A sample originally contaived 10^ 20 radioactive atoms, which emi… - Vidyadip

MCQ

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?

A
$3\times 10^{18}$
✓
$7\times 10^{19}$
C
$5\times 10^{18}$
D
$3\times 10^{19}$

✓

Answer

Correct option: B.

$7\times 10^{19}$

b
$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 $

$\therefore N_0 - N_1 = ?$

$= N_0 - N_0e^{-\lambda} = N_0(1 - e^{-\lambda})$

$= 10^{20} (1 - 0.3) = 0.7 \times 10^{20} = 7 \times 10^{19}$

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