a) How much time is required before a 2.50-mg sample of 14661Pm(t1/2 = 2020 days
ID: 1685801 • Letter: A
Question
a) How much time is required before a 2.50-mg sample of 14661Pm(t1/2 = 2020 days) is reduced to 1.25 mg? b) An isotope of krypton has a half-life of 3 minutes. A sample of this isotope produces 1000 counts per minute in a Geiger counter. Determine the number of counts per minute produced after 15 minutes. a) How much time is required before a 2.50-mg sample of 14661Pm(t1/2 = 2020 days) is reduced to 1.25 mg? b) An isotope of krypton has a half-life of 3 minutes. A sample of this isotope produces 1000 counts per minute in a Geiger counter. Determine the number of counts per minute produced after 15 minutes.Explanation / Answer
a) No = 2.50-mg, t1/2 = T = 2020 days, N = 1.25 mg, find t N = No*(1/2)^(t/T) N/No = (1/2)^(t/T) ln(N/No) = (t/T)*ln(1/2) so t = T*ln(N/No)/ln(1/2) = 2020 days Actually, N is half of No, by the definition of half life, t = t1/2 = 2020 days b) T = 3 min, dN/dt = 1000 counts/min, when t = 0. Find dN/dt when t = 15 min. N = No*(1/2)^(t/T) dN/dt = No*(1/2)^(t/T)*ln(1/2)/T when t = 0, dN/dt = No*ln(1/2)/T = 1000, so dN/dt = 1000*(1/2)^(t/T) when t = 15, dN/dt = 1000*(1/2)^(15/3) = 31.25 counts/min
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