Half Life of radioactive needle

[completenoob]

Homework Statement

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?

Homework Equations

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

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?

 

[Astronuc]

[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 N_b

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


Secular equilibrium occurs when dN_b/dt = 0, i.e. the activity of b is determined by the decay of a, i.e. N_b is proportional to N_a.

[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

 

[completenoob]

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?

 

[completenoob]

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?