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2 May, 18:51

An isotope has a half-life of 10 minutes. after 20 minutes, what percentage of the original nuclei remain?

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  1. 2 May, 19:38
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    To solve this, we can use two equations.

    t1/2 = ln 2 / λ = 0.693 / λ

    where, t1/2 is half-life and λ is the decay constant.

    t1/2 = 10 min = 0.693 / λ

    Hence, λ = 0.693 / 10 min - (1)

    Nt = Nο e∧ (-λt)

    Nt = amount of atoms at t = t time

    Nο = initial amount of atoms

    t = time taken

    by rearranging the equation,

    Nt/Nο = e∧ (-λt) - (2)

    From (1) and (2),

    Nt/Nο = e∧ ( - (0.693 / 10 min) x 20 min)

    Nt/Nο = 0.2500

    Percentage of remaining nuclei = (nuclei at t time / initial nuclei) x 100%

    = (Nt/Nο) x 100%

    = 0.2500 x 100%

    = 25.00%

    Hence, Percentage of remaining nuclei is 25.00%
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