Two radioactive substances X and Y originally have N 1 and N 2 nuclei respectively. Half life of X is half of the half life of Y. After three half lives of Y, number of nuclei of both are equal. The ratio ( N 1/ N 2) will be equal to:

Question

Two radioactive substances X and Y originally have N 1 and N 2 nuclei respectively. Half life of X is half of the half life of Y. After three half lives of Y, number of nuclei of both are equal. The ratio ( N 1/ N 2) will be equal to: (A) (1/8) (B) (3/1) (C) (8/1) (D) (1/3)

Tardigrade Admin · Accepted Answer

T x = t ; T y =2 t 3 T y =6 t N 1 prime= N 2 prime N 1 e -λ1 6 t = N 2 e -λ2 6 t ( N 1/ N 2)= e (λ1-λ2) 6 t = e ln 2((1/t)-(1/2 t )) × 6 t = e ( In 2) × 3= e ln 8=8 ( N 1/ N 2)=(8/1)

Q. Two radioactive substances $X$ and Y originally have $N_{1}$ and $N_{2}$ nuclei respectively. Half life of $X$ is half of the half life of $Y$ . After three half lives of $Y,$ number of nuclei of both are equal. The ratio $\frac{N _{1}}{N _{2}}$ will be equal to:

$\frac{1}{8}$

$\frac{3}{1}$

$\frac{8}{1}$

$\frac{1}{3}$

$T_{x} = t; T_{y} = 2 t$
$3 T_{y} = 6 t$
$N_{1}^{'} = N_{2}^{'}$
$N_{1} e^{- λ_{1} 6 t} = N_{2} e^{- λ_{2} 6 t}$
$\frac{N _{1}}{N _{2}} = e^{(λ_{1} - λ_{2}) 6 t} = e^{l n 2 (\frac{1}{t} - \frac{1}{2 t}) \times 6 t} = e^{(I n 2) \times 3} = e^{l n 8} = 8$
$\frac{N _{1}}{N _{2}} = \frac{8}{1}$

Q. Two radioactive substances X and Y originally have N1​ and N2​ nuclei respectively. Half life of X is half of the half life of Y. After three half lives of Y, number of nuclei of both are equal. The ratio N2​N1​​ will be equal to:

81​

13​

18​

31​