# Half Life of radioactive needle

1. Mar 9, 2009

### completenoob

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

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. Mar 9, 2009

### Astronuc

Staff Emeritus
$$A_a(t) = \frac{dN_a(t)}{dt} = -\lambda_a N_a(t)$$
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 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.

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

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

3. Mar 9, 2009

### completenoob

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?

4. Mar 10, 2009

### 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?