%%% asked in Science & MathematicsChemistry · 2 years ago

Initial stage you have 128.0 mg radioactive sample and after 24.0 days 2.00 mg of sample remain Calculate the half life of the sample?

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  • 2 years ago

    Half-life... t½

    Radioactive decay follows first-order kinetics. The integrated rate equation is:

    A = Ao e^(-kt) ..... A are the amounts (or activities), t is the elapsed time and k is the decay constant which is related to the half-life by

    k = ln2 / t½ ..... or ...... t½ = ln2 / k

    A = Ao e^(-kt) can be written as:

    ln A = -kt + ln Ao

    or

    ln (A/Ao) = -kt

    k = ln(A/Ao) / -t ......... solve for k

    k = ln(2.00 mg / 128 mg) / -24.0 days

    k = 0.1733 days⁻¹

    t½ = ln2 / 0.1733 days⁻¹ = 4.00 days

  • Dr W
    Lv 7
    2 years ago

    use this equation

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

    rearranging

    .. A(t) / A(o) = (1/2)^(t / h.l.)

    .. ln[ A(t) / A(o) ] = (t / h.l.) * ln(1/2)

    .. ln[ A(t) / A(o) ] = (t / h.l.) * (ln(1) - ln(2)).. .. . .recall that ln(a/b) = ln(a) - ln(b)

    .. ln[ A(t) / A(o) ] = (t / h.l.) * (-ln(2)).. .. . .recall that ln(1) = 0

    .. ln[ A(o) / A(t) ] = (t / h.l.) * ln(2).. .. . .recall that ln(a/b) = - ln(b/a)

    and finally, we get

    .. h.l = t * ln(2) / ln(Ao/At)

    solving

    .. h.l. = 24.0days * ln(2) / ln(64.0) = 4.0 days

  • 2 years ago

    (2.00 mg / 128.0 mg) = (1/2)^(24.0 days / x days)

    Do the division on the left:

    1/64 = (1/2)^(24.0 days / x days)

    Convert the fraction to powers of 2:

    (1/2)^6 = (1/2)^(24.0 days / x days)

    So:

    24.0 days / x days = 6

    6x days = 24 days

    x = 4.0 days

  • Anonymous
    2 years ago

    After 1 half life you have an activity of 64, after 2 half lives 32 etc.

    Divide 24 days by the number of half lives to reach 2 activity.

    (I'm not sure the mass change is the correct form of the question though, it should be REM, becquerels, curies or grays).

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