Understanding half-life decay through exponential formulas and practical examples

The image presents a half-life problem involving radioactive decay, with the following key elements:

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🔹 Problem Summary:

We are given a substance that decays over time, starting from 200 grams, and it halves every step:
- 200 g → 100 g (after 1 half-life)
- 100 g → 50 g (after 2 half-lives)
- 50 g → 25 g (after 3 half-lives)

This is classic exponential decay, and we're given the relevant formulas.

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🔹 Key Formulas Shown:

1. Exponential Decay Formula:
$$
A_f = A_0 e^{-kt}
$$
Where:
- $ A_f $ = final amount
- $ A_0 $ = initial amount
- $ k $ = decay constant
- $ t $ = time
- $ e $ = Euler’s number (~2.718)

2. Half-Life Formula:
$$
t_{1/2} = \frac{\ln 2}{k}
$$
This gives the time required for half of the substance to decay.

3. Logarithmic Form of Decay:
$$
\ln\left(\frac{A_f}{A_0}\right) = -kt
$$

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🔹 Step-by-Step Solution:

Let's solve a typical half-life problem using this information.

#### ✔ Example Problem:
> If a sample starts at 200 g and decays to 50 g in 6 years, what is the half-life?

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🧮 Step 1: Identify known values

- $ A_0 = 200 $ g
- $ A_f = 50 $ g
- $ t = 6 $ years

We want to find $ t_{1/2} $

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🧮 Step 2: Use the logarithmic form

$$
\ln\left(\frac{A_f}{A_0}\right) = -kt
$$
Plug in values:
$$
\ln\left(\frac{50}{200}\right) = -k(6)
$$
$$
\ln(0.25) = -6k
$$
$$
-1.3863 = -6k \quad \text{(since } \ln(0.25) = \ln(1/4) = -\ln(4) = -1.3863\text{)}
$$
$$
k = \frac{1.3863}{6} \approx 0.23105 \text{ per year}
$$

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🧮 Step 3: Find half-life using $ t_{1/2} = \frac{\ln 2}{k} $

$$
t_{1/2} = \frac{0.6931}{0.23105} \approx 3 \text{ years}
$$

✔ So, the half-life is 3 years.

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🔍 Interpretation:

- After 3 years: 200 g → 100 g
- After 6 years: 100 g → 50 g
- After 9 years: 50 g → 25 g

This matches the pattern shown in the image.

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✔ Final Answer:

> The half-life of the substance is 3 years.

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💡 Summary:

- The decay follows $ A_f = A_0 e^{-kt} $
- Half-life $ t_{1/2} = \frac{\ln 2}{k} $
- From data, we can compute $ k $, then $ t_{1/2} $
- In this case, since 200g → 50g in 6 years (two half-lives), each half-life is 3 years.

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Let me know if you'd like to solve a specific variation of this problem!