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A sample originally contaived $10^{20}$ 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?
$3\times 10^{18}$
$7\times 10^{19}$
$5\times 10^{18}$
$3\times 10^{19}$
Solution
$t = 0 $ $N_0 = 10^{20}$
$t = 1y$ $ N_1 = N_0e^{-\lambda(1)}$
$t = 2y$ $ N_2 =N_0e^{-\lambda(2)}$
$t = 3y$ $ N_3 = N_0e^{\lambda(3)}$
$\frac{{{N_2} – {N_3}}}{{{N_1} – {N_2}}} = 0.3\,\,\,$
$ \Rightarrow \frac{{{N_0}{e^{ – 2\lambda }} – {N_0}{e^{ – 3\lambda }}}}{{{N_0}{e^{ – \lambda }} – {N_0}{e^{ – 2\lambda }}}} = 0.3$
$\frac{{{N_0}{e^{ – 2\lambda }}(1 – {e^{ – \lambda }})}}{{{N_0}{e^{ – \lambda }}(1 – {e^{ – \lambda }})}} = 0.3$
$e^{-\lambda} = 0.3 $
$\therefore N_0 – N_1 = ?$
$= N_0 – N_0e^{-\lambda} = N_0(1 – e^{-\lambda})$
$= 10^{20} (1 – 0.3) = 0.7 \times 10^{20} = 7 \times 10^{19}$
