simple question about radioactivity equation...

StudioSaturn

In the radiactivity equation A = A₀e^{-ln(2)t/T_1/2} How do I get A₀? Is that just ln(2)N₀/T_1/2? What if I don't know the initial number of atoms in the sample? Thanks...

Astronuc

If one does not know N_o at t_o, one counts at A or N at t₁ and t₂, and then extrapolate back to t_o. One would also could also determine the relative amounts of decaying nuclide and daughter. Elements can be identified by chemical analysis, e.g., emission spectroscopy (perhaps with ICP) or mass spectrometry, and radionulides can be identified by characteristic radiation emissions. Usually one does a combination of analyses.

StudioSaturn

hmm... Ok so here's the question from my book then.
A sample X with Half-life 7.5min is measured from t1 = 3 min to t2=13 min. The total number of counts during those 10min is 34650. They want me to find the activity of the sample at t₀=0... Any thoughts? Thanks for your help!

Astronuc

If one is given the total counts between two times, then integrates the activity over time, i.e., between t₁ and t₂

N = [itex]\int_{t_1}^{t_2} A(t) dt[/itex], and one should know the expression for A(t) = λ N(t), and one know the expression for N(t) related to N_o.

edgepflow

Note that in general:

Quantity = Rate X Time

Shorthand,

Q = R t

or, in differential form:

dQ = R dt

And,

Q = Intergral [R dt]

On your case

R = A(t)

and you can find Ao.

StudioSaturn

Ok, so the A(t2) = A(t1)*e^{-[itex]\lambda[/itex]t₂} and solve for A(t1). But what is A(t2)? 34650/10min?
Then A(t1) = A(t₀)*e^{1[itex]\lambda[/itex]t₁} and solve for A(t₀) correct?

Astronuc

No.

One needs to work out the integral for the activity during the period from t1 to t2.
The counts = 34650 represents all the decays during that period, which is found by integrating the activity A(t) between the two times. Work out the integral.

Remember A(t) = λ N(t). But what is the expression for N(t)?