. If I have 5.7 g of the radioactive isotope Rn-222 (half-life of 3.8 days),?

Determine how much radon will I have after 11.4 days have passed? (Show or explain how you found your answer). THANK YOU

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  • Dr W
    Lv 7
    6 months ago
    Favorite Answer

    memorize and use this equation

    .. A(t) = A(o) * (1/2)^(t / half life)

    where

    .. A(t) = amount remaining after time = "t" has elapsed

    .. A(o) = initial amount

    .. t = elapsed time

    .. half life = ... half life

    this problem

    .. A(t) = 5.7g * (1/2)^(11.4days / 3.8days) = 0.71g

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  • 6 months ago

    Half-life problem ....

    Some of these problems require some math, and some are ridiculously easy. Your question is of the latter variety.

    11.4 days is three half-lives. At the end of each half-life interval half of the sample has decayed.

    0.............1 ..............2 ................3 .............. half-lives

    5.7g.....2.85g........1.425g.......0.7125g ....... mass remaining

    Round to two significant digits: 0.71g of Rn-222 remains

    =============

    Radioactive decay follows first-order kinetics and is described by the integrated rate equation:

    A = Ao e^(-kt) .....A and Ao are amounts, t is the elapsed time and k is the decay constant (sometimes expressed as λ) which is related to the half-life.

    .... k = ln2 / t½ = ln 2 / 3.8 days = 0.184 day⁻¹

    A = Ao e^(-kt)

    A = 5.7g x e^-(0.184 day⁻¹ x 11.4 day) = 0.71g

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  • david
    Lv 7
    6 months ago

    11.4 / 3,8 = 3 .. so 3 half-lives

    1/2 ^3 = 1/8 only 1/8 of the Rn remains

    1/8 X 5.7g = 0.71 g remains.

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  • 6 months ago

    5.7 g x (1/2)^(11.4 days / 3.8 days) = 0.71 g remains

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