(A) The number of undecayed nuclei at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.For sample $A_1$,$N_1 = N_0 e^{-(10\lambda_0)t}$.For sample $A_

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

(A) The number of undecayed nuclei at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.For sample $A_1$,$N_1 = N_0 e^{-(10\lambda_0)t}$.For sample $A_2$,$N_2 = N_0 e^{-\lambda_0 t}$.The ratio of undecayed nuclei is $N_1/N_2 = \frac{N_0 e^{-10\lambda_0 t}}{N_0 e^{-\lambda_0 t}} = e^{-9\lambda_0 t}$.Given $t = 1/(9\lambda_0)$,substitute this value into the ratio:$N_1/N_2 = e^{-9\lambda_0 (1/9\lambda_0)} = e^{-1} = 1/e$.

Class 12 Physics Nuclei Law of Radioactivity by Rutherford and Soddy and Half Life and Mean Life

Difficult

Two radioactive samples $A_1$ and $A_2$ have decay constants $10\lambda_0$ and $\lambda_0$ respectively. If they initially contain the same number of nuclei,find the ratio of the number of undecayed nuclei after time $t = 1/(9\lambda_0)$.

A
$1/e$
B
$1/e^2$
C
$1/e^9$
D
$1/e^{10}$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Physics Questions

Chapter

Nuclei

Subtopic

Law of Radioactivity by Rutherford and Soddy and Half Life and Mean Life

$A$ sample has a half-life of $10^{33}$ years. If the initial number of nuclei of the sample is $26 \times 10^{24}$,then the number of nuclei decayed in $1$ year is ........... $\times 10^{-7}$.

MediumAIIMS 2019

View Solution

For a substance,the fraction of its initial quantity $(N_0)$ which will disintegrate in its average lifetime is about $(e = 2.71)$.

MediumMHT CET 2022

View Solution

The activity of a radioactive sample is measured as $N_0$ counts per minute at $t = 0$ and $N_0/e$ counts per minute at $t = 5\, \text{minutes}$. The time (in minutes) at which the activity reduces to half of its initial value is

Difficult

View Solution

Obtain the amount of $_{27}^{60} Co$ necessary to provide a radioactive source of $8.0 \; mCi$ strength. The half-life of $_{27}^{60} Co$ is $5.3$ years.

Medium

View Solution

An active nucleus decays to one-third $\left(\frac{1}{3}\right)$ of its initial activity in $20 \text{ hours}$. The fraction of original activity remaining after $80 \text{ hours}$ is:

MediumTS EAMCET 2018

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

Two radioactive samples $A_1$ and $A_2$ have decay constants $10\lambda_0$ and $\lambda_0$ respectively. If they initially contain the same number of nuclei,find the ratio of the number of undecayed nuclei after time $t = 1/(9\lambda_0)$.

Similar Questions

$A$ sample has a half-life of $10^{33}$ years. If the initial number of nuclei of the sample is $26 \times 10^{24}$,then the number of nuclei decayed in $1$ year is ........... $\times 10^{-7}$.

For a substance,the fraction of its initial quantity $(N_0)$ which will disintegrate in its average lifetime is about $(e = 2.71)$.

The activity of a radioactive sample is measured as $N_0$ counts per minute at $t = 0$ and $N_0/e$ counts per minute at $t = 5\, \text{minutes}$. The time (in minutes) at which the activity reduces to half of its initial value is

Obtain the amount of $_{27}^{60} Co$ necessary to provide a radioactive source of $8.0 \; mCi$ strength. The half-life of $_{27}^{60} Co$ is $5.3$ years.

An active nucleus decays to one-third $\left(\frac{1}{3}\right)$ of its initial activity in $20 \text{ hours}$. The fraction of original activity remaining after $80 \text{ hours}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module