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

Explanation

To solve the problem, we need to find the time at which the amounts of two radioactive materials \( A_1 \) and \( A_2 \) become equal. We will use the concept of radioactive decay and the formula for the remaining quantity of a radioactive substance after a certain time.

### Step-by-Step Solution:

1. **Understand the Initial Conditions:**
- Initial amount of \( A_1 = 40 \, g \)
- Initial amount of \( A_2 = 160 \, g \)
- Half-life of \( A_1 = 20 \, s \)
- Half-life of \( A_2 = 10 \, s \)

2. **Write the Decay Formula:**
The amount of a radioactive substance remaining after time \( t \) can be expressed as:
\[ N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \]
where \( N_0 \) is the initial amount, \( T_{1/2} \) is the half-life, and \( t \) is the elapsed time.

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

4. **Set the Two Equations Equal to Each Other:**
We want to find the time \( t \) when \( N_1(t) = N_2(t) \):
\[ 40 \left( \frac{1}{2} \right)^{\frac{t}{20}} = 160 \left( \frac{1}{2} \right)^{\frac{t}{10}} \]

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:
\[ \left( \frac{1}{2} \right)^{\frac{t}{20}} = \left( \frac{1}{2} \right)^{-2} \left( \frac{1}{2} \right)^{\frac{t}{10}} \]

6. **Combine the Exponents:**
\[ \left( \frac{1}{2} \right)^{\frac{t}{20}} = \left( \frac{1}{2} \right)^{\frac{t}{10} - 2} \]
Since the bases are the same, we can set the exponents equal to each other:
\[ \frac{t}{20} = \frac{t}{10} - 2 \]

7. **Solve for \( t \):**
Multiply through by 20 to eliminate the fractions:
\[ t = 2t - 40 \]
Rearranging gives:
\[ 40 = t \]

8. **Conclusion:**
The time at which the amounts of \( A_1 \) and \( A_2 \) become equal is \( t = 40 \, s \).

### Final Answer:
The amount of the two in the mixture will become equal after **40 seconds**.