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(A) Let $N_0 = 10^{20}$ be the initial number of atoms at $t = 0$.The number of atoms remaining at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.Th

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$7 	imes 10^{19}$

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(A) Let $N_0 = 10^{20}$ be the initial number of atoms at $t = 0$.The number of atoms remaining at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.The number of $\alpha$-particles emitted during the $n$-th year is the difference in the number of atoms at the beginning and end of that year: $\Delta N_n = N(n-1) - N(n) = N_0 e^{-\lambda(n-1)} (1 - e^{-\lambda})$.For the second year $(n=2)$: $\Delta N_2 = N_0 e^{-\lambda} (1 - e^{-\lambda})$.For the third year $(n=3)$: $\Delta N_3 = N_0 e^{-2\lambda} (1 - e^{-\lambda})$.Given $\Delta N_3 = 0.3 	imes \Delta N_2$,we have:$N_0 e^{-2\lambda} (1 - e^{-\lambda}) = 0.3 	imes N_0 e^{-\lambda} (1 - e^{-\lambda})$.Dividing both sides by $N_0 (1 - e^{-\lambda}) e^{-\lambda}$,we get $e^{-\lambda} = 0.3$.The number of $\alpha$-particles emitted in the first year $(n=1)$ is:$\Delta N_1 = N_0 - N_1 = N_0 - N_0 e^{-\lambda} = N_0 (1 - e^{-\lambda})$.Substituting the values: $\Delta N_1 = 10^{20} (1 - 0.3) = 10^{20} (0.7) = 7 	imes 10^{19}$.

$A$ sample initially contains $10^{20}$ radioactive atoms. The number of $\alpha$-particles emitted in the third year is $0.3$ times the number of $\alpha$-particles emitted in the second year. How many $\alpha$-particles are emitted in the first year?

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