147861
The activity of 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 0.270 per minute
3 0.461 per minute
4 0.39 per minute
Explanation:
C Given that, Initial activity of the sample $\left(\mathrm{N}_{0}\right)=9750$ count $/ \mathrm{min}$ After $\mathrm{t}=5 \mathrm{~min}$ The final activity of the sample, $(\mathrm{N})=975 \mathrm{count} / \mathrm{min}$ We know that, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $975=9750 \mathrm{e}^{-\lambda \times 5}$ $1=10 \mathrm{e}^{-5 \lambda}$ $\frac{1}{10}=\mathrm{e}^{-5 \lambda}$ $10=\mathrm{e}^{5 \lambda}$ Taking logarithm both the side we get - $\ln (10)=\ln \mathrm{e}^{5 \lambda}$ $\ln (10)=5 \lambda$ $\lambda=\frac{\ln (10)}{5}$ $\lambda=\frac{2.303}{5}$ $\lambda=0.461 \text { per minutes }$
AIIMS-1998
NUCLEAR PHYSICS
147862
If the radioactive decay constant of radium is $1.07 \times 10^{-4}$ per year. Then its half life period approximately is equal to :
1 5000 years
2 6500 years
3 7000 years
4 8900 years
Explanation:
B Given that, Decay constant $(\lambda)=1.07 \times 10^{-4}$ per year. Half-life $\left(\mathrm{T}_{1 / 2}\right)=\frac{0.693}{\lambda}$ $\mathrm{T}_{1 / 2}=\frac{0.693}{1.07 \times 10^{-4}}$ $\mathrm{~T}_{1 / 2}=6.476 \times 10^{3}$ $\mathrm{~T}_{1 / 2}=6476$ $\mathrm{~T}_{1 / 2}\approx 6500 \text { year }$
AIIMS-1998
NUCLEAR PHYSICS
147863
The half-life of the isotope ${ }_{11} \mathrm{Na}^{24}$ is $15 \mathrm{~h}$. How much time does it take for $\frac{7}{8}$ th of a sample of this isotope to decay?
1 $75 \mathrm{~h}$
2 $65 \mathrm{~h}$
3 $55 \mathrm{~h}$
4 $45 \mathrm{~h}$
Explanation:
D According to question, undecayed isotope is $\frac{\mathrm{N}_{0}-\mathrm{N}}{\mathrm{N}_{0}}=\frac{7}{8}$ $1-\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{7}{8}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{8}$ We know that, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}} \quad\left(\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}}\right)$ $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{15}}$ Putting the value from equation (i), we get - $\frac{1}{8}=\left(\frac{1}{2}\right)^{\frac{t}{15}}$ $\left(\frac{1}{2}\right)^{3}=\left(\frac{1}{2}\right)^{\frac{t}{15}}$ On comparing both the side, we get - $\frac{t}{15}=3$ $t=3 \times 15=45 h$
Manipal UGET-2011
NUCLEAR PHYSICS
147865
Rn decays into Po by emitting an $\alpha$-particle with half-life of 4 days. A sample contains $6.4 \times 10^{10}$ atoms of $R$ after 12 days, the number of atoms of $\mathrm{Rn}$ left in the sample will be
1 $3.2 \times 10^{10}$
2 $0.53 \times 10^{10}$
3 $2.1 \times 10^{10}$
4 $0.8 \times 10^{10}$
Explanation:
D Given that, Half-life $\left(T_{1 / 2}\right)=4$ day Total time $(\mathrm{t})=12$ days Initial activity of sample $\left(\mathrm{N}_{\mathrm{o}}\right)=6.4 \times 10^{10}$ We know that, $\mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}=\frac{12}{4}=3$ From the radioactive equation, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{N}=6.4 \times 10^{10} \times\left(\frac{1}{2}\right)^{3}$ $\mathrm{~N}=6.4 \times 10^{10} \times \frac{1}{8}$ $\mathrm{~N}=0.8 \times 10^{10} \text { atoms }$
