Question 19.PS.5: Half-Life A sample of ^24Na initially undergoes 3.50 × 10^4 ......
Half-Life
A sample of ~^{24}Na initially undergoes 3.50 × 10^4 disintegrations per second (s^{-1}). After 24.0 h, its disintegration rate has fallen to 1.16 × 10^4 s^{-1}. What is the half-life of ~^{24}Na?
15.0 h
Strategy and Explanation We use Equation 19.2 relating activity (disintegration rate) at time zero and time t with the decay constant k. The experiment provided us with A, A_0, and the time.
\ln \frac{A}{A_0} = – kt [19.2]
\ln (\frac{1.16 × 10^4 s^{-1}}{3.50 × 10^4 s^{-1}}) = \ln(0.331) = – k(24.0 h)
k = – \frac{\ln(0.331)}{24 h} = – (\frac{- 1.104}{24.0 h}) = 0.0460 h^{-1}
From k we can determine t_{1/2}.
t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.0460 h^{-1}} = 15.0 h
Reasonable Answer Check The activity (disintegration rate) fell to between one half and one quarter of its initial value in 24.0 h, so the half-life must be less than 24.0 h, and this agrees with our more accurate calculation.