Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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:

60 s

80 s

20 s

40 s

Step-by-Step Solution

Formula: Radioactive decay follows first-order kinetics. The amount of substance remaining ( $N$ ) after time $t$ is given by $N = N_0 (1/2)^n$ , where $n = t/T_{1/2}$ is the number of half-lives .
Setup Equation: We want the time $t$ when the remaining amounts of $A_1$ and $A_2$ are equal ( $N_1 = N_2$ ). $40 \left(\frac{1}{2}\right)^{\frac{t}{20}} = 160 \left(\frac{1}{2}\right)^{\frac{t}{10}}$
Solve for t: $\frac{(1/2)^{t/20}}{(1/2)^{t/10}} = \frac{160}{40}$ $\left(\frac{1}{2}\right)^{\frac{t}{20} - \frac{t}{10}} = 4$ $\left(\frac{1}{2}\right)^{-\frac{t}{20}} = 2^2$ $2^{\frac{t}{20}} = 2^2$ $\frac{t}{20} = 2 \implies t = 40 \text{ s}$
Verification: At $t=40$ s:

$A_1$ (2 half-lives): $40 \to 20 \to 10$ g.
$A_2$ (4 half-lives): $160 \to 80 \to 40 \to 20 \to 10$ g.
Amounts are equal.

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