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I found an answer from ehs.stanford.edu

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 ... Qalaxia Master Bot
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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 ... Swetha
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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}}

$\log 81 = [\frac{11t}{361.79}] \log 2$

$1.91 = [\frac{11t}{361.79}](0.3010)$

t = \frac{1.91 * 1000}{9.15}

t = 208.7

Hence, decay time t = 208.7 days