147797
The half-life of a radioactive nuclide is $100 \mathrm{~h}$. The fraction of original activity that will remain after $150 \mathrm{~h}$ would be
1 $\frac{1}{2}$
2 $\frac{1}{2 \sqrt{2}}$
3 $\frac{2}{3}$
4 $\frac{2}{\sqrt{3} 2}$
Explanation:
B Given that, Half-life of radioactive nuclide $\left(\mathrm{T}_{1 / 2}\right)=100 \mathrm{~h}$ Total time taken $(t)=150 \mathrm{~h}$ We know that, $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{\frac{150}{100}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{3 / 2}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{8}\right)^{1 / 2}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{2 \sqrt{2}}$
NEET- 2021
NUCLEAR PHYSICS
147799
The mass of a radioactive sample is $10.38 \mathrm{~kg}$. If half-life of the sample is $\mathbf{3 . 8}$ days, then how much of the sample is retained after 19 days ?
1 $0.151 \mathrm{~kg}$
2 $0.16 \mathrm{~kg}$
3 $0.32 \mathrm{~kg}$
4 $1.51 \mathrm{~kg}$
Explanation:
C Given that, Initial number of radioactive sample $\left(\mathrm{N}_{\mathrm{o}}\right)=10.38 \mathrm{~kg}$ Half-life $\left(\mathrm{T}_{1 / 2}\right)=3.8$ day Total time $(\mathrm{t})=19$ day The number of sample after two half lives is $\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{T}_{1 / 2}} \quad\left(\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}\right)$ $\mathrm{N}=10.38\left(\frac{1}{2}\right)^{\frac{19}{3.8}}$ $\mathrm{~N}=10.38 \times\left(\frac{1}{2}\right)^{5}$ $\mathrm{~N}=\frac{10.38}{32}$ $\mathrm{~N}=0.324 \mathrm{~kg}$
Tripura-2021
NUCLEAR PHYSICS
147801
The half-life of a radioactive sample undergoing $\alpha$-decay is $1.4 \times 10^{17} \mathrm{~s}$. If the number of nuclei in the sample is $2.0 \times 10^{21}$. The activity of the sample is nearly
1 $10^{4} \mathrm{~Bq}$
2 $10^{5} \mathrm{~Bq}$
3 $10^{6} \mathrm{~Bq}$
4 $10^{3} \mathrm{~Bq}$
Explanation:
A Given that, Half life $\left(\mathrm{T}_{1 / 2}\right)=1.4 \times 10^{17} \mathrm{sec}$ Number of nuclei in a sample $(\mathrm{N})=2.0 \times 10^{21}$ We know that, Activity of the sample - $\mathrm{R}=\lambda \mathrm{N} \quad \because \lambda=\frac{0.623}{\mathrm{~T}_{1 / 2}}$ $\mathrm{R}=\frac{0.693}{\mathrm{~T}_{1 / 2}} \times 2 \times 10^{21}$ $\mathrm{R}=\frac{0.693}{1.4 \times 10^{17}} \times 2 \times 10^{21}$ $\mathrm{R}=\frac{2 \times 0.693}{1.4} \times 10^{4}$ $\mathrm{R}=9900$ $\mathrm{R} \approx 10000$ $\mathrm{R}=10^{4} \mathrm{~Bq}$
NEET- (Oct.) 2020
NUCLEAR PHYSICS
147798
The half-life period of a radioactive substance is 5 min. The amount of substance decayed in 20 min will be #[Qdiff: Hard, QCat: Numerical Based, examname: , [CG PET-2021, [UPSEE 2019]#
1 $6.25 \%$
2 $25 \%$
3 $75 \%$
4 $93.75 \%$
Explanation:
D Given that, Half-life of radioactive substance $\left(\mathrm{T}_{1 / 2}\right)=5 \mathrm{~min}$ Total time of decay $(\mathrm{t})=20 \mathrm{~min}$. We know that, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{\frac{20}{5}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{4}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{16}$ $Fraction decay $=1-\frac{1}{16}=\frac{15}{16}$ $Fraction decay in percentage $=\frac{15}{16} \times 100=93.75 \%$
147797
The half-life of a radioactive nuclide is $100 \mathrm{~h}$. The fraction of original activity that will remain after $150 \mathrm{~h}$ would be
1 $\frac{1}{2}$
2 $\frac{1}{2 \sqrt{2}}$
3 $\frac{2}{3}$
4 $\frac{2}{\sqrt{3} 2}$
Explanation:
B Given that, Half-life of radioactive nuclide $\left(\mathrm{T}_{1 / 2}\right)=100 \mathrm{~h}$ Total time taken $(t)=150 \mathrm{~h}$ We know that, $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{\frac{150}{100}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{3 / 2}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{8}\right)^{1 / 2}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{2 \sqrt{2}}$
NEET- 2021
NUCLEAR PHYSICS
147799
The mass of a radioactive sample is $10.38 \mathrm{~kg}$. If half-life of the sample is $\mathbf{3 . 8}$ days, then how much of the sample is retained after 19 days ?
