Ask Question
12 August, 12:40

Fluorine-21 has a half life of approximately 5 seconds. what fraction of the original nuclei would remain after 1 minute?

+3
Answers (1)
  1. 12 August, 16:27
    0
    We can use two equations for this problem.

    t1/2 = ln 2 / λ = 0.693 / λ

    Where t1/2 is the half-life of the element and λ is decay constant.

    5 s = 0.693 / λ

    λ = 0.693 / 5 s (1)

    Nt = Nο eΛ (-λt)

    Nt/No = eΛ (-λt) (2)

    Where Nt is atoms at t time, No is the initial amount of substance, λ is decay constant and t is the time taken.

    t = 1 minute = 60 s

    From (1) and (2),

    Nt/No = eΛ ( - (0.693 / 5 s) 60 s)

    Nt/No = 2.445 x 10⁻⁴

    Hence, the fraction of the original nuclei would remain after 1 minute is 2.445 x 10⁻⁴.
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “Fluorine-21 has a half life of approximately 5 seconds. what fraction of the original nuclei would remain after 1 minute? ...” in 📘 Chemistry if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers