The disintegration of radioactive phosphorus to silicon follows the first order kinetics with rate constant k\(_1\) = 3.85 x 10\(^{-3}\). Determine the half...

Answer Details

For a reaction that follows first-order kinetics, the half-life is related to the rate constant by the formula:

\[t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{k}\]

Given \(k_1 = 3.85 \times 10^{-3}\, \text{s}^{-1}\):

\[t_{1/2} = \frac{0.693}{3.85 \times 10^{-3}}\]

\[t_{1/2} = \frac{0.693}{0.00385}\]

\[t_{1/2} = 180\, \text{s}\]

The half-life of radioactive phosphorus is 180 s.

An important feature of first-order kinetics is that the half-life is independent of the initial concentration - it depends only on the rate constant. This is why radioactive decay (which always follows first-order kinetics) has a constant half-life regardless of how much of the substance remains.