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(B) Let $N_0 = 10^{20}$ be the initial number of radioactive atoms.Let $\lambda$ be the decay constant.The number of atoms remaining after time $t$ is

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$7 	imes 10^{19}$

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(B) Let $N_0 = 10^{20}$ be the initial number of radioactive atoms.Let $\lambda$ be the decay constant.The number of atoms remaining after time $t$ is $N(t) = N_0 e^{-\lambda t}$.The number of $\alpha$-particles emitted in the $n$-th year is the difference in the number of atoms at the start and end of that year: $\Delta N_n = N(n-1) - N(n) = N_0 e^{-\lambda(n-1)} - N_0 e^{-\lambda n} = N_0 e^{-\lambda(n-1)}(1 - e^{-\lambda})$.The ratio of particles emitted in the third year to the second year is given by:$\frac{\Delta N_3}{\Delta N_2} = \frac{N_0 e^{-2\lambda}(1 - e^{-\lambda})}{N_0 e^{-\lambda}(1 - e^{-\lambda})} = e^{-\lambda} = 0.3$.The number of $\alpha$-particles emitted in the first year is $\Delta N_1 = N(0) - N(1) = N_0(1 - e^{-\lambda})$.Substituting the values: $\Delta N_1 = 10^{20}(1 - 0.3) = 10^{20}(0.7) = 7 	imes 10^{19}$.

$t (h)$	$0$	$1$	$2$	$3$	$4$
$R (MBq)$	$100$	$35.36$	$12.51$	$4.42$	$1.56$

$A$ sample originally contained $10^{20}$ radioactive atoms,which emit $\alpha$-particles. The ratio of $\alpha$-particles emitted in the third year to that emitted during the second year is $0.3$. How many $\alpha$-particles were emitted in the first year?

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