# Problem: 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 99Tc: Disintegrations per Minute Time (h) 18001302.51045.0777.55910.04612.52417.5Using these data, make an appropriate graph and curve fit to determine the half-life.

###### FREE Expert Solution

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:

where:

t1/2 = 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
###### Problem Details
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 99Tc:

 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.