The radioactive materials x_1 and x_2 have de | vedclass

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

Question

(D) Let $N_0$ be the initial number of nuclei for both materials.The number of nuclei remaining at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.Fo

Vedclass · Accepted Answer

$\frac{1}{9 \lambda}$

Vedclass · Answer

(D) Let $N_0$ be the initial number of nuclei for both materials.The number of nuclei remaining at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.For material $x_1$,$N_1(t) = N_0 e^{-(10\lambda)t}$.For material $x_2$,$N_2(t) = N_0 e^{-\lambda t}$.Given the ratio $\frac{N_1}{N_2} = \frac{1}{e}$,we have:$\frac{N_0 e^{-10\lambda t}}{N_0 e^{-\lambda t}} = e^{-1}$$e^{-10\lambda t + \lambda t} = e^{-1}$$e^{-9\lambda t} = e^{-1}$Comparing the exponents: $-9\lambda t = -1$Therefore,$t = \frac{1}{9\lambda}$.

The radioactive materials $x_1$ and $x_2$ have decay constants $10 \lambda$ and $\lambda$ respectively. If initially they have the same number of nuclei,then the ratio of the number of nuclei of $x_1$ to that of $x_2$ will be $1/e$ after a time $t$ equal to:

Similar Questions

The activity of a radioactive substance can be represented by various units. Select the correct option.

The decay constant of a radioactive element is $0.01 \ s^{-1}$. Its half-life period is ....... $s$.

Half lives of two radioactive substances $A$ and $B$ are respectively $20$ minutes and $40$ minutes. Initially,the sample of $A$ and $B$ have equal number of nuclei. After $80$ minutes,the ratio of the remaining numbers of $A$ and $B$ nuclei is

For a radioactive element,the mean life is $\tau$. At time $t = 0$,the number of nuclei decaying per unit time is $n$. The number of nuclei that have decayed between time $0$ and $t$ is:

The activity of a radioactive sample falls from $700 \; s^{-1}$ to $500 \; s^{-1}$ in $30 \; min$. Its half-life is close to ......... $min$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module