. . . Radioactive decay The nucleus of an atom consists of combinations of protons and neutrons.
Many of these combinations of protons and neutrons are unstable, that is the atoms decay or transmute into the atoms of another substance. Such nuclei are said to be radioactive . It is assumed that the rate d dt A at which the nuclei of a substance decays is proportional to the amount A( ) t of the substance remaining at time t . Thus, the required differential equation is d dt A A ∝ or d dt k A A …( ), where k is the constant of proportionality.
Here k < , since decay occurs. Remarks From equations ( ) and ( ), we see that the differential equations are the same, but the difference is only in the interpretations of the symbols and the constants of proportionality. For growth as we expect in ( ), k > and in the case of ( ) for decay, k < . A single differential equation can serve as a mathematical model for many different phenomena.
Example . A radioactive isotope has an initial mass 200mg , After two years it is decreased by 50mg . Find the expression for the amount of the isotope remaining at any time. What is its half-life?
(half-life means the time taken for the radioactivity of a specified isotope to fall to half its original value). Let A be the mass of the isotope remaining after t years, and let − k be the constant of proportionality, where k > . Then the rate of decomposition is modeled by d dt k A A = − , where the minus sign indicates that the mass is decreasing. It is a separable equation.
Separating the variables, we get d kdt A A = − Integrating on both sides, we get log A = − kt C or A Ce kt − . Given that the initial mass is 200mg. That is, A = when t = and thus, C = . Thus, we get A = e kt .
Also, A = when t = and therefore, k = Hence, A( ) is the mass of isotope remaining after t years. - - The half-life t h is the time corresponding to A = 100mg . Thus, t h =