91Ó°ÊÓ

In the context of radioactive decay, 'half-life' is a critical concept. It refers to the amount of time it takes for half of the radioactive atoms in a sample to decay. It is symbolized as 'T'.
Understanding half-life helps us predict how long it takes for a certain amount of radioactive material to decrease to a specific fraction of its initial amount.
For Iodine-125, the half-life is 60.1 days. This means that every 60.1 days, the quantity of Iodine-125 in a sample is reduced to half. If we start with 100 grams, after 60.1 days, we'd have 50 grams left.
Think of it like this:

• After one half-life: 100 grams becomes 50 grams.
• After two half-lives: 50 grams becomes 25 grams.
• And so on.

The half-life remains constant, which simplifies calculations in radioactive decay problems. So, when you know the half-life of a substance, you can calculate the amount left after any period.

The radioactive decay formula is fundamental in calculating how much of a radioactive substance remains over time. The formula is given by:
\( A = A_0 \times \frac{1}{2}^{\frac{t}{T}} \)
Here's a breakdown of the variables:

• \( A \) is the amount remaining after time \( t \).
• \( A_0 \) is the initial amount of the substance.
• \( t \) is the time elapsed.
• \( T \) is the half-life of the substance.

To solve our problem where Iodine-125 decreased by 25%, and we want to know the duration, we first determine the remaining amount. If 25% is decayed, 75% remains, i.e., \( A = 0.75 \times A_0 \)
Plugging the values in, we have:
\( 0.75 = \frac{1}{2}^{\frac{t}{60.1}} \)
By rearranging to solve for \( t \), taking the natural logarithm of both sides simplifies the equation and allows us to isolate \( t \).
This formula helps us quantify changes in radioactive substances over various periods.

Natural logarithms, often abbreviated as 'ln', are a special kind of logarithm with the base 'e' (approximately 2.718). They are useful in solving radioactive decay problems because they can linearize exponential equations.
When we rearranged our decay formula, we got:
\( 0.75 = \frac{1}{2}^{\frac{t}{60.1}} \)
Taking the natural logarithm of both sides allows us to handle the exponent effectively:
\( \text{ln}(0.75) = \text{ln}\bigg(\frac{1}{2}^{\frac{t}{60.1}}\bigg) \)
Using properties of logarithms, we know:
\( \text{ln}(a^b) = b \times \text{ln}(a) \)
Applying this property, we get:
\( \frac{t}{60.1} \times \text{ln}(0.5) = \text{ln}(0.75) \)
Hence, solving for \( t \) involves basic algebra after knowing the values of \( \text{ln}(0.75) \) and \( \text{ln}(0.5) \).
Natural logarithms transform complex exponential decay problems into more solvable linear forms. They are a key tool in mathematics, especially for growth and decay processes.