A source contains two phosphorous radio nuclides ^{32} _{15} P (T_{\frac{1}{2}} = 14.3d) and ^{33}_{15} P (T_{\frac{1}{2}} = 25.3d). Initially, 10% of the decays come from ^{33}_{15} P . How long one must wait until 90% do so?

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Radiation Safety Manual
Almost all radionuclides that decay by alpha emission have atomic number ... For example, the beta dose rate at 3 cm from a 1 mCi vial of P-32 is: 2.7 x 105 x ...
For more information, see Radiation Safety Manual
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Law of Radioactive Decay, Decay Rate, Half-Mean Life, Q&A
Next, let's find the relation between the mean life τ and the disintegration constant λ. For this, let's consider equation (5),. The number of nuclei which decay in ...
For more information, see Law of Radioactive Decay, Decay Rate, Half-Mean Life, Q&A
Rate of disintegration
Amount of radioactive material left after n half life periods N = N_0 (\frac{1}{2})^n = N_0 (\frac{1}{2})^{\frac{t}{T_{\frac{1}{2}}}} = N_0 2^{\frac{-t}{T_{\frac{1}{2}}}} Where, T_{\frac{1}{2}} - half life and t - time
Given that
Half life of ^{32}_{15} P, T_{\frac{1}{2}} = 14.3d
Half life of ^{33}_{15} P, T_{\frac{1}{2}} = 25.3d
A source contains two radio nuclides initial decay process
10% of the decay come from ^{32}_{15} P, N_0 = x
90% of the decay come from ^{33}_{15} P, N_0' = 9x
Finally decay process
90% of the decay come from ^{32}_{15} P N = 9y
10% of the decay come from ^{33}_{15} P N' = y
Step 1: Set up a time equation for the decay of radioactive material [( ^{32}_{15} P ) and (_{15}^{33}P)]
Amount of radioactive material left N = N_0 2^{\frac{-t}{T_{\frac{1}{2}}}}
9y = x 2^{\frac{-t}{14.3d}} ....................................(1)
Amount of radioactive material left N' = N_0' 2^{\frac{-t}{T_{\frac{1}{2}}}}
y = (9x)*2^{\frac{-t}{25.3d}} ............................................(2)
Step 2: Calculating the time for the decay of radioactive material
Dividing equation (1) and (2)
\frac{9y}{y}=\frac{x\ \ \cdot2^{\frac{-t}{14.3d}}}{(9x)\cdot2^{\frac{-t}{25.3d}}}
9\cdot9=2^{\frac{-t}{14.3d}}\cdot2^{\frac{t}{25.3d}}
81 = 2^{\frac{11t}{361.79}}
Applying the log on both sides
\log 81 = \log 2^{\frac{11t}{361.79}}
[math] \log 81 = [\frac{11t}{361.79}] \log 2 [/math]
[math] 1.91 = [\frac{11t}{361.79}](0.3010) [/math]
t = \frac{1.91 * 1000}{9.15}
t = 208.7
Hence, decay time t = 208.7 days