A radioactive isotope of Bismuth undergoes Beta Decay with a half-life of 22 days. How long (in days) will it take the Activity (A) of the sample to decrease to one-ninth of its original value? Assume that the original activity is A0.

just find t where

(1/2)^(t/22) = 1/9

The value of A0 does not matter, as it will be on both sides of the equation.

(t/22) log(1/2) = log(1/9)
Now just churn the numbers.

To solve this problem, we'll need to use the radioactive decay equation:

A = A0 * (1/2)^(t/T)

Where:
A is the current activity
A0 is the initial activity
t is the time that has passed
T is the half-life of the isotope

We want to find the time it takes for the activity to decrease to one-ninth of its original value, which means A = A0/9. We can substitute these values into the equation:

A0/9 = A0 * (1/2)^(t/T)

We can simplify this equation by canceling out A0 on both sides:

1/9 = (1/2)^(t/T)

To isolate the exponential term, we'll take the logarithm (base 1/2) of both sides:

log(1/9) = log[(1/2)^(t/T)]

Using logarithm properties, we can bring down the exponent:

log(1/9) = (t/T) * log(1/2)

Now, we can solve for t by dividing both sides by log(1/2):

(t/T) = log(1/9) / log(1/2)

Finally, we multiply both sides by T to solve for t:

t = T * (log(1/9) / log(1/2))

Now we can plug in the given half-life of 22 days and calculate the value of t:

t = 22 * (log(1/9) / log(1/2))

Using a calculator, we find:

t ≈ 65.605 days

Therefore, it will take approximately 65.605 days for the activity of the sample to decrease to one-ninth of its original value.