A mixture consists of two radioactive materials `A_1` and `A_2` with half-lives of 20 s and 10 s respectively. Initially the mixture has 40 g of `A_1` of 160 g of `A_2`. After what time of amount of the two in the mixture will become equal ?

Options

A. 60 s
B. 80 s
C. 20 s
D. 40s

Answer

To solve the problem of when the amounts of two radioactive materials \( A_1 \) and \( A_2 \) will become equal, we can follow these steps:

### Step 1: Understand the decay of radioactive materials
Radioactive materials decay over time, and the amount remaining after a certain period can be calculated using the half-life formula. The mass remaining after \( n \) half-lives is given by:

\[ m = m_0 \left( \frac{1}{2} \right)^n \]

where:
- \( m_0 \) is the initial mass,
- \( m \) is the remaining mass after \( n \) half-lives,
- \( n \) is the number of half-lives that have passed.

### Step 2: Identify initial conditions and half-lives
For our problem:
- Initial mass of \( A_1 (m_{A_1}) = 40 \, \text{g} \)
- Initial mass of \( A_2 (m_{A_2}) = 160 \, \text{g} \)
- Half-life of \( A_1 (t_{1/2, A_1}) = 20 \, \text{s} \)
- Half-life of \( A_2 (t_{1/2, A_2}) = 10 \, \text{s} \)

### Step 3: Set up equations for remaining masses
Let \( t \) be the time in seconds. The number of half-lives that have passed for each material can be calculated as:

\[ n_{A_1} = \frac{t}{t_{1/2, A_1}} = \frac{t}{20} \]
\[ n_{A_2} = \frac{t}{t_{1/2, A_2}} = \frac{t}{10} \]

Now we can express the remaining masses after time \( t \):

\[ m_{A_1}(t) = 40 \left( \frac{1}{2} \right)^{\frac{t}{20}} = 40 \cdot 2^{-\frac{t}{20}} \]
\[ m_{A_2}(t) = 160 \left( \frac{1}{2} \right)^{\frac{t}{10}} = 160 \cdot 2^{-\frac{t}{10}} \]

### Step 4: Set the masses equal to each other
We want to find the time \( t \) when the masses are equal:

\[ 40 \cdot 2^{-\frac{t}{20}} = 160 \cdot 2^{-\frac{t}{10}} \]

### Step 5: Simplify the equation
Dividing both sides by 40 gives:

\[ 2^{-\frac{t}{20}} = 4 \cdot 2^{-\frac{t}{10}} \]

Since \( 4 = 2^2 \), we can rewrite the equation:

\[ 2^{-\frac{t}{20}} = 2^2 \cdot 2^{-\frac{t}{10}} \]

This simplifies to:

\[ 2^{-\frac{t}{20}} = 2^{2 - \frac{t}{10}} \]

### Step 6: Set the exponents equal
Since the bases are the same, we can equate the exponents:

\[ -\frac{t}{20} = 2 - \frac{t}{10} \]

### Step 7: Solve for \( t \)
To eliminate the fractions, multiply the entire equation by 20:

\[ -t = 40 - 2t \]

Rearranging gives:

\[ -t + 2t = 40 \]
\[ t = 40 \, \text{s} \]

### Conclusion
The time after which the amounts of the two materials will become equal is \( t = 40 \, \text{s} \).

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