1 $0.151 \mathrm{~kg}$
2 $0.16 \mathrm{~kg}$
3 $0.32 \mathrm{~kg}$
4 $1.51 \mathrm{~kg}$
Explanation:
C Given that, Initial number of radioactive sample $\left(\mathrm{N}_{\mathrm{o}}\right)=10.38 \mathrm{~kg}$ Half-life $\left(\mathrm{T}_{1 / 2}\right)=3.8$ day Total time $(\mathrm{t})=19$ day The number of sample after two half lives is $\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{T}_{1 / 2}} \quad\left(\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}\right)$ $\mathrm{N}=10.38\left(\frac{1}{2}\right)^{\frac{19}{3.8}}$ $\mathrm{~N}=10.38 \times\left(\frac{1}{2}\right)^{5}$ $\mathrm{~N}=\frac{10.38}{32}$ $\mathrm{~N}=0.324 \mathrm{~kg}$
Tripura-2021
NUCLEAR PHYSICS
147801
The half-life of a radioactive sample undergoing $\alpha$-decay is $1.4 \times 10^{17} \mathrm{~s}$. If the number of nuclei in the sample is $2.0 \times 10^{21}$. The activity of the sample is nearly
1 $10^{4} \mathrm{~Bq}$
2 $10^{5} \mathrm{~Bq}$
3 $10^{6} \mathrm{~Bq}$
4 $10^{3} \mathrm{~Bq}$
Explanation:
A Given that, Half life $\left(\mathrm{T}_{1 / 2}\right)=1.4 \times 10^{17} \mathrm{sec}$ Number of nuclei in a sample $(\mathrm{N})=2.0 \times 10^{21}$ We know that, Activity of the sample - $\mathrm{R}=\lambda \mathrm{N} \quad \because \lambda=\frac{0.623}{\mathrm{~T}_{1 / 2}}$ $\mathrm{R}=\frac{0.693}{\mathrm{~T}_{1 / 2}} \times 2 \times 10^{21}$ $\mathrm{R}=\frac{0.693}{1.4 \times 10^{17}} \times 2 \times 10^{21}$ $\mathrm{R}=\frac{2 \times 0.693}{1.4} \times 10^{4}$ $\mathrm{R}=9900$ $\mathrm{R} \approx 10000$ $\mathrm{R}=10^{4} \mathrm{~Bq}$
NEET- (Oct.) 2020
NUCLEAR PHYSICS
147798
The half-life period of a radioactive substance is 5 min. The amount of substance decayed in 20 min will be #[Qdiff: Hard, QCat: Numerical Based, examname: , [CG PET-2021, [UPSEE 2019]#
1 $6.25 \%$
2 $25 \%$
3 $75 \%$
4 $93.75 \%$
Explanation:
D Given that, Half-life of radioactive substance $\left(\mathrm{T}_{1 / 2}\right)=5 \mathrm{~min}$ Total time of decay $(\mathrm{t})=20 \mathrm{~min}$. We know that, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{\frac{20}{5}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{4}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{16}$ $Fraction decay $=1-\frac{1}{16}=\frac{15}{16}$ $Fraction decay in percentage $=\frac{15}{16} \times 100=93.75 \%$
147797
The half-life of a radioactive nuclide is $100 \mathrm{~h}$. The fraction of original activity that will remain after $150 \mathrm{~h}$ would be
1 $\frac{1}{2}$
2 $\frac{1}{2 \sqrt{2}}$
3 $\frac{2}{3}$
4 $\frac{2}{\sqrt{3} 2}$
Explanation:
B Given that, Half-life of radioactive nuclide $\left(\mathrm{T}_{1 / 2}\right)=100 \mathrm{~h}$ Total time taken $(t)=150 \mathrm{~h}$ We know that, $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{\frac{150}{100}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{3 / 2}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{8}\right)^{1 / 2}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{2 \sqrt{2}}$
NEET- 2021
NUCLEAR PHYSICS
147799
The mass of a radioactive sample is $10.38 \mathrm{~kg}$. If half-life of the sample is $\mathbf{3 . 8}$ days, then how much of the sample is retained after 19 days ?
