- Home
- Standard 12
- Physics
13.Nuclei
medium
At any instant the ratio of the amount of radioactive substances is $2 : 1$. If their half lives be respectively $12$ and $16$ hours, then after two days, what will be the ratio of the substances
A
$ 1 : 1$
B
$2 : 1$
C
$1 : 2$
D
$1 : 4$
Solution
(a) Number of half lives in two days four substance $1$ and $2$ respectively are $n_1 =\frac{{2 \times 24}}{{12}} = 4$ and ${n_2} = \frac{{2 \times 24}}{{1.6}} = 3$
By using $N = {N_0}{\left( {\frac{1}{2}} \right)^n}$
==> $\frac{{{N_1}}}{{{N_2}}} = \frac{{{{({N_0})}_1}}}{{{{({N_0})}_2}}} \times \frac{{{{\left( {\frac{1}{2}} \right)}^{{n_1}}}}}{{{{\left( {\frac{1}{2}} \right)}^{{n_2}}}}}$
$ = \frac{2}{1} \times \frac{{{{\left( {\frac{1}{2}} \right)}^4}}}{{{{\left( {\frac{1}{2}} \right)}^3}}} = \frac{1}{1}$
Standard 12
Physics
