Asked by Hana
The activity of radioactive source falls to one sixteenth of originally value in 32 minutes.what is the half life,in minutes, of this decay process
Answers
Answered by
Bot
The formula for radioactive decay is:
N = N0*(1/2)^(t/t1/2)
where N is the final amount of radioactive material, N0 is the initial amount, t is the time elapsed, and t1/2 is the half-life.
From the problem, we know that N/N0 = 1/16 and t = 32 minutes. Substituting these values into the formula:
1/16 = (1/2)^(32/t1/2)
Take the natural logarithm of both sides:
ln(1/16) = ln(1/2)^(32/t1/2)
ln(1/16) = (32/t1/2)*ln(1/2)
Solve for t1/2:
t1/2 = (32/ln(1/2))*ln(1/16)
t1/2 = 69.3 minutes
Therefore, the half-life of this radioactive decay process is 69.3 minutes.
N = N0*(1/2)^(t/t1/2)
where N is the final amount of radioactive material, N0 is the initial amount, t is the time elapsed, and t1/2 is the half-life.
From the problem, we know that N/N0 = 1/16 and t = 32 minutes. Substituting these values into the formula:
1/16 = (1/2)^(32/t1/2)
Take the natural logarithm of both sides:
ln(1/16) = ln(1/2)^(32/t1/2)
ln(1/16) = (32/t1/2)*ln(1/2)
Solve for t1/2:
t1/2 = (32/ln(1/2))*ln(1/16)
t1/2 = 69.3 minutes
Therefore, the half-life of this radioactive decay process is 69.3 minutes.
Answered by
oobleck
AAAaannndd the bot gets it wrong yet again!
1/16 = (1/2)^4
so 32 minutes is 4 half-lives.
Half-life is 8 minutes
1/16 = (1/2)^4
so 32 minutes is 4 half-lives.
Half-life is 8 minutes
Answered by
Bot
You are correct! I apologize for the incorrect answer. The activity of a radioactive source falling to one-sixteenth of its original value in 32 minutes indicates that four half-lives have passed.
Using the formula for radioactive decay, we can write:
N/N0 = (1/2)^(t/T)
where N is the final amount of radioactive material, N0 is the initial amount, t is the time elapsed, and T is the half-life.
Since the activity has fallen to one-sixteenth of its original value, we have N/N0 = 1/16. We also know that four half-lives have passed, so t/T = 4. Substituting these values into the formula above, we get:
1/16 = (1/2)^(4)
Taking the natural logarithm of both sides, we get:
ln(1/16) = ln(1/2^4)
Simplifying the right-hand side, we get:
ln(1/16) = -4*ln(2)
Solving for T, we get:
T = t/4 = 32 minutes / 4 = 8 minutes
Therefore, the half-life of this radioactive decay process is 8 minutes. Thank you for bringing the error to my attention!
Using the formula for radioactive decay, we can write:
N/N0 = (1/2)^(t/T)
where N is the final amount of radioactive material, N0 is the initial amount, t is the time elapsed, and T is the half-life.
Since the activity has fallen to one-sixteenth of its original value, we have N/N0 = 1/16. We also know that four half-lives have passed, so t/T = 4. Substituting these values into the formula above, we get:
1/16 = (1/2)^(4)
Taking the natural logarithm of both sides, we get:
ln(1/16) = ln(1/2^4)
Simplifying the right-hand side, we get:
ln(1/16) = -4*ln(2)
Solving for T, we get:
T = t/4 = 32 minutes / 4 = 8 minutes
Therefore, the half-life of this radioactive decay process is 8 minutes. Thank you for bringing the error to my attention!
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