(C) The ratio of ${}^{14}C$ to ${}^{12}C$ in the fossil bone is given as $\frac{N}{N_0} = \frac{1}{16}$.The law of radioactive decay states that $\fra

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

(C) The ratio of ${}^{14}C$ to ${}^{12}C$ in the fossil bone is given as $\frac{N}{N_0} = \frac{1}{16}$.The law of radioactive decay states that $\frac{N}{N_0} = \left(\frac{1}{2}\right)^n$,where $n$ is the number of half-lives.Equating the two,we get $\left(\frac{1}{2}\right)^n = \frac{1}{16} = \left(\frac{1}{2}\right)^4$.Therefore,the number of half-lives $n = 4$.The age of the fossil $t$ is given by $t = n \times T_{1/2}$,where $T_{1/2} = 5730 \, years$.$t = 4 \times 5730 = 22920 \, years$.

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

Medium

The fossil bone has a ${}^{14}C:{}^{12}C$ ratio,which is $\frac{1}{16}$ of that in a living animal bone. If the half-life of ${}^{14}C$ is $5730 \, years$,then the age of the fossil bone is .......... $years$.

AIIMS 2006 All AIIMS

A
$11460$
B
$17190$
C
$22920$
D
$45840$

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

Previous Year

AIIMS 2006 Paper All AIIMS years →

The half-life of a radioactive substance is $5$ years. After $x$ years,a given sample of the radioactive substance gets reduced to $6.25 \%$ of its initial value. The value of $x$ is ...............

MediumJEE MAIN 2022

View Solution

Activity of a radioactive substance is $R_1$ at time $t_1$ and $R_2$ at time $t_2$ $(t_2 > t_1)$. Then the ratio $\frac{R_2}{R_1}$ is:

View Solution

$C^{14}$ has a half-life of $5700$ years. At the end of $11400$ years,the actual amount left is:

Easy

View Solution

The decay constant for the nuclide $^{64}Cu$ is $1.6 \times 10^{-5} \, s^{-1}$. Find the activity of a $1 \, mg$ sample of $^{64}Cu$ in $Ci$. (Given: Atomic mass of Copper = $64 \, g/mol$,$1 \, Ci = 3.7 \times 10^{10} \, Bq$)

Medium

View Solution

Two radioactive substances $A$ and $B$ have decay constants $5\lambda$ and $\lambda$ respectively. At $t = 0$,a sample has the same number of the two nuclei. The time taken for the ratio of the number of nuclei to become $(1/e)^2$ will be

MediumJEE MAIN 2019

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

The fossil bone has a ${}^{14}C:{}^{12}C$ ratio,which is $\frac{1}{16}$ of that in a living animal bone. If the half-life of ${}^{14}C$ is $5730 \, years$,then the age of the fossil bone is .......... $years$.

Similar Questions

The half-life of a radioactive substance is $5$ years. After $x$ years,a given sample of the radioactive substance gets reduced to $6.25 \%$ of its initial value. The value of $x$ is ...............

Activity of a radioactive substance is $R_1$ at time $t_1$ and $R_2$ at time $t_2$ $(t_2 > t_1)$. Then the ratio $\frac{R_2}{R_1}$ is:

$C^{14}$ has a half-life of $5700$ years. At the end of $11400$ years,the actual amount left is:

The decay constant for the nuclide $^{64}Cu$ is $1.6 \times 10^{-5} \, s^{-1}$. Find the activity of a $1 \, mg$ sample of $^{64}Cu$ in $Ci$. (Given: Atomic mass of Copper = $64 \, g/mol$,$1 \, Ci = 3.7 \times 10^{10} \, Bq$)

Two radioactive substances $A$ and $B$ have decay constants $5\lambda$ and $\lambda$ respectively. At $t = 0$,a sample has the same number of the two nuclei. The time taken for the ratio of the number of nuclei to become $(1/e)^2$ will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module