147846
An archaeologist analyses the wood in a prehistoric structure and finds that $\mathrm{C}^{14}$ (Half life $=5700$ years) to $C^{12}$ is only one-forth of that found in the cells of buried plants. The age of the wood is about
1 5700 years
2 2850 years
3 11,400 years
4 22, 800 years
Explanation:
C Given that, half-life $\left(\mathrm{T}_{1 / 2}\right)=5700$ year According to question - $\frac{\mathrm{C}_{14}}{\mathrm{C}_{12}}=\frac{1}{4}=\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{T}_{1 / 2}}$ $\frac{1}{4}=\left(\frac{1}{2}\right)^{\mathrm{t} / 5700}$ $\left(\frac{1}{2}\right)^{2}=\left(\frac{1}{2}\right)^{\mathrm{t} / 5700}$ On comparing both side, we get- $\frac{t}{5700}=2$ $t=2 \times 5700$ $t=11,400 \text { years }$
AIIMS-2013
NUCLEAR PHYSICS
147848
A nucleus ${ }_{\mathrm{Z}} \mathrm{X}^{\mathrm{A}}$ emits an $\alpha$-particle with velocity $v$. The recoil speed of the daughter nucleus is :
1 $\frac{\mathrm{A}-4}{4 \mathrm{v}}$
2 $\frac{4 \mathrm{~V}}{\mathrm{~A}-4}$
3 $\mathrm{V}$
4 $\frac{\mathrm{V}}{4}$
Explanation:
B According to the question, $\mathrm{Z}^{\mathrm{A}} \stackrel{\alpha}{\longrightarrow}{ }_{\mathrm{Z}-2} \mathrm{Y}^{\mathrm{A}-4}+{ }_{2} \mathrm{He}^{4}$ Let the recoil speed of daughter nucleus is $v^{\prime}$. By the law of conservation of momentum - Initial momentum $\left(\mathrm{P}_{\mathrm{i}}\right)=$ Final momentum $\left(\mathrm{P}_{\mathrm{f}}\right)$ $0=(A-4) v^{\prime}-4 v$ $(A-4) v^{\prime}=4 v$ $v^{\prime}=\frac{4 v}{A-4}$
Karnataka CET-2013
NUCLEAR PHYSICS
147849
A radioactive nuclide is produced at the constant rate of $n$ per second (say, by bombarding a target with neutrons). The expected number $N$ of nuclei in existence $t$ seconds after the number is $N_{0}$ is given by
4 $\mathrm{N}=\frac{\mathrm{n}}{\lambda}+\left(\mathrm{N}_{0}+\frac{\mathrm{n}}{\lambda}\right) \mathrm{e}^{-\lambda \mathrm{t}}$ Where $\lambda$ is the decay constant of sample.
147850
The fossil bone has a ${ }^{14} \mathrm{C}:{ }^{12} \mathrm{C}$ ratio, which is $\left[\frac{1}{16}\right]$ of that in a living animal bone. If the halflife of ${ }^{14} \mathrm{C}$ is 5730 years, then the age of the fossil bone is
1 11460 years
2 17190 years
3 22920 years
4 45840 years
Explanation:
C Given that, half-life $\left(\mathrm{T}_{1 / 2}\right)=5730$ year According to question - $\frac{{ }^{14} \mathrm{C}}{{ }^{12} \mathrm{C}}=\frac{1}{16}=\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{T}_{\mathrm{I} / 2}}$ $\frac{1}{16}=\left(\frac{1}{2}\right)^{\mathrm{t} / 5730}$ $\left(\frac{1}{2}\right)^{4}=\left(\frac{1}{2}\right)^{\mathrm{t} / 5730}$ On comparing both side, we get - $\frac{t}{5730}=4$ $t=4 \times 5730$ $t=22920 \text { years }$
147846
An archaeologist analyses the wood in a prehistoric structure and finds that $\mathrm{C}^{14}$ (Half life $=5700$ years) to $C^{12}$ is only one-forth of that found in the cells of buried plants. The age of the wood is about
1 5700 years
2 2850 years
3 11,400 years
4 22, 800 years
Explanation:
C Given that, half-life $\left(\mathrm{T}_{1 / 2}\right)=5700$ year According to question - $\frac{\mathrm{C}_{14}}{\mathrm{C}_{12}}=\frac{1}{4}=\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{T}_{1 / 2}}$ $\frac{1}{4}=\left(\frac{1}{2}\right)^{\mathrm{t} / 5700}$ $\left(\frac{1}{2}\right)^{2}=\left(\frac{1}{2}\right)^{\mathrm{t} / 5700}$ On comparing both side, we get- $\frac{t}{5700}=2$ $t=2 \times 5700$ $t=11,400 \text { years }$
AIIMS-2013
NUCLEAR PHYSICS
147848
A nucleus ${ }_{\mathrm{Z}} \mathrm{X}^{\mathrm{A}}$ emits an $\alpha$-particle with velocity $v$. The recoil speed of the daughter nucleus is :
1 $\frac{\mathrm{A}-4}{4 \mathrm{v}}$
2 $\frac{4 \mathrm{~V}}{\mathrm{~A}-4}$
3 $\mathrm{V}$
4 $\frac{\mathrm{V}}{4}$
Explanation:
B According to the question, $\mathrm{Z}^{\mathrm{A}} \stackrel{\alpha}{\longrightarrow}{ }_{\mathrm{Z}-2} \mathrm{Y}^{\mathrm{A}-4}+{ }_{2} \mathrm{He}^{4}$ Let the recoil speed of daughter nucleus is $v^{\prime}$. By the law of conservation of momentum - Initial momentum $\left(\mathrm{P}_{\mathrm{i}}\right)=$ Final momentum $\left(\mathrm{P}_{\mathrm{f}}\right)$ $0=(A-4) v^{\prime}-4 v$ $(A-4) v^{\prime}=4 v$ $v^{\prime}=\frac{4 v}{A-4}$
Karnataka CET-2013
NUCLEAR PHYSICS
147849
A radioactive nuclide is produced at the constant rate of $n$ per second (say, by bombarding a target with neutrons). The expected number $N$ of nuclei in existence $t$ seconds after the number is $N_{0}$ is given by
4 $\mathrm{N}=\frac{\mathrm{n}}{\lambda}+\left(\mathrm{N}_{0}+\frac{\mathrm{n}}{\lambda}\right) \mathrm{e}^{-\lambda \mathrm{t}}$ Where $\lambda$ is the decay constant of sample.
