The ratio of the amount of two elements X and | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) At radioactive equilibrium,the rate of decay of $X$ is equal to the rate of decay of $Y$. Thus,$\lambda_X N_X = \lambda_Y N_Y$. This implies $\fra

Vedclass · Accepted Answer

$245$

Vedclass · Answer

(B) At radioactive equilibrium,the rate of decay of $X$ is equal to the rate of decay of $Y$. Thus,$\lambda_X N_X = \lambda_Y N_Y$. This implies $\frac{N_X}{N_Y} = \frac{\lambda_Y}{\lambda_X} = \frac{t_{1/2}(X)}{t_{1/2}(Y)}$. Given $\frac{N_X}{N_Y} = 1 : 2 	imes 10^{-6}$ and $t_{1/2}(Y) = 4.9 	imes 10^{-4} \ days$. Therefore,$t_{1/2}(X) = \frac{N_X}{N_Y} 	imes t_{1/2}(Y) = \frac{1}{2 	imes 10^{-6}} 	imes 4.9 	imes 10^{-4} \ days$. $t_{1/2}(X) = 0.5 	imes 10^6 	imes 4.9 	imes 10^{-4} \ days = 245 \ days$.

The ratio of the amount of two elements $X$ and $Y$ at radioactive equilibrium is $1 : 2 \times 10^{-6}$. If the half-life period of element $Y$ is $4.9 \times 10^{-4} \ days$,then the half-life period of element $X$ will be .......... $days$.

Similar Questions

The Carbon-$14$ dating method is based on the fact that:

The half-life of $^{14}C$ is about $.........$ $years$.

The $T_{1/2}$ of $C^{14}$ isotope is $5770$ years. The time after which $72\%$ of the isotope is left is .......... years.

The average life period of a radioactive element is the reciprocal of its

The rate of radioactive disintegration at an instant for a radioactive sample of half-life $2.2 \times 10^{9} \ s$ is $10^{10} \ s^{-1}$. The number of radioactive atoms in that sample at that instant is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module