1 $0.151 \mathrm{~kg}$
2 $0.16 \mathrm{~kg}$
3 $0.32 \mathrm{~kg}$
4 $1.51 \mathrm{~kg}$
Explanation:
C Given that, Initial number of radioactive sample $\left(\mathrm{N}_{\mathrm{o}}\right)=10.38 \mathrm{~kg}$ Half-life $\left(\mathrm{T}_{1 / 2}\right)=3.8$ day Total time $(\mathrm{t})=19$ day The number of sample after two half lives is $\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{T}_{1 / 2}} \quad\left(\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}\right)$ $\mathrm{N}=10.38\left(\frac{1}{2}\right)^{\frac{19}{3.8}}$ $\mathrm{~N}=10.38 \times\left(\frac{1}{2}\right)^{5}$ $\mathrm{~N}=\frac{10.38}{32}$ $\mathrm{~N}=0.324 \mathrm{~kg}$
Tripura-2021
NUCLEAR PHYSICS
147801
The half-life of a radioactive sample undergoing $\alpha$-decay is $1.4 \times 10^{17} \mathrm{~s}$. If the number of nuclei in the sample is $2.0 \times 10^{21}$. The activity of the sample is nearly
1 $10^{4} \mathrm{~Bq}$
2 $10^{5} \mathrm{~Bq}$
3 $10^{6} \mathrm{~Bq}$
4 $10^{3} \mathrm{~Bq}$
Explanation:
A Given that, Half life $\left(\mathrm{T}_{1 / 2}\right)=1.4 \times 10^{17} \mathrm{sec}$ Number of nuclei in a sample $(\mathrm{N})=2.0 \times 10^{21}$ We know that, Activity of the sample - $\mathrm{R}=\lambda \mathrm{N} \quad \because \lambda=\frac{0.623}{\mathrm{~T}_{1 / 2}}$ $\mathrm{R}=\frac{0.693}{\mathrm{~T}_{1 / 2}} \times 2 \times 10^{21}$ $\mathrm{R}=\frac{0.693}{1.4 \times 10^{17}} \times 2 \times 10^{21}$ $\mathrm{R}=\frac{2 \times 0.693}{1.4} \times 10^{4}$ $\mathrm{R}=9900$ $\mathrm{R} \approx 10000$ $\mathrm{R}=10^{4} \mathrm{~Bq}$
NEET- (Oct.) 2020
NUCLEAR PHYSICS
147798
The half-life period of a radioactive substance is 5 min. The amount of substance decayed in 20 min will be #[Qdiff: Hard, QCat: Numerical Based, examname: , [CG PET-2021, [UPSEE 2019]#
1 $6.25 \%$
2 $25 \%$
3 $75 \%$
4 $93.75 \%$
Explanation:
D Given that, Half-life of radioactive substance $\left(\mathrm{T}_{1 / 2}\right)=5 \mathrm{~min}$ Total time of decay $(\mathrm{t})=20 \mathrm{~min}$. We know that, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{\frac{20}{5}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{4}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{16}$ $Fraction decay $=1-\frac{1}{16}=\frac{15}{16}$ $Fraction decay in percentage $=\frac{15}{16} \times 100=93.75 \%$
147797
The half-life of a radioactive nuclide is $100 \mathrm{~h}$. The fraction of original activity that will remain after $150 \mathrm{~h}$ would be
1 $\frac{1}{2}$
2 $\frac{1}{2 \sqrt{2}}$
3 $\frac{2}{3}$
4 $\frac{2}{\sqrt{3} 2}$
Explanation:
B Given that, Half-life of radioactive nuclide $\left(\mathrm{T}_{1 / 2}\right)=100 \mathrm{~h}$ Total time taken $(t)=150 \mathrm{~h}$ We know that, $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{\frac{150}{100}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{3 / 2}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{8}\right)^{1 / 2}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{2 \sqrt{2}}$
NEET- 2021
NUCLEAR PHYSICS
147799
The mass of a radioactive sample is $10.38 \mathrm{~kg}$. If half-life of the sample is $\mathbf{3 . 8}$ days, then how much of the sample is retained after 19 days ?
