When a nuclide is unstable (radioactive) is can **decay** by various nuclear processes to a lower (stable) energy state.
Each nuclide randomly decays following a poisson distribution at a decay rate \(\lambda\). Since there is a very large number of individual nuclides, we can calculate the average
number remaining. Let's assume a radioactive substance decays into a stable daughter element.

\(A^* \rightarrow B \)

The number of radioactive nuclides is \(N\) and decreases with time as they decay into the daughter nuclude. Thus, \(N=N(t)\). The rate of change is \(\frac{dN}{dt} = -\lambda \cdot N \) since the quantity decreases at a constant rate and is proportional to the amount of radioactive nuclides available to decay. Solving the differential equation yields:

\(\frac{dN}{dt} = -\lambda \cdot N \)

\(\int\frac{dN}{N} = \int-\lambda \cdot dt \)

\( ln(N) = -\lambda \cdot t + C\)

\( N(t) = e^{-\lambda \cdot t + C}\)

\( N(t) = e^{-\lambda \cdot t} \cdot e^{C}\)

\( N(t) = e^{-\lambda \cdot t} \cdot C^{'}\)

The initial boundary condition is \(N(t=0)=N_0\). At time = 0 we have the initial amount of radioactive nucludes \(N_0\).

\( N(t) = e^{-\lambda \cdot t} \cdot C^{'}\)

\( N(0)=N_0 = e^{-\lambda \cdot (0)} \cdot C^{'}\)

\( N_0 = C^{'}\)

This gives the final equation describing the exponential decay of a radioactive element.

\( N(t) = N_0 \cdot e^{-\lambda \cdot t} \)

There are still \(N_0\) total nuclide after some time passes but some are now daughter atoms. The number of daughter atoms increases with time:

\( N_D(t) = N_0 - N(t) \)

\( N_D(t) = N_0 - N_0 \cdot e^{-\lambda \cdot t} \)

The relationship between parent and daughter nuclides is shown below.

A radioactive sample (10 g) decays for 5 hr. Your lab partner calculates that 3 g of radioactive material remains. What is the mass of the stable nuclide?

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A common quantity in radioactive decay is the **half-life**. The half-life is the amount of time
it takes for half the radioactive substance to decay. After 2 half-lives, 25% of the substance remains and so on.
After about 5 half-lives, the radioactive material has mostly decayed.

# Half-Lives | Radioactive Material (%) | Stable Material (%) |

1 | 50 | 50 |

2 | 25 | 75 |

3 | 12.5 | 87.5 |

4 | 6.25 | 93.75 |

5 | 3.125 | 96.875 |

We can express the formula above in terms of the half-life, \(t_{1/2}\), accordingly:

\( N(t) = N_0 \cdot e^{-\lambda \cdot t} \)

\( N(t_{1/2})=\frac{N_0}{2} = N_0 \cdot e^{-\lambda \cdot t_{1/2}} \)

\( \frac{N_0}{2} = N_0 \cdot e^{-\lambda \cdot t_{1/2}} \)

\( \frac{1}{2} = e^{-\lambda \cdot t_{1/2}} \)

\( -ln(\frac{1}{2}) = -\lambda \cdot t_{1/2} \)

\( \frac{-ln(\frac{1}{2})}{\lambda} = t_{1/2} \)

\( t_{1/2} = \frac{ln(2)}{\lambda}\)

\( \lambda = \frac{ln(2)}{t_{1/2}}\)

\( N(t) = N_0 \cdot e^{-ln(2) \frac{t}{t_{1/2}}} \)

A radioactive element has a half-life of A mins. If the daughter element is stable, what percentage exists after 3 half-lives.

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