147850
The fossil bone has a ${ }^{14} \mathrm{C}:{ }^{12} \mathrm{C}$ ratio, which is $\left[\frac{1}{16}\right]$ of that in a living animal bone. If the halflife of ${ }^{14} \mathrm{C}$ is 5730 years, then the age of the fossil bone is
1 11460 years
2 17190 years
3 22920 years
4 45840 years
Explanation:
C Given that, half-life $\left(\mathrm{T}_{1 / 2}\right)=5730$ year According to question - $\frac{{ }^{14} \mathrm{C}}{{ }^{12} \mathrm{C}}=\frac{1}{16}=\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{T}_{\mathrm{I} / 2}}$ $\frac{1}{16}=\left(\frac{1}{2}\right)^{\mathrm{t} / 5730}$ $\left(\frac{1}{2}\right)^{4}=\left(\frac{1}{2}\right)^{\mathrm{t} / 5730}$ On comparing both side, we get - $\frac{t}{5730}=4$ $t=4 \times 5730$ $t=22920 \text { years }$
147846
An archaeologist analyses the wood in a prehistoric structure and finds that $\mathrm{C}^{14}$ (Half life $=5700$ years) to $C^{12}$ is only one-forth of that found in the cells of buried plants. The age of the wood is about
1 5700 years
2 2850 years
3 11,400 years
4 22, 800 years
Explanation:
C Given that, half-life $\left(\mathrm{T}_{1 / 2}\right)=5700$ year According to question - $\frac{\mathrm{C}_{14}}{\mathrm{C}_{12}}=\frac{1}{4}=\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{T}_{1 / 2}}$ $\frac{1}{4}=\left(\frac{1}{2}\right)^{\mathrm{t} / 5700}$ $\left(\frac{1}{2}\right)^{2}=\left(\frac{1}{2}\right)^{\mathrm{t} / 5700}$ On comparing both side, we get- $\frac{t}{5700}=2$ $t=2 \times 5700$ $t=11,400 \text { years }$
AIIMS-2013
NUCLEAR PHYSICS
147848
A nucleus ${ }_{\mathrm{Z}} \mathrm{X}^{\mathrm{A}}$ emits an $\alpha$-particle with velocity $v$. The recoil speed of the daughter nucleus is :
1 $\frac{\mathrm{A}-4}{4 \mathrm{v}}$
2 $\frac{4 \mathrm{~V}}{\mathrm{~A}-4}$
3 $\mathrm{V}$
4 $\frac{\mathrm{V}}{4}$
Explanation:
B According to the question, $\mathrm{Z}^{\mathrm{A}} \stackrel{\alpha}{\longrightarrow}{ }_{\mathrm{Z}-2} \mathrm{Y}^{\mathrm{A}-4}+{ }_{2} \mathrm{He}^{4}$ Let the recoil speed of daughter nucleus is $v^{\prime}$. By the law of conservation of momentum - Initial momentum $\left(\mathrm{P}_{\mathrm{i}}\right)=$ Final momentum $\left(\mathrm{P}_{\mathrm{f}}\right)$ $0=(A-4) v^{\prime}-4 v$ $(A-4) v^{\prime}=4 v$ $v^{\prime}=\frac{4 v}{A-4}$
Karnataka CET-2013
NUCLEAR PHYSICS
147849
A radioactive nuclide is produced at the constant rate of $n$ per second (say, by bombarding a target with neutrons). The expected number $N$ of nuclei in existence $t$ seconds after the number is $N_{0}$ is given by
4 $\mathrm{N}=\frac{\mathrm{n}}{\lambda}+\left(\mathrm{N}_{0}+\frac{\mathrm{n}}{\lambda}\right) \mathrm{e}^{-\lambda \mathrm{t}}$ Where $\lambda$ is the decay constant of sample.
