If `80%` of a radioactive element undergoing decay is left over after a certain periof of time `t` form the start, how many such periofs should elapse form the start for just over `50%` of the element to be left over?

Explanation

To solve the problem step by step, we will use the principles of radioactive decay, which follows first-order kinetics.

### Step 1: Understand the given information
We know that 80% of a radioactive element is left after a certain time \( t \). This means that 20% has decayed.

### Step 2: Set up the initial and final concentrations
Let's assume the initial amount of the radioactive element is 100%. After time \( t \), the amount left is 80%.

- Initial concentration (A₀) = 100%
- Final concentration (A) = 80%

### Step 3: Use the first-order kinetics formula
For first-order reactions, the relationship between the time, rate constant \( k \), and concentrations is given by:
\[ t = \frac{2.303}{k} \log \left( \frac{A_0}{A} \right) \]
Substituting the values we have:
\[ t = \frac{2.303}{k} \log \left( \frac{100}{80} \right) \]

### Step 4: Calculate the logarithm
Now we calculate the logarithm:
\[ \log \left( \frac{100}{80} \right) = \log(1.25) \]
Using a calculator, we find:
\[ \log(1.25) \approx 0.0969 \]

### Step 5: Substitute back into the equation
Now substituting this value back into the equation for \( t \):
\[ t = \frac{2.303}{k} \times 0.0969 \]

### Step 6: Determine the half-life
The half-life \( t_{1/2} \) for a first-order reaction is given by:
\[ t_{1/2} = \frac{0.693}{k} \]
From the equation for \( t \), we can express \( k \) in terms of \( t \):
\[ k = \frac{0.693}{t_{1/2}} \]
Substituting this into the equation for \( t \):
\[ t = \frac{2.303 \cdot t_{1/2}}{0.693} \cdot 0.0969 \]
Calculating \( \frac{2.303}{0.693} \approx 3.32 \):
\[ t \approx 3.32 \cdot t_{1/2} \cdot 0.0969 \]

### Step 7: Calculate the number of periods for 50% remaining
To find how many periods \( n \) it takes for just over 50% of the element to be left, we need to know how many half-lives it takes to reach just over 50%.

After 1 half-life, 50% remains. After 2 half-lives, 25% remains. Therefore, to have just over 50% remaining, we need to wait for 1 half-life.

### Final Answer
Thus, just over 50% of the element will remain after **1 half-life**.

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