1 $0.151 \mathrm{~kg}$
2 $0.16 \mathrm{~kg}$
3 $0.32 \mathrm{~kg}$
4 $1.51 \mathrm{~kg}$
Explanation:
C Given that, Initial number of radioactive sample $\left(\mathrm{N}_{\mathrm{o}}\right)=10.38 \mathrm{~kg}$ Half-life $\left(\mathrm{T}_{1 / 2}\right)=3.8$ day Total time $(\mathrm{t})=19$ day The number of sample after two half lives is $\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{T}_{1 / 2}} \quad\left(\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}\right)$ $\mathrm{N}=10.38\left(\frac{1}{2}\right)^{\frac{19}{3.8}}$ $\mathrm{~N}=10.38 \times\left(\frac{1}{2}\right)^{5}$ $\mathrm{~N}=\frac{10.38}{32}$ $\mathrm{~N}=0.324 \mathrm{~kg}$
Tripura-2021
NUCLEAR PHYSICS
147801
The half-life of a radioactive sample undergoing $\alpha$-decay is $1.4 \times 10^{17} \mathrm{~s}$. If the number of nuclei in the sample is $2.0 \times 10^{21}$. The activity of the sample is nearly
1 $10^{4} \mathrm{~Bq}$
2 $10^{5} \mathrm{~Bq}$
3 $10^{6} \mathrm{~Bq}$
4 $10^{3} \mathrm{~Bq}$
Explanation:
A Given that, Half life $\left(\mathrm{T}_{1 / 2}\right)=1.4 \times 10^{17} \mathrm{sec}$ Number of nuclei in a sample $(\mathrm{N})=2.0 \times 10^{21}$ We know that, Activity of the sample - $\mathrm{R}=\lambda \mathrm{N} \quad \because \lambda=\frac{0.623}{\mathrm{~T}_{1 / 2}}$ $\mathrm{R}=\frac{0.693}{\mathrm{~T}_{1 / 2}} \times 2 \times 10^{21}$ $\mathrm{R}=\frac{0.693}{1.4 \times 10^{17}} \times 2 \times 10^{21}$ $\mathrm{R}=\frac{2 \times 0.693}{1.4} \times 10^{4}$ $\mathrm{R}=9900$ $\mathrm{R} \approx 10000$ $\mathrm{R}=10^{4} \mathrm{~Bq}$
NEET- (Oct.) 2020
NUCLEAR PHYSICS
147798
The half-life period of a radioactive substance is 5 min. The amount of substance decayed in 20 min will be #[Qdiff: Hard, QCat: Numerical Based, examname: , [CG PET-2021, [UPSEE 2019]#
1 $6.25 \%$
2 $25 \%$
3 $75 \%$
4 $93.75 \%$
Explanation:
D Given that, Half-life of radioactive substance $\left(\mathrm{T}_{1 / 2}\right)=5 \mathrm{~min}$ Total time of decay $(\mathrm{t})=20 \mathrm{~min}$. We know that, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{\frac{20}{5}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{4}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{16}$ $Fraction decay $=1-\frac{1}{16}=\frac{15}{16}$ $Fraction decay in percentage $=\frac{15}{16} \times 100=93.75 \%$
