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1 February, 01:50

A 1.00 * 10^-6 - g sample of nobelium, 254/102 No, has a half-life of 55 seconds after it is formed. What is the percentage of 254/102 No remaining at the following times? a) 5.0 min after it formsb) 1.0 h after it forms

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  1. 1 February, 02:29
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    2.2 % and 0 %

    Explanation:

    The equation we will be using to solve this question is:

    N/N₀ = e⁻λ t

    where N₀ : Number of paricles at t = 0

    N = Number of particles after time t

    λ = Radioactive decay constant

    e = Euler's constant

    We are not given λ, but it can be determined from the half life with the equation:

    λ = 0.693 / t 1/2 where t 1/2 is the half-life

    Substituting our values:

    λ = 0.693 / 55 s = 0.0126/s

    a) For t = 5 min = 300 s

    N / N₀ = e^ - (0.0126/s x 300 s) = e^-3.8 = 0.022 = 2.2 %

    b) For t = 1 hr = 3600 s

    N / N₀ = e^ - (0.0126/s x 3600 s) = 2.9 x 10 ⁻²⁰ = 0 % (For all practical purposes)
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