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20 July, 06:23

In 29 days, the number of radioactive nuclei decreases to one-sixteenth the number present initially. What is the half-life (in days) of the material?

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Answers (2)
  1. 20 July, 07:35
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    The half life of the material approximately 7years

    Explanation:

    Half life is defined as the time taken by a radioactive material to decay to half of its original substance.

    Half life t1/2 = ln2/λ

    λ is the decay constant.

    To get the decay constant, we will use the relationship

    N = Noe^-λt

    No is the initial value of the substance

    N is the final value of the substance after decay

    t is the time taken by the substance to decay.

    From the formula

    N/No = e^-λt ... (1)

    If radioactive nuclei decreases to one-sixteenth the number present initially, this means N = 1/16No

    N/No = 1/16

    t = 29days

    Substituting this values into equation 1 we have:

    1/16 = e^-29λ

    Taking the ln of both sides

    ln (1/16) = lne^-29λ

    ln (1/16) = - 29λ

    λ = ln (1/16) / -29

    λ = ln0.0625/-29

    λ = - 2.77/-29

    λ = 0.0955

    Substituting λ = 0.0955 into the half life formula, we have;

    t1/2 = ln2/0.0955

    t1/2 = 7.26days

    The half life of the materia is 7.26years
  2. 20 July, 09:51
    0
    In 29 days, the number of radioactive nuclei decreases to one-sixteenth the number present initially. What is the half-life (in days) of the material

    Explanation:

    The radioactive decay law says

    N (t) = No•2^ (-t/t½)

    Then,

    In 29 days

    t = 29

    The radioactive element has decreased to one-sixteenth of it's original

    Then, N (t) / No = 1/16i

    Now,

    N (t) = No•2^ (-t/t½)

    N (t) / No = 2^ (-t/t½)

    1/16 = 2^ (-29/t½)

    Take In base 2 of both sides

    In (1/16) = In•2^ (-29/t½)

    In (2^-4) = - 29/t½

    -4In (2) = - 29 / t½

    Since the ln is in base 2 then, In (2) = 1

    -4 = - 29 / t½

    Therefore

    t½ = - 29 / - 4

    t½ = 7.25 days

    t½ ≈ 7 days

    So, the half life of the radioactive element is approximately 7 days
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