A certain substance has a half-life of 5.0 hours. How many nuclei of the substance are required to give an initial activity of 6.0 μCi?
(1 Ci = 3.7 × $10^{10}$ Bq)

Question

A certain substance has a half-life of 5.0 hours. How many nuclei of the substance are required to give an initial activity of 6.0 μCi? 
(1 Ci = 3.7 × $10^{10}$ Bq)

Quizplus · Accepted Answer

The correct answer is A.Explanation:The activity of a radioactive substance is given by the equation:$$A = -\frac{dN}{dt} = \lambda N$$where:* A is the activity in Bq* N is the number of radioactive nuclei* \lambda is the decay constant in s^-1The half-life of a radioactive substance is the time it takes for half of the nuclei to decay. The decay constant is related to the half-life by the equation:$$\lambda = \frac{ln 2}{t_{1/2}}$$where:* t_{1/2} is the half-life in sIn this problem, we are given that the half-life of the substance is 5.0 hours, which is 18000 s. We are also given that the initial activity of the substance is 6.0 μCi, which is 2.22 × 10^4 Bq. We can use the equation for activity to solve for the number of nuclei:$$N = \frac{A}{\lambda} = \frac{2.22 	imes 10^4 Bq}{\frac{ln 2}{18000 s}} = 5.8 	imes 10^9 nuclei$$Therefore, the correct answer is A.

A Certain Substance Has a Half-Life of 5 101010^{10}1010 Bq) A) 5

A Certain Substance Has a Half-Life of 5 $10^{10}$ Bq)
A) 5