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How much time will it take for 30 grams of 222Rn (222 as in the atomic Mass of Rn) to decay to 7.5 g?

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  • Dr W
    Lv 7
    1 month ago

    you need to know the half life of 222Rn to solve this

    https://en.wikipedia.org/wiki/Isotopes_of_radon

    .. half life = 3.8235(3) days

    fyi.. that (3) means the 3.8235 is a measured number and 1 standard deviation of the measurements is 3 in the rightmost digit... .i.e.. 67% of the time, the measurement falls between 3.8232 and 3.8238 days.  

    anyway, you can either

    .. (1) eyeball this and see that 30g/2 = 15g (1 half life).. 15g/2 = 7.5g (2nd

    .. . . .half life) so that 2 half lives have passed so that

    .. . .. . .t = 2*3.8235 days = 7.647 days

    OR... 

    .. (2) the more general approach.. . showing all the steps 1 at at time

    .. . .. ..... A(t) = A(o) * (1/2)^(t / half life).. .. .<----- memorize this

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

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

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

    .. . .. ..... ln(A(t) / A(o)) = (t / half life) *(0 - ln(2))

    .. . .. ..... ln(A(t) / A(o)) = (t / half life) * (-ln(2))

    .. . .. ..... ln(A(t) / A(o)) / -ln(2) = (t / half life)

    .. . .. ..... ln(A(o) / A(t)) / ln(2) = (t / half life)

    . . .. . and finally

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

    .. .. ... .. .t = half life * (ln(30 / 7.5) / ln(2) 

    .. .. ... .. .t = half life * ln(4) / ln(2)

    .. .. ... .. .t = half life * ln(2² ) / ln(2)

    .. .. ... .. .t = half life * 2*ln(2 ) / ln(2)

    .. .. ... .. .t = half life * 2

    .. .. ... .. .t = 3.8235 days * 2

    .. .. ... .. .t = 7.647 days

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  • 1 month ago

    2 half lives:

    30 * 1/2 = 15 * 1/2 = 7.5

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  • 1 month ago

    (7.5 g) / (30 g) = 0.25 = 1/4 = (1/2)^2

    So takes two half-lives, but that's not really a good measure of time.

    So looking up the half-life of 222Rn in the source below:

    3.82 days x 2 = 7.64 days

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  • 1 month ago

    ²²²Rn has a half life of 3.82 d

    Half life

    N(t) = N₀(1/2)^(t/th) 

    N₀ is initial amount 

    N(t) is the amount remaining after time t 

    th is the half life time, ie, time for half the amount to decay 

    7.5 = 20(1/2)^(t/3.82) 

    (1/2)^(t/3.82)  = 7.5/20 = 0.375

    t/3.82 = log (base0.5)0.375 = 1.415

    t = 5.41 days

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  • Anonymous
    1 month ago

    7.5 g is 34% of the original amount

     

    0.34 = 0.5^n, solve for n which is the number of half lives

     

    time = n * half life for Radon.

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