147752
In a sample of radioactive material, what percentage of the initial number of active nuclei will decay during one mean life is
1 $37 \%$
2 $50 \%$
3 $63 \%$
4 $69.3 \%$
Explanation:
C Let number of active nuclei initially be $\mathrm{N}_{0}$ Mean life $=\frac{1}{\lambda}$ Using the law of radioactive decay, $\mathrm{N}=\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-\lambda \mathrm{t}}$ We need to find the number of nuclei that decayed, which can be given by $\mathrm{N}^{\prime}=\mathrm{N}_{\mathrm{o}}-\mathrm{N}$ $\mathrm{N}^{\prime}=\mathrm{N}-\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-\lambda \mathrm{t}}$ \(\mathrm{N}^{\prime}=\mathrm{N}_{\mathrm{o}}\left(1-\mathrm{e}^{-\lambda \mathrm{t}}\right)\) \(\mathrm{N}^{\prime}=\mathrm{N}_{\mathrm{o}}\left(1-\mathrm{e}^{-\lambda(1 / \lambda)}\right)\) \(\mathrm{N}^{\prime}=\mathrm{N}_{\mathrm{o}}\left(1-\mathrm{e}^{-1}\right)\) \(\mathrm{N}^{\prime}=0.63 \mathrm{~N}_{\mathrm{o}}\) \(\therefore\) Percentage decayed \(=63 \%\)
UP CPMT-2005
NUCLEAR PHYSICS
147753
If a sample of $16 \mathrm{~g}$ radioactive substance disintegrate to $1 \mathrm{~g}$ in 120 days, then what will be the half-life of the sample -
1 15 days
2 7.5 days
3 30 days
4 60 days
Explanation:
C Given, $\mathrm{N}=1 \mathrm{~g}, \mathrm{~N}_{0}=16 \mathrm{~g}, \mathrm{t}=120$ days We know that, $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{1}{16}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\left(\frac{1}{2}\right)^{4}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{n}=4$ Also we know that, $\mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}$ $\mathrm{~T}_{1 / 2}=\frac{\mathrm{t}}{\mathrm{n}}=\frac{120}{4}=30 \text { days }$
UP CPMT-2003
NUCLEAR PHYSICS
147754
During a mean life of a radioactive element the fraction that disintegrates is
1 $\mathrm{e}$
2 $\frac{1}{\mathrm{e}}$
3 $\frac{\mathrm{e}-1}{\mathrm{e}}$
4 $\frac{\mathrm{e}}{\mathrm{e}-1}$
Explanation:
C The mean lifetime $(\mathrm{t})=\frac{1}{\lambda}$ Fraction that disintegrates is, $\frac{\mathrm{N}_{0}-\mathrm{N}_{\mathrm{t}}}{\mathrm{N}_{0}}=\frac{\mathrm{N}_{0}-\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}}{\mathrm{N}_{0}}$ $=1-\mathrm{e}^{-\lambda \times \frac{1}{\lambda}}$ $=1-\frac{1}{\mathrm{e}}$ $=\frac{\mathrm{e}-1}{\mathrm{e}}$
UP CPMT-2001
NUCLEAR PHYSICS
147760
If ${ }_{92} U^{238}$ emits $8 \alpha$-particles and $6 \beta$-particle then the resulting nucleus is
1 ${ }_{82} \mathrm{U}^{206}$
2 ${ }_{82} \mathrm{~Pb}^{206}$
3 ${ }_{82} \mathrm{U}^{210}$
4 ${ }_{82} \mathrm{U}^{214}$
Explanation:
B After emitting $1 \alpha$ - particles, the atomic number decreases by 2 and mass number decreases by 4. $1 \beta$-particle, the atomic number increases by 1 and mass number is unaffected. So, in this situation, the mass number of the daughter nucleus is - $238-8 \times 4+0=206$ And the atomic number is $92-8 \times 2+6=82$ Thus, the answer is ${ }_{82} \mathrm{~Pb}^{206}$.
