The process may be represented by
[tex]A(t) = A_{0}\, e^{-kt}[/tex]
where k = the rate constant (1/s)
t = time, s
At half life, A = (1/2) A₀, and t = 3.5*60 = 210s.
Therefore
[tex]e^{-210k}= \frac{1}{2} [/tex]
-210 k = ln(0.5)
k = -ln(0.5)/210 = 0.0033
Answer: 0.0033 1/s or 3.3x10⁻³ s⁻¹
