147611
In the given reaction ${ }_{z} \mathrm{X}^{\mathrm{A}} \rightarrow_{\mathrm{z}-2} \mathrm{Y}^{\mathrm{A}-4} \rightarrow_{\mathrm{z}-1} \mathrm{~K}^{\mathrm{A}-4} \rightarrow_{\mathrm{z}-1} \mathrm{~K}^{\mathrm{A}-4}$ Radioactive radiations are emitted in the sequence.
1 $\alpha, \beta, \gamma$
2 $\beta, \alpha, \gamma$
3 $\alpha, \gamma, \beta$
4 $\beta, \gamma, \alpha$
Explanation:
A In first step, ${ }_{\mathrm{Z}} \mathrm{X}^{\mathrm{A}} \rightarrow{ }_{\mathrm{Z}-2} \mathrm{Y}^{\mathrm{A}-4}$ The first atomic number $Z$ decreases by 2 and mass number A decreases by 4 . So, $\alpha$-particle is emitted. In second steps - \(\mathrm{Z}-2_{-2} \mathrm{Y}^{\mathrm{A}-4} \rightarrow \mathrm{Z}-1_{1} \mathrm{~K}^{\mathrm{A}-4}\) The atomic number $(Z-2)$ increases by 1 and mass number $(\mathrm{A}-4)$ remain same. Hence, $\beta$-particle emitted In third steps - $\mathrm{Z}-1 \mathrm{~K}^{\mathrm{A}-4} \rightarrow \mathrm{Z}-1_{1} \mathrm{~K}^{\mathrm{A}-4}$ The atomic number $(Z-1)$ and mass number $(A-4)$ remain same. So, $\gamma$-particle emitted. Hence, radioactive radiation are emitted in sequence of $\alpha, \beta, \gamma$.
CG PET 2019
NUCLEAR PHYSICS
147612
For a radioactive material, half-life is $\mathbf{1 0}$ minutes. If initially there are 600 number of nuclei, the time taken (in minutes) for the disintegration of 450 nuclei is
1 30
2 10
3 20
4 15
Explanation:
C Given that, half life $\left(t_{1 / 2}\right)=10$ minute Nuclei disintegrate $=450$ Nuclei left $=600-450=150$ As we know that, $\mathrm{t}_{1 / 2}=\frac{\ln 2}{\lambda}$ $\lambda=\frac{\ln 2}{10}$ According to radioactive decay law, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $150=600 \mathrm{e}^{-(\ln 2 / 10) \mathrm{t}}$ $\frac{1}{4}=\mathrm{e}^{-(\ln 2 / 10) \mathrm{t}}$ $\mathrm{t}=20 \text { minutes }$
NEET- 2018
NUCLEAR PHYSICS
147614
The activity of a radioactive sample is measured as 9750 counts per minute at $t=0$ and as 975 counts per minute at $t=5$ minutes. The decay constant is approximately
1 0.922 per minute
2 0.691 per minute
3 0.461 per minute
4 0.230 per minute
Explanation:
C Initial activity of the sample $\left(\mathrm{N}_{\mathrm{o}}\right)=9750$ counts/min Activity after $\mathrm{t}=5 \mathrm{~min}, \mathrm{~N}=975$ counts $/ \mathrm{min}$ According to radioactivity decay law, $N=N_{o} e^{-\lambda t}$ $975=9750 e^{-5 \lambda}$ $\frac{1}{10}=\mathrm{e}^{-5 \lambda}$ Taking $\log$ both side, $-\ln 10=-5 \lambda$ $2.303=5 \lambda$ $\lambda=0.461 \mathrm{~min}^{-1}$
VITEEE-2018
NUCLEAR PHYSICS
147615
A radioactive sample $S_{1}$ having activity $A_{1}$ has twice the number of nuclei as another sample $S_{2}$ activity $A_{2}$. If $A_{2}=2 A_{1}$, then ratio of half-life of $S_{1}$ to the half-life of $S_{2}$ is-
147611
In the given reaction ${ }_{z} \mathrm{X}^{\mathrm{A}} \rightarrow_{\mathrm{z}-2} \mathrm{Y}^{\mathrm{A}-4} \rightarrow_{\mathrm{z}-1} \mathrm{~K}^{\mathrm{A}-4} \rightarrow_{\mathrm{z}-1} \mathrm{~K}^{\mathrm{A}-4}$ Radioactive radiations are emitted in the sequence.
