The nuclide ⁹⁶Nb decays by a first-order process with a rate constant of 2.96 × 10^-2 h^-1.How long will it take for 86.0% of the initial amount of ⁹⁶Nb to be consumed?

Question

Quizplus · Accepted Answer

To solve this problem, we can use the first-order decay equation:

N(t) = N0e^(-kt)

Where N(t) is the amount of the nuclide remaining after time t, N0 is the initial amount, k is the rate constant, and e is the base of natural logarithms (approximately equal to 2.718).

We want to find the time it takes for 86.0% of the initial amount to be consumed, which means we need to solve for t when N(t) = 0.14N0:

0.14N0 = N0e^(-kt)

Taking the natural logarithm of both sides and solving for t gives:

t = ln(0.14)/(-k) = 66.4 hours

Therefore, the answer is B, 66.4 h.

The Nuclide 96Nb Decays by a First-Order Process with a Rate

The Nuclide ⁹⁶Nb Decays by a First-Order Process with a Rate