Question

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 of the two in the mixture will become equal after

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

Answer 1

The Correct Option is D

 To solve the problem of determining when the amounts of the radioactive materials \( A_1 \) and \( A_2 \) in the mixture will become equal, we need to apply the formula for radioactive decay and use their respective half-lives.

The formula for the remaining quantity \( N \) of a radioactive substance after time \( t \) is given by:

\[ N = N_0 \times \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \]

where \( N_0 \) is the initial quantity and \( T_{1/2} \) is the half-life of the substance.

Let's solve for both substances:

1. For \( A_1 \) (initially 40 g, \( T_{1/2} = 20 \, s \)):
2. For \( A_2 \) (initially 160 g, \( T_{1/2} = 10 \, s \)):
3. We need to find the time \( t \) when \( N_{A_1} = N_{A_2} \).
4. Simplify the equation:
5. Equating powers of 1/2 from both sides:
6. Solve for \( t \):

Therefore, after 40 seconds, the amounts of \( A_1 \) and \( A_2 \) will be equal.