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NUCLEAR PHYSICS
147861
The activity of 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 0.270 per minute
3 0.461 per minute
4 0.39 per minute
Explanation:
C Given that, Initial activity of the sample $\left(\mathrm{N}_{0}\right)=9750$ count $/ \mathrm{min}$ After $\mathrm{t}=5 \mathrm{~min}$ The final activity of the sample, $(\mathrm{N})=975 \mathrm{count} / \mathrm{min}$ We know that, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $975=9750 \mathrm{e}^{-\lambda \times 5}$ $1=10 \mathrm{e}^{-5 \lambda}$ $\frac{1}{10}=\mathrm{e}^{-5 \lambda}$ $10=\mathrm{e}^{5 \lambda}$ Taking logarithm both the side we get - $\ln (10)=\ln \mathrm{e}^{5 \lambda}$ $\ln (10)=5 \lambda$ $\lambda=\frac{\ln (10)}{5}$ $\lambda=\frac{2.303}{5}$ $\lambda=0.461 \text { per minutes }$
AIIMS-1998
NUCLEAR PHYSICS
147862
If the radioactive decay constant of radium is $1.07 \times 10^{-4}$ per year. Then its half life period approximately is equal to :
1 5000 years
2 6500 years
3 7000 years
4 8900 years
Explanation:
B Given that, Decay constant $(\lambda)=1.07 \times 10^{-4}$ per year. Half-life $\left(\mathrm{T}_{1 / 2}\right)=\frac{0.693}{\lambda}$ $\mathrm{T}_{1 / 2}=\frac{0.693}{1.07 \times 10^{-4}}$ $\mathrm{~T}_{1 / 2}=6.476 \times 10^{3}$ $\mathrm{~T}_{1 / 2}=6476$ $\mathrm{~T}_{1 / 2}\approx 6500 \text { year }$
AIIMS-1998
NUCLEAR PHYSICS
147863
The half-life of the isotope ${ }_{11} \mathrm{Na}^{24}$ is $15 \mathrm{~h}$. How much time does it take for $\frac{7}{8}$ th of a sample of this isotope to decay?
1 $75 \mathrm{~h}$
2 $65 \mathrm{~h}$
3 $55 \mathrm{~h}$
4 $45 \mathrm{~h}$
Explanation:
D According to question, undecayed isotope is $\frac{\mathrm{N}_{0}-\mathrm{N}}{\mathrm{N}_{0}}=\frac{7}{8}$ $1-\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{7}{8}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{8}$ We know that, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}} \quad\left(\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}}\right)$ $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{15}}$ Putting the value from equation (i), we get - $\frac{1}{8}=\left(\frac{1}{2}\right)^{\frac{t}{15}}$ $\left(\frac{1}{2}\right)^{3}=\left(\frac{1}{2}\right)^{\frac{t}{15}}$ On comparing both the side, we get - $\frac{t}{15}=3$ $t=3 \times 15=45 h$
Manipal UGET-2011
NUCLEAR PHYSICS
147865
Rn decays into Po by emitting an $\alpha$-particle with half-life of 4 days. A sample contains $6.4 \times 10^{10}$ atoms of $R$ after 12 days, the number of atoms of $\mathrm{Rn}$ left in the sample will be
1 $3.2 \times 10^{10}$
2 $0.53 \times 10^{10}$
3 $2.1 \times 10^{10}$
4 $0.8 \times 10^{10}$
Explanation:
D Given that, Half-life $\left(T_{1 / 2}\right)=4$ day Total time $(\mathrm{t})=12$ days Initial activity of sample $\left(\mathrm{N}_{\mathrm{o}}\right)=6.4 \times 10^{10}$ We know that, $\mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}=\frac{12}{4}=3$ From the radioactive equation, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{N}=6.4 \times 10^{10} \times\left(\frac{1}{2}\right)^{3}$ $\mathrm{~N}=6.4 \times 10^{10} \times \frac{1}{8}$ $\mathrm{~N}=0.8 \times 10^{10} \text { atoms }$
