1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Staff Emeritus
    Science Advisor

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

    [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
  4. Mar 9, 2009 #3
    Ok. So
    N_a(t)=N_a(0) e^{-\lambda t}

    Then set
    N_b \longrightarrow \frac{1}{2}N_b

    Plug that into
    N_b = \frac{\lambda_a}{\lambda_b} N_a
    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?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook