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Half Life of radioactive needle

  1. Mar 9, 2009 #1
    1. The problem statement, all variables and given/known data
    A radioactive needle contains 222 Rn (t1/2=3.83 d) in secular equilibrium with 226 Ra(t1/2=1600 a). How long does is it required for 222 Rn to decay to half of its original activity?


    2. Relevant equations
    [tex] A(t) = -\frac{dN(t)}{dt} = \lambda N(t) [/tex]



    3. The attempt at a solution
    Secular equilibrium occurs when the activity of the daughter is approximately equal to the activity of the parent.
    Do I just work the above equation for the Daughter Rn?
     
    Last edited: Mar 9, 2009
  2. jcsd
  3. Mar 9, 2009 #2

    Astronuc

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    Staff: Mentor

    [tex] A_a(t) = \frac{dN_a(t)}{dt} = -\lambda_a N_a(t) [/tex]
    and

    [tex] A_b(t) = \frac{dN_b(t)}{dt} = \lambda_a N_a(t) - \lambda_b N_b(t) [/tex]

    one must derive an expression for Nb

    Has one not done so in class or is there not such a derivation in one's textbook?


    Secular equilibrium occurs when dNb/dt = 0, i.e. the activity of b is determined by the decay of a, i.e. Nb is proportional to Na.

    [tex] N_b = \frac{\lambda_a}{\lambda_b} N_a[/tex]

    Reference - http://jnm.snmjournals.org/cgi/reprint/20/2/162.pdf
    http://en.wikipedia.org/wiki/Secular_equilibrium
     
  4. Mar 9, 2009 #3
    Ok. So
    [tex]
    N_a(t)=N_a(0) e^{-\lambda t}
    [/tex]

    Then set
    [tex]
    N_b \longrightarrow \frac{1}{2}N_b
    [/tex]

    Plug that into
    [tex]
    N_b = \frac{\lambda_a}{\lambda_b} N_a
    [/tex]
    And solve for t. Right?
     
  5. Mar 10, 2009 #4
    Correction, I should take the derivative of
    [tex] N_b(t) = \frac{\lambda_a}{lambda_b} N_a(0) e^{-\lambda_a t} [/tex]
    Put in [tex] \frac{A(t)}{2} [/tex] equal to that stuff
    Then solve for t correct?
     
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