Question

# Half lives of two radioactive nuclei $$A$$ and $$B$$ are $$10$$ minutes and $$20$$ minutes, respectively. If, initially a sample has equal number of nuclei, then after $$60$$ minutes, the ratio of decayed numbers of nuclei A and B will be:

A
$$9:8$$
B
$$1:8$$
C
$$8:1$$
D
$$3:8$$
Solution
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#### Correct option is A. $$9:8$$$$N_A = N_{OA} e^{-\lambda t} = \dfrac{N_{OA}}{2^{t/t}\dfrac{1}{2}} = \dfrac{N_{OA}}{2^6}$$$$\therefore$$ Number of nuclei decayed$$= N_{OA} - \dfrac{N_{OA}}{2^6} = \dfrac{63N_{OA}}{64}$$$$N_B = N_{OB}e^{-\lambda t} = \dfrac{N_{OB}}{2^{t/t} \dfrac{1}{2}} = \dfrac{N_{OB}}{2^3}$$$$\therefore$$ Number of nuclei decayed$$= N_{OB} - \dfrac{N_{OB}}{2^3} = \dfrac{7N_{OB}}{8}$$Since $$N_{OA} = N_{OB}$$$$\therefore$$ ratio of decayed numbers of nuclei$$A \& B = \dfrac{63 N_{OA} \times 8}{64 \times 7N_{OB}} = \dfrac{9}{8}$$

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