HW_Chap_20a t Radioactive Decay Calculations 22 of 22 Constants Periodic Table I
ID: 303696 • Letter: H
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HW_Chap_20a t Radioactive Decay Calculations 22 of 22 Constants Periodic Table If a substance is radioactive, this means that the nucleus is unstable and will therefore decay by any number of processes (alpha decay, beta decay, etc.). The decay of radioactive elements follows first-order kinetics. Therefore the rate of decay can be described by the same integrated rate equations and half-life equations that are used to describe the rate of first-order chemical reactions In At and 1/20.693 where Ao is the initial amount or activity, At is the amount or activity at time t, and k is the rate constant By manipulation of these equations (substituting 0.693/t1/2 for k in the integrated rate equation), we can arrive at the following formula fraction remaining (0.5 where n is the number of half-lives. The equation relating the number of half-lives to time t is 1/2 where t1/2 is the length of one half-lifeExplanation / Answer
given At/AO= exp(-Kt), (1)
given A= 100 and At= 400, t= 54.5 min, substituting these values in Eq.1 give
100/400= exp(-k*54.5), K= 0.02544/min
half life (t1/2) is related to K as t1/2=0.693/K= 0.693/0.02544= 27.24 min
2. given t1/2=3.2 hrs, K=0.693/t1/2= 0.693/3.2 =0.216/hr
given M= 47.8 gm at t= 8 hrs, from Eq.1, M/Mo = exp(-0.216*8)
Mo= 47.8/exp(-0.216*8)= 269 gm , mass present 8 hours ago
3. from t1/2= 432, K= 0.693/432/yr= 0.001604/yr
given M/MO=0.45
from Eq.1, 0.45= exp(-0.001604*t), t= 497.8 years
4. half lfie of C-14, t1/2 = 5730 years, K=0.693/5730 =0.000121/yr
from eq.1, given M/MO= 0.4 = exp(-0.000121*t), t= 7576years
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