The half-life of ³⁰P is 2.50 minutes.If a sample of ³⁰P has 1.60 $\times$ 10¹⁰ nuclei present,what is its activity?

Question

Quizplus · Accepted Answer

The activity of a sample is given by: Activity = λN Where λ is the decay constant and N is the number of nuclei present in the sample. The decay constant is related to the half-life by: λ = ln(2)/t1/2 Substituting the given values, we get: λ = ln(2)/2.50 min = 0.2778 min-1 Activity = λN = (0.2778 min-1)(1.60 F1F1F1S1F1F1F10 10¹⁰) = 7.4 F1F1F1S1F1F1F10 10⁷ Bq Therefore, the best choice is B) 7.4 F1F1F1S1F1F1F10 10⁷ Bq.

The Half-Life of 30P Is 2 ×\times× 1010 Nuclei Present,what Is Its Activity? A) 4

The Half-Life of ³⁰P Is 2 $\times$ 10¹⁰ Nuclei Present,what Is Its Activity?
A) 4