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:

Question

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Answer 1

$N_{A} = N_{O A} e^{- λ t} = \frac{N _{O A}}{2 ^{t / t} \frac{1}{2}} = \frac{N _{O A}}{2 ^{6}}$
$∴$ Number of nuclei decayed

$= N_{O A} - \frac{N _{O A}}{2 ^{6}} = \frac{6 3 N _{O A}}{6 4}$

$N_{B} = N_{O B} e^{- λ t} = \frac{N _{O B}}{2 ^{t / t} \frac{1}{2}} = \frac{N _{O B}}{2 ^{3}}$

$∴$ Number of nuclei decayed
$= N_{O B} - \frac{N _{O B}}{2 ^{3}} = \frac{7 N _{O B}}{8}$

Since $N_{O A} = N_{O B}$

$∴$ ratio of decayed numbers of nuclei
$A & B = \frac{6 3 N _{O A} \times 8}{6 4 \times 7 N _{O B}} = \frac{9}{8}$

Answer 2

N_{A} = N_{O A} e^{- λ t} = \frac{N _{O A}}{2 ^{t / t} \frac{1}{2}} = \frac{N _{O A}}{2 ^{6}}

∴

Number of nuclei decayed

= N_{O A} - \frac{N _{O A}}{2 ^{6}} = \frac{6 3 N _{O A}}{6 4}

N_{B} = N_{O B} e^{- λ t} = \frac{N _{O B}}{2 ^{t / t} \frac{1}{2}} = \frac{N _{O B}}{2 ^{3}}

∴

Number of nuclei decayed

= N_{O B} - \frac{N _{O B}}{2 ^{3}} = \frac{7 N _{O B}}{8}

Since

N_{O A} = N_{O B}

∴

ratio of decayed numbers of nuclei

A & B = \frac{6 3 N _{O A} \times 8}{6 4 \times 7 N _{O B}} = \frac{9}{8}