The radioactive decay of a certain sample produced 995 disintegrations per minut

Question

The radioactive decay of a certain sample produced 995 disintegrations per minute. Exactly 6.00 days later, the rate of decay was found to be 513 disintegrations per minute. Calculate the half-life, in days, for the decay of this sample. Answer: By mass spectral analysis, a sample of strontium is known to contain 2.64x1010 atoms of Sr-90 as the only radioactive element. The absolute disintegration rate of this sample is measured as 1238 disintegrations per minute. Calculate the half-life (in years) of Sr-90 Answer How long will it take (in years) for the disintegration rate of this sample to drop to 795 disintegrations per minute? Answer: In a living organism, the decay of C-14 produces 15.3 disintegrations per minute per gram of carbon. The half-life of C-14 is 5730 years. A bone sample with 3.7 g of carbon has 18.2 disintegrations per minute. How old is the bone sample in years? Answer:

Explanation / Answer

first order kinetics,

k = (1/t)ln(a/x)

a = initial amount = 995 dps

x = amount of isomer after t, time = 513 dps

t= time = 6 days

k = (1/6)ln(995/513)

k = 0.11 day-1

T1/2 = 0.693/0.11 = 6.3 days

2)

first order kinetics,

k = (1/t)ln(a/x)

a = initial amount = 2.64*10^10 atoms

x = amount of isomer after t, time = (2.64*10^10)-(1238) atoms

t= time = 1 min

k = (1/1)ln((2.64*10^10)/((2.64*10^10)-(1238)))

k = 4.69*10^-8 min-1

T1/2 = 0.693/(4.69*10^-8) = 1.48*10^7 min

3) (4.69*10^-8) = (1/t)ln(1238/795)

t = 17.97 years

4)

first order kinetics,

k = 0.693/t/12

k= 0.693/(5730) = 0.000121 y-1

k = (1/t)ln(a/x)

a = living carbon ACTIVITY = 15.3 dps

x = current activity wooden beam = 18.2/3.7 = 4.92 dps

0.000121 = (1/t)ln(15.3/4.92)

t = age of the bone = 9376.4 years

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