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
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
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!
To find the half-life of the decay process, we can use the formula:
t(1/2) = (t * log(2)) / log(N₀ / N₁)
where:
- t(1/2) is the half-life
- t is the time in minutes (32 minutes in this case)
- N₀ is the initial value
- N₁ is the final value (in this case, one sixteenth of N₀)
Let's plug in the values and calculate the half-life:
t(1/2) = (32 * log(2)) / log(N₀ / (1/16))
Since N₁ is one sixteenth of N₀, we can simplify the equation:
t(1/2) = (32 * log(2)) / log(16)
t(1/2) = (32 * log(2)) / 4
Now, we can calculate the half-life:
t(1/2) = 8 * log(2)
t(1/2) ≈ 8 * 0.6931
t(1/2) ≈ 5.54 minutes
Therefore, the half-life of this decay process is approximately 5.54 minutes.
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.