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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