91Ó°ÊÓ

Radioactive decay is a process by which an unstable atomic nucleus loses energy by emitting radiation. This is a random and spontaneous process.
The rate of decay of a radioactive substance is proportional to the amount of substance currently present.
In equations, radioactive decay is often modeled using exponential functions because the amount of substance decreases continuously over time.
For example, if you start with 60 grams of a radioactive material and after 3 hours only 20 grams remain, you can describe this decay using an exponential function like: \( y = y_{0} e^{kt} \), where:

• \( y \) is the amount of substance left.
• \( y_{0} \) is the initial amount.
• \( k \) is the decay constant.
• \( t \) is time.

This function helps predict how much of the substance will remain after any given period.

Natural logarithms (ln) are logarithms with the base \( e \), where \( e \approx 2.71828 \).
The natural logarithm of a number is the power to which \( e \) must be raised to equal that number.
This is especially useful in problems involving exponential decay, like radioactive decay.
Logarithms turn multiplication into addition and exponents into multiplication, simplifying the equations. For example, if \( e^{x} = a \), then \( \text{ln}(a) = x \).
In our exercise, we use this property to solve for the decay constant \( k \) by taking the natural logarithm of both sides.

Exponential functions are mathematical functions of the form \( f(x) = a e^{bx} \), where

• \( a \) and \( b \) are constants.
• \( e \) is the base of the natural logarithm.

These functions describe processes that grow or decrease at a rate proportional to their current value, making them ideal for modeling growth and decay.
In radioactive decay, the function \( y = y_{0} e^{kt} \) represents how the quantity of a radioactive substance diminishes over time.
The decay constant \( k \) determines the rate of decay. A negative \( k \) value indicates decay, where the substance decreases over time.

To solve exponential equations, follow these steps:
1. Set up the equation with given values. For instance, from our previous example: \( 20 = 60 e^{3k} \).
2. Isolate the exponential term by dividing both sides by the initial amount: \(\frac{20}{60} = e^{3k} \), which simplifies to \(\frac{1}{3} = e^{3k} \).
3. Apply natural logarithms to both sides to employ the property \( \text{ln}(e^{x}) = x \): \( \text{ln}\big(\frac{1}{3}\big) = \text{ln}(e^{3k}) \), thus simplifying to \( \text{ln}\big(\frac{1}{3}\big) = 3k \).
4. Solve for \( k \) by dividing by 3: \( k = \frac{\text{ln}\big(\frac{1}{3}\big)}{3} \).
These steps simplify complex-looking expressions into manageable calculations to find the decay constant.