The value 0.693 in the half-life formula comes from the natural logarithm of 2 (ln 2). It is used in the derivation of the first-order half-life equation.

Why is 0.693 Used in Half-Life Calculations?

For a first-order reaction, the concentration of a drug or substance decreases exponentially over time. The rate equation is:

C=C0e−ktC = C_0 e^{-kt}

At half-life (T1/2T_{1/2}), the concentration is reduced to half:

C02=C0e−kT1/2\frac{C_0}{2} = C_0 e^{-k T_{1/2}}

Dividing both sides by C0C_0:

12=e−kT1/2\frac{1}{2} = e^{-k T_{1/2}}

Taking the natural logarithm (ln) on both sides:

ln⁡12=−kT1/2\ln \frac{1}{2} = -k T_{1/2}

Since ln⁡(1/2)=−ln⁡2=−0.693\ln(1/2) = -\ln 2 = -0.693, we get:

−0.693=−kT1/2-0.693 = -k T_{1/2} T1/2=0.693kT_{1/2} = \frac{0.693}{k}

Key Takeaways:

• 0.693 is just the natural logarithm of 2 (ln⁡2\ln 2).
• It comes from the exponential decay equation used for first-order reactions.
• It makes calculating half-life simpler and universal for first-order kinetics.

This formula is widely used in pharmacology, radioactive decay, and chemical kinetics to determine how long it takes for a substance to reduce to half its original amount.