The radioactive nuclide ⁶³Ni decays by a first-order process via the emission of a beta particle.The ⁶³Ni nuclide has a half-life of 100.years.How long will it take for 71% of ⁶³Ni to decay?

Question

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The decay of ⁶³Ni follows a first-order process, which means that the rate of decay is proportional to the amount of ⁶³Ni present. The half-life of the nuclide is given as 100 years, which means that every 100 years, half of the initial amount of ⁶³Ni will decay. To determine how long it will take for 71% of ⁶³Ni to decay, we can use the following equation: ln(Nt/N0) = -kt where Nt/N0 is the fraction of ⁶³Ni remaining (in this case, 0.71), k is the rate constant for the decay process, and t is the time elapsed. We can rearrange this equation to solve for t: t = -ln(Nt/N0)/k We know that the half-life of ⁶³Ni is 100 years, so we can use this information to calculate the rate constant: ln(0.5) = -k(100) k = ln(0.5)/(-100) ≈ 0.00693/year Now we can plug in the values for Nt/N0 and k: t = -ln(0.71)/(0.00693) t ≈ 177.7 years Therefore, it will take approximately 178 years for 71% of ⁶³Ni to decay. The closest answer choice is D) 178 years.

The Radioactive Nuclide 63Ni Decays by a First-Order Process Via

The Radioactive Nuclide ⁶³Ni Decays by a First-Order Process Via