1 $\alpha, \beta, \gamma$
2 $\beta, \alpha, \gamma$
3 $\alpha, \gamma, \beta$
4 $\beta, \gamma, \alpha$
Explanation:
A In first step, ${ }_{\mathrm{Z}} \mathrm{X}^{\mathrm{A}} \rightarrow{ }_{\mathrm{Z}-2} \mathrm{Y}^{\mathrm{A}-4}$ The first atomic number $Z$ decreases by 2 and mass number A decreases by 4 . So, $\alpha$-particle is emitted. In second steps - \(\mathrm{Z}-2_{-2} \mathrm{Y}^{\mathrm{A}-4} \rightarrow \mathrm{Z}-1_{1} \mathrm{~K}^{\mathrm{A}-4}\) The atomic number $(Z-2)$ increases by 1 and mass number $(\mathrm{A}-4)$ remain same. Hence, $\beta$-particle emitted In third steps - $\mathrm{Z}-1 \mathrm{~K}^{\mathrm{A}-4} \rightarrow \mathrm{Z}-1_{1} \mathrm{~K}^{\mathrm{A}-4}$ The atomic number $(Z-1)$ and mass number $(A-4)$ remain same. So, $\gamma$-particle emitted. Hence, radioactive radiation are emitted in sequence of $\alpha, \beta, \gamma$.
CG PET 2019
NUCLEAR PHYSICS
147612
For a radioactive material, half-life is $\mathbf{1 0}$ minutes. If initially there are 600 number of nuclei, the time taken (in minutes) for the disintegration of 450 nuclei is
1 30
2 10
3 20
4 15
Explanation:
C Given that, half life $\left(t_{1 / 2}\right)=10$ minute Nuclei disintegrate $=450$ Nuclei left $=600-450=150$ As we know that, $\mathrm{t}_{1 / 2}=\frac{\ln 2}{\lambda}$ $\lambda=\frac{\ln 2}{10}$ According to radioactive decay law, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $150=600 \mathrm{e}^{-(\ln 2 / 10) \mathrm{t}}$ $\frac{1}{4}=\mathrm{e}^{-(\ln 2 / 10) \mathrm{t}}$ $\mathrm{t}=20 \text { minutes }$
NEET- 2018
NUCLEAR PHYSICS
147614
The activity of a radioactive sample is measured as 9750 counts per minute at $t=0$ and as 975 counts per minute at $t=5$ minutes. The decay constant is approximately
1 0.922 per minute
2 0.691 per minute
3 0.461 per minute
4 0.230 per minute
Explanation:
C Initial activity of the sample $\left(\mathrm{N}_{\mathrm{o}}\right)=9750$ counts/min Activity after $\mathrm{t}=5 \mathrm{~min}, \mathrm{~N}=975$ counts $/ \mathrm{min}$ According to radioactivity decay law, $N=N_{o} e^{-\lambda t}$ $975=9750 e^{-5 \lambda}$ $\frac{1}{10}=\mathrm{e}^{-5 \lambda}$ Taking $\log$ both side, $-\ln 10=-5 \lambda$ $2.303=5 \lambda$ $\lambda=0.461 \mathrm{~min}^{-1}$
VITEEE-2018
NUCLEAR PHYSICS
147615
A radioactive sample $S_{1}$ having activity $A_{1}$ has twice the number of nuclei as another sample $S_{2}$ activity $A_{2}$. If $A_{2}=2 A_{1}$, then ratio of half-life of $S_{1}$ to the half-life of $S_{2}$ is-
147611
In the given reaction ${ }_{z} \mathrm{X}^{\mathrm{A}} \rightarrow_{\mathrm{z}-2} \mathrm{Y}^{\mathrm{A}-4} \rightarrow_{\mathrm{z}-1} \mathrm{~K}^{\mathrm{A}-4} \rightarrow_{\mathrm{z}-1} \mathrm{~K}^{\mathrm{A}-4}$ Radioactive radiations are emitted in the sequence.
