Two radioactive substances $\mathrm{X}$ and $\mathrm{Y}$ 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 :
Correct Option: , 3
$\mathrm{T}_{\mathrm{x}}=\mathrm{t} ; \mathrm{T}_{\mathrm{y}}=2 \mathrm{t}$
$3 \mathrm{~T}_{\mathrm{y}}=6 \mathrm{t}$
$\mathrm{N}_{1}^{\prime}=\mathrm{N}_{2}^{\prime}$
$\mathrm{N}_{1} \mathrm{e}^{-\lambda_{1} 6 \mathrm{t}}=\mathrm{N}_{2} \mathrm{e}^{-\lambda_{2} 6 \mathrm{t}}$
$\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\mathrm{e}^{\left(\lambda_{1}-\lambda_{2}\right) 6 \mathrm{t}}=\mathrm{e}^{\ln 2\left(\frac{1}{\mathrm{t}}-\frac{1}{2 \mathrm{t}}\right) \times 6 \mathrm{t}}=\mathrm{e}^{(\ln 2) \times 3}=\mathrm{e}^{\ln 8}=8$
$\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{8}{1}$