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

Answer 1

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.

Answer 2

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

Answer 3

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!