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(B) The number of atoms $N$ in a mass $m$ with atomic weight $M$ is given by $N = \frac{m}{M} N_A$,where $N_A$ is Avogadro's number.Initial number of

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$50$

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(B) The number of atoms $N$ in a mass $m$ with atomic weight $M$ is given by $N = \frac{m}{M} N_A$,where $N_A$ is Avogadro's number.Initial number of atoms: $N_{A,0} = \frac{m}{M_A} N_A$ and $N_{B,0} = \frac{m}{M_B} N_A$.Given $m_A = m_B = 10^{-2} \ kg$ and $\frac{M_A}{M_B} = \frac{1}{2}$,so $M_B = 2M_A$.Thus,$\frac{N_{A,0}}{N_{B,0}} = \frac{M_B}{M_A} = 2$.After time $t = 16 \ s$,the number of remaining atoms is $N(t) = N_0 (0.5)^{\frac{t}{T_{1/2}}}$.For substance $A$: $N_A(16) = N_{A,0} (0.5)^{\frac{16}{4}} = N_{A,0} (0.5)^4 = \frac{N_{A,0}}{16}$.For substance $B$: $N_B(16) = N_{B,0} (0.5)^{\frac{16}{8}} = N_{B,0} (0.5)^2 = \frac{N_{B,0}}{4}$.The ratio is $\frac{N_A(16)}{N_B(16)} = \frac{N_{A,0}}{16} 	imes \frac{4}{N_{B,0}} = \frac{1}{4} 	imes \frac{N_{A,0}}{N_{B,0}} = \frac{1}{4} 	imes 2 = \frac{1}{2} = \frac{50}{100}$.Therefore,$x = 50$.

$A$ sample contains $10^{-2} \ kg$ each of two substances $A$ and $B$ with half-lives $4 \ s$ and $8 \ s$ respectively. The ratio of their atomic weights is $1:2$. The ratio of the amounts (number of atoms) of $A$ and $B$ remaining after $16 \ s$ is $\frac{x}{100}$. The value of $x$ is:

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