = 209 days