Suppose that the activity of a radioactive substance is initially 398 disintegra
ID: 1475773 • Letter: S
Question
Suppose that the activity of a radioactive substance is initially 398 disintegrations/min and two days later it is 285 disintegrations/min. What is the activity four days later still, or six days after the start? Give your answer in disintegrations/min. The half-life for the _ decay of uranium is 4.47 × 109 yr. Determine the age (in years) of a rock specimen that contains 60.0% ofits original number of atoms. Suppose that the activity of a radioactive substance is initially 398 disintegrations/min and two days later it is 285 disintegrations/min. What is the activity four days later still, or six days after the start? Give your answer in disintegrations/min. The half-life for the _ decay of uranium is 4.47 × 109 yr. Determine the age (in years) of a rock specimen that contains 60.0% ofits original number of atoms. Suppose that the activity of a radioactive substance is initially 398 disintegrations/min and two days later it is 285 disintegrations/min. What is the activity four days later still, or six days after the start? Give your answer in disintegrations/min. The half-life for the _ decay of uranium is 4.47 × 109 yr. Determine the age (in years) of a rock specimen that contains 60.0% ofits original number of atoms.Explanation / Answer
The formula for the amount of radioactivity over time is
A(t) = Ao * e^(-kt)
A(t) is the activity (amount) at time t
k is the decay constant =ln(2)/t1/2
A(0) = 398 disintegrations/min =398DPM
A(2) = 285 DPM
285 = 398 e^(-2*k)
Take natural logs of both sides of the "=" sign:
ln (285) = ln(398) -2k
5.652 = 5.986 - 2k
-0.3344 = -2k
0.1672 = k
Now we can plug k in to the equation for values of t other than 2:
A(6) = A(0)* e(-0.1672 * 6) = 398 * e^(-1.0032)
A(6) = 0.3667 * 398 = 145.9 DPM
b) t1/2 = 4.47*10^7yr
60/100*398DPM=398*e^(-ln(2)/(4.47 × 10^9)*t)
=>0.6=e^(-1.5506*10^-10t)
=>-0.5108=-1.5506*10^-10*t=>t=3.294*10^9year
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