The rate of decay of a radioactive substance is proportional...

The rate of decay of a radioactive substance is proportional to the amount A remaining at any instant. If $A=A_{0}$ at $t=0$, prove that, if the time taken for the amount of the substance to become $\frac{1}{2} A_{0}$ is $T$, then $A=A_{0} e^{-(t \ln 2) / T}$. Prove also that the time taken for the amount remaining to be reduced to $\frac{1}{20} A_{0}$ is $4 \cdot 32 T$.

Key Concepts

Exponential Decay Model

The radioactive decay process is modeled by the differential equation dA/dt = -kA, where k is a positive constant representing the decay rate. This model captures the idea that the decay rate at any moment is directly proportional to the amount of substance remaining, leading naturally to an exponential decrease over time.

Separation of Variables

To solve the differential equation for exponential decay, one employs the technique of separation of variables. This involves rewriting the equation in the form dA/A = -k dt and then integrating both sides. The process yields a logarithmic expression that can be exponentiated to obtain the general solution A = A0 e^(-kt).

Half-Life and Decay Constant

The half-life, denoted by T, is defined as the time it takes for the substance to decay to half its initial amount. By substituting A = A0/2 at t = T into the solution A = A0 e^(-kt), one can solve for the decay constant k, arriving at the relation k = ln2/T. This connection is fundamental for expressing the decay law in terms of the half-life.

Logarithmic and Exponential Relationships

Once the decay constant is determined, logarithmic properties are used to solve for the time required for the substance to diminish to other specified fractions of the initial amount. For instance, to find the time when A = A0/20, one sets up the equation A0 e^(-(t ln2)/T) = A0/20. Taking logarithms allows for isolating t, demonstrating the utility of logarithmic manipulation in solving exponential decay problems.

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