147861
The activity of 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 0.270 per minute
3 0.461 per minute
4 0.39 per minute
Explanation:
C Given that, Initial activity of the sample $\left(\mathrm{N}_{0}\right)=9750$ count $/ \mathrm{min}$ After $\mathrm{t}=5 \mathrm{~min}$ The final activity of the sample, $(\mathrm{N})=975 \mathrm{count} / \mathrm{min}$ We know that, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $975=9750 \mathrm{e}^{-\lambda \times 5}$ $1=10 \mathrm{e}^{-5 \lambda}$ $\frac{1}{10}=\mathrm{e}^{-5 \lambda}$ $10=\mathrm{e}^{5 \lambda}$ Taking logarithm both the side we get - $\ln (10)=\ln \mathrm{e}^{5 \lambda}$ $\ln (10)=5 \lambda$ $\lambda=\frac{\ln (10)}{5}$ $\lambda=\frac{2.303}{5}$ $\lambda=0.461 \text { per minutes }$
AIIMS-1998
NUCLEAR PHYSICS
147862
If the radioactive decay constant of radium is $1.07 \times 10^{-4}$ per year. Then its half life period approximately is equal to :
1 5000 years
2 6500 years
3 7000 years
4 8900 years
Explanation:
B Given that, Decay constant $(\lambda)=1.07 \times 10^{-4}$ per year. Half-life $\left(\mathrm{T}_{1 / 2}\right)=\frac{0.693}{\lambda}$ $\mathrm{T}_{1 / 2}=\frac{0.693}{1.07 \times 10^{-4}}$ $\mathrm{~T}_{1 / 2}=6.476 \times 10^{3}$ $\mathrm{~T}_{1 / 2}=6476$ $\mathrm{~T}_{1 / 2}\approx 6500 \text { year }$
AIIMS-1998
NUCLEAR PHYSICS
147863
The half-life of the isotope ${ }_{11} \mathrm{Na}^{24}$ is $15 \mathrm{~h}$. How much time does it take for $\frac{7}{8}$ th of a sample of this isotope to decay?
1 $75 \mathrm{~h}$
2 $65 \mathrm{~h}$
3 $55 \mathrm{~h}$
4 $45 \mathrm{~h}$
Explanation:
D According to question, undecayed isotope is $\frac{\mathrm{N}_{0}-\mathrm{N}}{\mathrm{N}_{0}}=\frac{7}{8}$ $1-\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{7}{8}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{8}$ We know that, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}} \quad\left(\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}}\right)$ $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{15}}$ Putting the value from equation (i), we get - $\frac{1}{8}=\left(\frac{1}{2}\right)^{\frac{t}{15}}$ $\left(\frac{1}{2}\right)^{3}=\left(\frac{1}{2}\right)^{\frac{t}{15}}$ On comparing both the side, we get - $\frac{t}{15}=3$ $t=3 \times 15=45 h$
Manipal UGET-2011
NUCLEAR PHYSICS
147865
Rn decays into Po by emitting an $\alpha$-particle with half-life of 4 days. A sample contains $6.4 \times 10^{10}$ atoms of $R$ after 12 days, the number of atoms of $\mathrm{Rn}$ left in the sample will be
1 $3.2 \times 10^{10}$
2 $0.53 \times 10^{10}$
3 $2.1 \times 10^{10}$
4 $0.8 \times 10^{10}$
Explanation:
D Given that, Half-life $\left(T_{1 / 2}\right)=4$ day Total time $(\mathrm{t})=12$ days Initial activity of sample $\left(\mathrm{N}_{\mathrm{o}}\right)=6.4 \times 10^{10}$ We know that, $\mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}=\frac{12}{4}=3$ From the radioactive equation, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{N}=6.4 \times 10^{10} \times\left(\frac{1}{2}\right)^{3}$ $\mathrm{~N}=6.4 \times 10^{10} \times \frac{1}{8}$ $\mathrm{~N}=0.8 \times 10^{10} \text { atoms }$