1 $\alpha, \beta, \gamma$
2 $\beta, \alpha, \gamma$
3 $\alpha, \gamma, \beta$
4 $\beta, \gamma, \alpha$
Explanation:
A In first step, ${ }_{\mathrm{Z}} \mathrm{X}^{\mathrm{A}} \rightarrow{ }_{\mathrm{Z}-2} \mathrm{Y}^{\mathrm{A}-4}$ The first atomic number $Z$ decreases by 2 and mass number A decreases by 4 . So, $\alpha$-particle is emitted. In second steps - \(\mathrm{Z}-2_{-2} \mathrm{Y}^{\mathrm{A}-4} \rightarrow \mathrm{Z}-1_{1} \mathrm{~K}^{\mathrm{A}-4}\) The atomic number $(Z-2)$ increases by 1 and mass number $(\mathrm{A}-4)$ remain same. Hence, $\beta$-particle emitted In third steps - $\mathrm{Z}-1 \mathrm{~K}^{\mathrm{A}-4} \rightarrow \mathrm{Z}-1_{1} \mathrm{~K}^{\mathrm{A}-4}$ The atomic number $(Z-1)$ and mass number $(A-4)$ remain same. So, $\gamma$-particle emitted. Hence, radioactive radiation are emitted in sequence of $\alpha, \beta, \gamma$.
CG PET 2019
NUCLEAR PHYSICS
147612
For a radioactive material, half-life is $\mathbf{1 0}$ minutes. If initially there are 600 number of nuclei, the time taken (in minutes) for the disintegration of 450 nuclei is
1 30
2 10
3 20
4 15
Explanation:
C Given that, half life $\left(t_{1 / 2}\right)=10$ minute Nuclei disintegrate $=450$ Nuclei left $=600-450=150$ As we know that, $\mathrm{t}_{1 / 2}=\frac{\ln 2}{\lambda}$ $\lambda=\frac{\ln 2}{10}$ According to radioactive decay law, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $150=600 \mathrm{e}^{-(\ln 2 / 10) \mathrm{t}}$ $\frac{1}{4}=\mathrm{e}^{-(\ln 2 / 10) \mathrm{t}}$ $\mathrm{t}=20 \text { minutes }$
NEET- 2018
NUCLEAR PHYSICS
147614
The activity of a radioactive sample is measured as 9750 counts per minute at $t=0$ and as 975 counts per minute at $t=5$ minutes. The decay constant is approximately
1 0.922 per minute
2 0.691 per minute
3 0.461 per minute
4 0.230 per minute
Explanation:
C Initial activity of the sample $\left(\mathrm{N}_{\mathrm{o}}\right)=9750$ counts/min Activity after $\mathrm{t}=5 \mathrm{~min}, \mathrm{~N}=975$ counts $/ \mathrm{min}$ According to radioactivity decay law, $N=N_{o} e^{-\lambda t}$ $975=9750 e^{-5 \lambda}$ $\frac{1}{10}=\mathrm{e}^{-5 \lambda}$ Taking $\log$ both side, $-\ln 10=-5 \lambda$ $2.303=5 \lambda$ $\lambda=0.461 \mathrm{~min}^{-1}$
VITEEE-2018
NUCLEAR PHYSICS
147615
A radioactive sample $S_{1}$ having activity $A_{1}$ has twice the number of nuclei as another sample $S_{2}$ activity $A_{2}$. If $A_{2}=2 A_{1}$, then ratio of half-life of $S_{1}$ to the half-life of $S_{2}$ is-
147611
In the given reaction ${ }_{z} \mathrm{X}^{\mathrm{A}} \rightarrow_{\mathrm{z}-2} \mathrm{Y}^{\mathrm{A}-4} \rightarrow_{\mathrm{z}-1} \mathrm{~K}^{\mathrm{A}-4} \rightarrow_{\mathrm{z}-1} \mathrm{~K}^{\mathrm{A}-4}$ Radioactive radiations are emitted in the sequence.
