We’re being asked to **determine the half-life of cobalt–60**.

**We shall follow these steps in solving this problem.**

*Step 1: *Plot the time and natural logarithm (ln) of disintegration per minute as x and y respectively.

*Step 2:* Obtain and get the negative value of the slope (i.e. the answer should be positive). This is the decay constant.

*Step 3: *Solve for the half-life.

Recall that ** half-life** is the time needed for the amount of a reactant to decrease by 50% or one-half.

The half-life of a first-order reaction is given by:

$\overline{){{\mathbf{t}}}_{\mathbf{1}\mathbf{/}\mathbf{2}}{\mathbf{=}}\frac{\mathbf{ln}\mathbf{}\mathbf{2}}{\mathbf{k}}}$

where:

**t**_{1/2} = half-life

**k** = decay constant

Recall that ** radioactive/nuclear decay of isotopes** follows first-order kinetics, and the integrated rate law for first-order reactions is:

$\overline{){\mathbf{ln}}{{\mathbf{\left[}}{\mathbf{N}}{\mathbf{\right]}}}_{{\mathbf{t}}}{\mathbf{=}}{\mathbf{-}}{\mathbf{kt}}{\mathbf{+}}{\mathbf{ln}}{{\mathbf{\left[}}{\mathbf{N}}{\mathbf{\right]}}}_{{\mathbf{o}}}}$

**We are given the table below:**

Disintegrations per Minute | Time (h) |

180 | 0 |

130 | 2.5 |

104 | 5.0 |

77 | 7.5 |

59 | 10.0 |

46 | 12.5 |

24 | 17.5 |

The synthetic radioisotope technetium-99, which decays by beta emission, is the most widely used isotope in nuclear medicine. The following data were collected on a sample of ^{99}Tc:

Disintegrations per Minute | Time (h) |

180 | 0 |

130 | 2.5 |

104 | 5.0 |

77 | 7.5 |

59 | 10.0 |

46 | 12.5 |

24 | 17.5 |

Using these data, make an appropriate graph and curve fit to determine the half-life.

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