147850
The fossil bone has a ${ }^{14} \mathrm{C}:{ }^{12} \mathrm{C}$ ratio, which is $\left[\frac{1}{16}\right]$ of that in a living animal bone. If the halflife of ${ }^{14} \mathrm{C}$ is 5730 years, then the age of the fossil bone is
1 11460 years
2 17190 years
3 22920 years
4 45840 years
Explanation:
C Given that, half-life $\left(\mathrm{T}_{1 / 2}\right)=5730$ year According to question - $\frac{{ }^{14} \mathrm{C}}{{ }^{12} \mathrm{C}}=\frac{1}{16}=\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{T}_{\mathrm{I} / 2}}$ $\frac{1}{16}=\left(\frac{1}{2}\right)^{\mathrm{t} / 5730}$ $\left(\frac{1}{2}\right)^{4}=\left(\frac{1}{2}\right)^{\mathrm{t} / 5730}$ On comparing both side, we get - $\frac{t}{5730}=4$ $t=4 \times 5730$ $t=22920 \text { years }$
147846
An archaeologist analyses the wood in a prehistoric structure and finds that $\mathrm{C}^{14}$ (Half life $=5700$ years) to $C^{12}$ is only one-forth of that found in the cells of buried plants. The age of the wood is about
1 5700 years
2 2850 years
3 11,400 years
4 22, 800 years
Explanation:
C Given that, half-life $\left(\mathrm{T}_{1 / 2}\right)=5700$ year According to question - $\frac{\mathrm{C}_{14}}{\mathrm{C}_{12}}=\frac{1}{4}=\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{T}_{1 / 2}}$ $\frac{1}{4}=\left(\frac{1}{2}\right)^{\mathrm{t} / 5700}$ $\left(\frac{1}{2}\right)^{2}=\left(\frac{1}{2}\right)^{\mathrm{t} / 5700}$ On comparing both side, we get- $\frac{t}{5700}=2$ $t=2 \times 5700$ $t=11,400 \text { years }$
AIIMS-2013
NUCLEAR PHYSICS
147848
A nucleus ${ }_{\mathrm{Z}} \mathrm{X}^{\mathrm{A}}$ emits an $\alpha$-particle with velocity $v$. The recoil speed of the daughter nucleus is :
1 $\frac{\mathrm{A}-4}{4 \mathrm{v}}$
2 $\frac{4 \mathrm{~V}}{\mathrm{~A}-4}$
3 $\mathrm{V}$
4 $\frac{\mathrm{V}}{4}$
Explanation:
B According to the question, $\mathrm{Z}^{\mathrm{A}} \stackrel{\alpha}{\longrightarrow}{ }_{\mathrm{Z}-2} \mathrm{Y}^{\mathrm{A}-4}+{ }_{2} \mathrm{He}^{4}$ Let the recoil speed of daughter nucleus is $v^{\prime}$. By the law of conservation of momentum - Initial momentum $\left(\mathrm{P}_{\mathrm{i}}\right)=$ Final momentum $\left(\mathrm{P}_{\mathrm{f}}\right)$ $0=(A-4) v^{\prime}-4 v$ $(A-4) v^{\prime}=4 v$ $v^{\prime}=\frac{4 v}{A-4}$
Karnataka CET-2013
NUCLEAR PHYSICS
147849
A radioactive nuclide is produced at the constant rate of $n$ per second (say, by bombarding a target with neutrons). The expected number $N$ of nuclei in existence $t$ seconds after the number is $N_{0}$ is given by
4 $\mathrm{N}=\frac{\mathrm{n}}{\lambda}+\left(\mathrm{N}_{0}+\frac{\mathrm{n}}{\lambda}\right) \mathrm{e}^{-\lambda \mathrm{t}}$ Where $\lambda$ is the decay constant of sample.
147850
The fossil bone has a ${ }^{14} \mathrm{C}:{ }^{12} \mathrm{C}$ ratio, which is $\left[\frac{1}{16}\right]$ of that in a living animal bone. If the halflife of ${ }^{14} \mathrm{C}$ is 5730 years, then the age of the fossil bone is
1 11460 years
2 17190 years
3 22920 years
4 45840 years
Explanation:
C Given that, half-life $\left(\mathrm{T}_{1 / 2}\right)=5730$ year According to question - $\frac{{ }^{14} \mathrm{C}}{{ }^{12} \mathrm{C}}=\frac{1}{16}=\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{T}_{\mathrm{I} / 2}}$ $\frac{1}{16}=\left(\frac{1}{2}\right)^{\mathrm{t} / 5730}$ $\left(\frac{1}{2}\right)^{4}=\left(\frac{1}{2}\right)^{\mathrm{t} / 5730}$ On comparing both side, we get - $\frac{t}{5730}=4$ $t=4 \times 5730$ $t=22920 \text { years }$
