A sample of phosphorus-32 is found to have an activity of 885 disintegrations per hour (dph).After 25 hours the activity has decreased to 841 dph.What is the rate constant for the decay of phosphorus-32?

Question

Quizplus · Accepted Answer

The decay of phosphorus-32 can be described by the first-order kinetics equation: $A = A_0 e^{-kt}$, where $A$ is the activity at time $t$, $A_0$ is the initial activity, $k$ is the rate constant, and $t$ is the time. Rearranging for $k$ and substituting the given values ($A_0 = 885$ dph, $A = 841$ dph, $t = 25$ hours), we can solve for $k$. The correct answer is found by solving for $k$ using the formula $k = \frac{\ln(A_0/A)}{t}$, which yields a value close to 0.0020 hr^-1.

A Sample of Phosphorus-32 Is Found to Have an Activity