147861
The activity of 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 0.270 per minute
3 0.461 per minute
4 0.39 per minute
Explanation:
C Given that, Initial activity of the sample $\left(\mathrm{N}_{0}\right)=9750$ count $/ \mathrm{min}$ After $\mathrm{t}=5 \mathrm{~min}$ The final activity of the sample, $(\mathrm{N})=975 \mathrm{count} / \mathrm{min}$ We know that, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $975=9750 \mathrm{e}^{-\lambda \times 5}$ $1=10 \mathrm{e}^{-5 \lambda}$ $\frac{1}{10}=\mathrm{e}^{-5 \lambda}$ $10=\mathrm{e}^{5 \lambda}$ Taking logarithm both the side we get - $\ln (10)=\ln \mathrm{e}^{5 \lambda}$ $\ln (10)=5 \lambda$ $\lambda=\frac{\ln (10)}{5}$ $\lambda=\frac{2.303}{5}$ $\lambda=0.461 \text { per minutes }$
AIIMS-1998
NUCLEAR PHYSICS
147862
If the radioactive decay constant of radium is $1.07 \times 10^{-4}$ per year. Then its half life period approximately is equal to :
1 5000 years
2 6500 years
3 7000 years
4 8900 years
Explanation:
B Given that, Decay constant $(\lambda)=1.07 \times 10^{-4}$ per year. Half-life $\left(\mathrm{T}_{1 / 2}\right)=\frac{0.693}{\lambda}$ $\mathrm{T}_{1 / 2}=\frac{0.693}{1.07 \times 10^{-4}}$ $\mathrm{~T}_{1 / 2}=6.476 \times 10^{3}$ $\mathrm{~T}_{1 / 2}=6476$ $\mathrm{~T}_{1 / 2}\approx 6500 \text { year }$
AIIMS-1998
NUCLEAR PHYSICS
147863
The half-life of the isotope ${ }_{11} \mathrm{Na}^{24}$ is $15 \mathrm{~h}$. How much time does it take for $\frac{7}{8}$ th of a sample of this isotope to decay?
1 $75 \mathrm{~h}$
2 $65 \mathrm{~h}$
3 $55 \mathrm{~h}$
4 $45 \mathrm{~h}$
Explanation:
D According to question, undecayed isotope is $\frac{\mathrm{N}_{0}-\mathrm{N}}{\mathrm{N}_{0}}=\frac{7}{8}$ $1-\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{7}{8}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{8}$ We know that, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}} \quad\left(\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}}\right)$ $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{15}}$ Putting the value from equation (i), we get - $\frac{1}{8}=\left(\frac{1}{2}\right)^{\frac{t}{15}}$ $\left(\frac{1}{2}\right)^{3}=\left(\frac{1}{2}\right)^{\frac{t}{15}}$ On comparing both the side, we get - $\frac{t}{15}=3$ $t=3 \times 15=45 h$
Manipal UGET-2011
NUCLEAR PHYSICS
147865
Rn decays into Po by emitting an $\alpha$-particle with half-life of 4 days. A sample contains $6.4 \times 10^{10}$ atoms of $R$ after 12 days, the number of atoms of $\mathrm{Rn}$ left in the sample will be
1 $3.2 \times 10^{10}$
2 $0.53 \times 10^{10}$
3 $2.1 \times 10^{10}$
4 $0.8 \times 10^{10}$
Explanation:
D Given that, Half-life $\left(T_{1 / 2}\right)=4$ day Total time $(\mathrm{t})=12$ days Initial activity of sample $\left(\mathrm{N}_{\mathrm{o}}\right)=6.4 \times 10^{10}$ We know that, $\mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}=\frac{12}{4}=3$ From the radioactive equation, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{N}=6.4 \times 10^{10} \times\left(\frac{1}{2}\right)^{3}$ $\mathrm{~N}=6.4 \times 10^{10} \times \frac{1}{8}$ $\mathrm{~N}=0.8 \times 10^{10} \text { atoms }$