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

A(t) = -\frac{dN(t)}{dt} = \lambda N(t)

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]

A_a(t) = \frac{dN_a(t)}{dt} = -\lambda_a N_a(t)
and

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

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.

N_b = \frac{\lambda_a}{\lambda_b} N_a

Reference - http://jnm.snmjournals.org/cgi/reprint/20/2/162.pdf
http://en.wikipedia.org/wiki/Secular_equilibrium

 

[completenoob]

Ok. So
<br /> N_a(t)=N_a(0) e^{-\lambda t} <br />

Then set
<br /> N_b \longrightarrow \frac{1}{2}N_b <br />

Plug that into
<br /> N_b = \frac{\lambda_a}{\lambda_b} N_a<br />
And solve for t. Right?

 

[completenoob]

Correction, I should take the derivative of
N_b(t) = \frac{\lambda_a}{lambda_b} N_a(0) e^{-\lambda_a t}
Put in \frac{A(t)}{2} equal to that stuff
Then solve for t correct?