1 $\alpha, \beta, \gamma$
2 $\beta, \alpha, \gamma$
3 $\alpha, \gamma, \beta$
4 $\beta, \gamma, \alpha$
Explanation:
A In first step, ${ }_{\mathrm{Z}} \mathrm{X}^{\mathrm{A}} \rightarrow{ }_{\mathrm{Z}-2} \mathrm{Y}^{\mathrm{A}-4}$ The first atomic number $Z$ decreases by 2 and mass number A decreases by 4 . So, $\alpha$-particle is emitted. In second steps - \(\mathrm{Z}-2_{-2} \mathrm{Y}^{\mathrm{A}-4} \rightarrow \mathrm{Z}-1_{1} \mathrm{~K}^{\mathrm{A}-4}\) The atomic number $(Z-2)$ increases by 1 and mass number $(\mathrm{A}-4)$ remain same. Hence, $\beta$-particle emitted In third steps - $\mathrm{Z}-1 \mathrm{~K}^{\mathrm{A}-4} \rightarrow \mathrm{Z}-1_{1} \mathrm{~K}^{\mathrm{A}-4}$ The atomic number $(Z-1)$ and mass number $(A-4)$ remain same. So, $\gamma$-particle emitted. Hence, radioactive radiation are emitted in sequence of $\alpha, \beta, \gamma$.
CG PET 2019
NUCLEAR PHYSICS
147612
For a radioactive material, half-life is $\mathbf{1 0}$ minutes. If initially there are 600 number of nuclei, the time taken (in minutes) for the disintegration of 450 nuclei is
1 30
2 10
3 20
4 15
Explanation:
C Given that, half life $\left(t_{1 / 2}\right)=10$ minute Nuclei disintegrate $=450$ Nuclei left $=600-450=150$ As we know that, $\mathrm{t}_{1 / 2}=\frac{\ln 2}{\lambda}$ $\lambda=\frac{\ln 2}{10}$ According to radioactive decay law, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $150=600 \mathrm{e}^{-(\ln 2 / 10) \mathrm{t}}$ $\frac{1}{4}=\mathrm{e}^{-(\ln 2 / 10) \mathrm{t}}$ $\mathrm{t}=20 \text { minutes }$
NEET- 2018
NUCLEAR PHYSICS
147614
The activity of a radioactive sample is measured as 9750 counts per minute at $t=0$ and as 975 counts per minute at $t=5$ minutes. The decay constant is approximately
1 0.922 per minute
2 0.691 per minute
3 0.461 per minute
4 0.230 per minute
Explanation:
C Initial activity of the sample $\left(\mathrm{N}_{\mathrm{o}}\right)=9750$ counts/min Activity after $\mathrm{t}=5 \mathrm{~min}, \mathrm{~N}=975$ counts $/ \mathrm{min}$ According to radioactivity decay law, $N=N_{o} e^{-\lambda t}$ $975=9750 e^{-5 \lambda}$ $\frac{1}{10}=\mathrm{e}^{-5 \lambda}$ Taking $\log$ both side, $-\ln 10=-5 \lambda$ $2.303=5 \lambda$ $\lambda=0.461 \mathrm{~min}^{-1}$
VITEEE-2018
NUCLEAR PHYSICS
147615
A radioactive sample $S_{1}$ having activity $A_{1}$ has twice the number of nuclei as another sample $S_{2}$ activity $A_{2}$. If $A_{2}=2 A_{1}$, then ratio of half-life of $S_{1}$ to the half-life of $S_{2}$ is-
