The number of radioactive nuclei present at the start of an experiment is 3.27 x 1015. The number present twenty days later is 7.66 x 1014. What is the half-life (in days) of the nuclei?
To find the half-life of the radioactive nuclei, we need to determine the time it takes for half of the nuclei to decay.
In this case, the number of nuclei at the start of the experiment is 3.27 x 10^15, and the number present twenty days later is 7.66 x 10^14.
We can set up a proportion to find the ratio of the number of nuclei after 20 days to the number of nuclei at the start:
(7.66 x 10^14) / (3.27 x 10^15) = (1/2) (since it is half of the original value)
Now, let's solve for the half-life (t):
(7.66 x 10^14) / (3.27 x 10^15) = 1 / (2^t)
To simplify the calculation, let's express both sides of the equation with the same base, such as 10:
log((7.66 x 10^14) / (3.27 x 10^15)) = log(1 / (2^t))
Using logarithmic properties, we can express the right side of the equation as:
log((7.66 x 10^14) / (3.27 x 10^15)) = -t * log(2)
Now, we can solve for t by rearranging the equation:
t = -log((7.66 x 10^14) / (3.27 x 10^15)) / log(2)
Plugging in the values and calculating:
t = -log(7.66 x 10^14 / 3.27 x 10^15) / log(2)
After evaluating this expression, we find that the half-life of the nuclei is approximately 13.17 days.
To find the half-life of the nuclei, we can use the formula:
N = N₀ * (1/2)^(t/T)
Where:
N is the final number of nuclei (7.66 x 10^14)
N₀ is the initial number of nuclei (3.27 x 10^15)
t is the time elapsed (20 days)
T is the half-life we want to find
Substituting the given values into the formula, we get:
7.66 x 10^14 = 3.27 x 10^15 * (1/2)^(20/T)
Divide both sides of the equation by 3.27 x 10^15:
(7.66 x 10^14) / (3.27 x 10^15) = (1/2)^(20/T)
Now, take the logarithm (base 1/2) of both sides:
log₁/₂((7.66 x 10^14) / (3.27 x 10^15)) = log₁/₂((1/2)^(20/T))
Using the logarithmic property log(a / b) = log(a) - log(b), we can simplify the equation:
log₁/₂(7.66 x 10^14) - log₁/₂(3.27 x 10^15) = (20/T)
The left side of the equation can be calculated using the change of base formula:
log₁/₂(7.66 x 10^14) ≈ ln(7.66 x 10^14) / ln(1/2)
log₁/₂(3.27 x 10^15) ≈ ln(3.27 x 10^15) / ln(1/2)
Substituting these values into the equation:
ln(7.66 x 10^14) / ln(1/2) - ln(3.27 x 10^15) / ln(1/2) = (20/T)
Now we can solve for T by rearranging the equation:
(20/T) = ln(7.66 x 10^14) / ln(1/2) - ln(3.27 x 10^15) / ln(1/2)
Multiply both sides by T:
20 = T * [ln(7.66 x 10^14) / ln(1/2) - ln(3.27 x 10^15) / ln(1/2)]
Finally, divide both sides by [ln(7.66 x 10^14) / ln(1/2) - ln(3.27 x 10^15) / ln(1/2)] to isolate T:
T = 20 / [ln(7.66 x 10^14) / ln(1/2) - ln(3.27 x 10^15) / ln(1/2)]
Calculating this expression will give us the value of the half-life (T) in days.