Cesium-137 is part of the nuclear waste produced by uranium-235 fission. The half-life of cesium-137 is 30.2 years. How much time is required for the activity of a sample of cesium-137 to fall to 25.1 percent of its original value?
ln(100/25.1) = kt
k = 0.693/t1/2
sUBSTITUTE AND SOLVE FOR t.
To find the time required for the activity of a sample of cesium-137 to fall to 25.1 percent of its original value, we need to use the concept of half-life.
The half-life of cesium-137 is given as 30.2 years. This means that in 30.2 years, the activity of a sample of cesium-137 will be reduced to half of its original value.
Let's say the initial activity of the sample is A₀. We want to find the time required for the activity to fall to 25.1 percent of A₀, which is equivalent to 0.251A₀.
We can use the formula for exponential decay:
A(t) = A₀ * (1/2)^(t / half-life)
Where A(t) is the remaining activity at time t, A₀ is the initial activity, and the half-life is the half-life of the substance.
Now, let's substitute the values into the equation:
0.251A₀ = A₀ * (1/2)^(t / 30.2)
Divide both sides of the equation by A₀:
0.251 = (1/2)^(t / 30.2)
To solve for t, we need to take the logarithm of both sides. Taking the logarithm base 2 on both sides gives:
log₂(0.251) = t / 30.2
Multiply both sides by 30.2:
t = 30.2 * log₂(0.251)
Using a calculator or software, we can find the value of log₂(0.251) and calculate t.
Note: The result may be negative since the activity has fallen below 25.1 percent of its original value. In this case, the absolute value of t represents the time it took to reach that percentage.