A 7.85 × 10^-⁵ mol sample of copper-61 emits 1.47 × 10¹⁹ positrons in 90.0 minutes. What is the decay constant for copper-61?

Question

Quizplus · Accepted Answer

The decay constant (λ) can be calculated using the formula λ = ln(2)/t1/2, where t1/2 is the half-life of the isotope. 
From the information given, we know that the sample of copper-61 emitted 1.47 × 1019 positrons in 90.0 minutes. 
The half-life of copper-61 is 3.333 hours, which can be calculated using the formula t1/2 = ln(2)/λ. 
Substituting the half-life into this formula, we get λ = ln(2)/t1/2 = ln(2)/3.333 hours = 0.207 h-1. 
Therefore, the decay constant for copper-61 is 0.207 h-1, which is choice C.

A 785 × 10-5 Mol Sample of Copper-61 Emits 1

A 785 × 10^-⁵ Mol Sample of Copper-61 Emits 1