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` and `160 g` of `a_2`. The amount the two in the mixture will become equal after

Options

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

Explanation

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 Problem
We have two radioactive materials:
- \( A_1 \) with an initial amount of \( 40 \, \text{g} \) and a half-life of \( 20 \, \text{s} \).
- \( A_2 \) with an initial amount of \( 160 \, \text{g} \) and a half-life of \( 10 \, \text{s} \).

We need to find the time \( t \) when the amounts of \( A_1 \) and \( A_2 \) will be equal.

### Step 2: Write the Decay Formula
The amount of a radioactive substance remaining after time \( t \) can be calculated using the formula:
\[ N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \]
where:
- \( N(t) \) is the remaining quantity after time \( t \),
- \( N_0 \) is the initial quantity,
- \( T_{1/2} \) is the half-life of the substance.

### Step 3: Set Up the Equations
For \( A_1 \):
\[ N_1(t) = 40 \left( \frac{1}{2} \right)^{\frac{t}{20}} \]
For \( A_2 \):
\[ N_2(t) = 160 \left( \frac{1}{2} \right)^{\frac{t}{10}} \]

### Step 4: Set the Amounts Equal
We want to find \( t \) when \( N_1(t) = N_2(t) \):
\[ 40 \left( \frac{1}{2} \right)^{\frac{t}{20}} = 160 \left( \frac{1}{2} \right)^{\frac{t}{10}} \]

### Step 5: Simplify the Equation
Divide both sides by \( 40 \):
\[ \left( \frac{1}{2} \right)^{\frac{t}{20}} = 4 \left( \frac{1}{2} \right)^{\frac{t}{10}} \]
Since \( 4 = \left( \frac{1}{2} \right)^{-2} \), we can rewrite the equation as:
\[ \left( \frac{1}{2} \right)^{\frac{t}{20}} = \left( \frac{1}{2} \right)^{-2} \left( \frac{1}{2} \right)^{\frac{t}{10}} \]
This simplifies to:
\[ \left( \frac{1}{2} \right)^{\frac{t}{20}} = \left( \frac{1}{2} \right)^{\frac{t}{10} - 2} \]

### Step 6: Equate the Exponents
Since the bases are the same, we can equate the exponents:
\[ \frac{t}{20} = \frac{t}{10} - 2 \]

### Step 7: Solve for \( t \)
Multiply through by \( 20 \) to eliminate the fractions:
\[ t = 2t - 40 \]
Rearranging gives:
\[ t - 2t = -40 \implies -t = -40 \implies t = 40 \, \text{s} \]

### Conclusion
The time after which the amounts of \( A_1 \) and \( A_2 \) will become equal is \( 40 \, \text{s} \).

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