147752
In a sample of radioactive material, what percentage of the initial number of active nuclei will decay during one mean life is
1 $37 \%$
2 $50 \%$
3 $63 \%$
4 $69.3 \%$
Explanation:
C Let number of active nuclei initially be $\mathrm{N}_{0}$ Mean life $=\frac{1}{\lambda}$ Using the law of radioactive decay, $\mathrm{N}=\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-\lambda \mathrm{t}}$ We need to find the number of nuclei that decayed, which can be given by $\mathrm{N}^{\prime}=\mathrm{N}_{\mathrm{o}}-\mathrm{N}$ $\mathrm{N}^{\prime}=\mathrm{N}-\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-\lambda \mathrm{t}}$ \(\mathrm{N}^{\prime}=\mathrm{N}_{\mathrm{o}}\left(1-\mathrm{e}^{-\lambda \mathrm{t}}\right)\) \(\mathrm{N}^{\prime}=\mathrm{N}_{\mathrm{o}}\left(1-\mathrm{e}^{-\lambda(1 / \lambda)}\right)\) \(\mathrm{N}^{\prime}=\mathrm{N}_{\mathrm{o}}\left(1-\mathrm{e}^{-1}\right)\) \(\mathrm{N}^{\prime}=0.63 \mathrm{~N}_{\mathrm{o}}\) \(\therefore\) Percentage decayed \(=63 \%\)
UP CPMT-2005
NUCLEAR PHYSICS
147753
If a sample of $16 \mathrm{~g}$ radioactive substance disintegrate to $1 \mathrm{~g}$ in 120 days, then what will be the half-life of the sample -
1 15 days
2 7.5 days
3 30 days
4 60 days
Explanation:
C Given, $\mathrm{N}=1 \mathrm{~g}, \mathrm{~N}_{0}=16 \mathrm{~g}, \mathrm{t}=120$ days We know that, $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{1}{16}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\left(\frac{1}{2}\right)^{4}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{n}=4$ Also we know that, $\mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}$ $\mathrm{~T}_{1 / 2}=\frac{\mathrm{t}}{\mathrm{n}}=\frac{120}{4}=30 \text { days }$
UP CPMT-2003
NUCLEAR PHYSICS
147754
During a mean life of a radioactive element the fraction that disintegrates is
1 $\mathrm{e}$
2 $\frac{1}{\mathrm{e}}$
3 $\frac{\mathrm{e}-1}{\mathrm{e}}$
4 $\frac{\mathrm{e}}{\mathrm{e}-1}$
Explanation:
C The mean lifetime $(\mathrm{t})=\frac{1}{\lambda}$ Fraction that disintegrates is, $\frac{\mathrm{N}_{0}-\mathrm{N}_{\mathrm{t}}}{\mathrm{N}_{0}}=\frac{\mathrm{N}_{0}-\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}}{\mathrm{N}_{0}}$ $=1-\mathrm{e}^{-\lambda \times \frac{1}{\lambda}}$ $=1-\frac{1}{\mathrm{e}}$ $=\frac{\mathrm{e}-1}{\mathrm{e}}$
UP CPMT-2001
NUCLEAR PHYSICS
147760
If ${ }_{92} U^{238}$ emits $8 \alpha$-particles and $6 \beta$-particle then the resulting nucleus is
1 ${ }_{82} \mathrm{U}^{206}$
2 ${ }_{82} \mathrm{~Pb}^{206}$
3 ${ }_{82} \mathrm{U}^{210}$
4 ${ }_{82} \mathrm{U}^{214}$
Explanation:
B After emitting $1 \alpha$ - particles, the atomic number decreases by 2 and mass number decreases by 4. $1 \beta$-particle, the atomic number increases by 1 and mass number is unaffected. So, in this situation, the mass number of the daughter nucleus is - $238-8 \times 4+0=206$ And the atomic number is $92-8 \times 2+6=82$ Thus, the answer is ${ }_{82} \mathrm{~Pb}^{206}$.
147752
In a sample of radioactive material, what percentage of the initial number of active nuclei will decay during one mean life is
1 $37 \%$
2 $50 \%$
3 $63 \%$
4 $69.3 \%$
Explanation:
C Let number of active nuclei initially be $\mathrm{N}_{0}$ Mean life $=\frac{1}{\lambda}$ Using the law of radioactive decay, $\mathrm{N}=\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-\lambda \mathrm{t}}$ We need to find the number of nuclei that decayed, which can be given by $\mathrm{N}^{\prime}=\mathrm{N}_{\mathrm{o}}-\mathrm{N}$ $\mathrm{N}^{\prime}=\mathrm{N}-\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-\lambda \mathrm{t}}$ \(\mathrm{N}^{\prime}=\mathrm{N}_{\mathrm{o}}\left(1-\mathrm{e}^{-\lambda \mathrm{t}}\right)\) \(\mathrm{N}^{\prime}=\mathrm{N}_{\mathrm{o}}\left(1-\mathrm{e}^{-\lambda(1 / \lambda)}\right)\) \(\mathrm{N}^{\prime}=\mathrm{N}_{\mathrm{o}}\left(1-\mathrm{e}^{-1}\right)\) \(\mathrm{N}^{\prime}=0.63 \mathrm{~N}_{\mathrm{o}}\) \(\therefore\) Percentage decayed \(=63 \%\)
UP CPMT-2005
NUCLEAR PHYSICS
147753
If a sample of $16 \mathrm{~g}$ radioactive substance disintegrate to $1 \mathrm{~g}$ in 120 days, then what will be the half-life of the sample -
1 15 days
2 7.5 days
3 30 days
4 60 days
Explanation:
C Given, $\mathrm{N}=1 \mathrm{~g}, \mathrm{~N}_{0}=16 \mathrm{~g}, \mathrm{t}=120$ days We know that, $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{1}{16}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\left(\frac{1}{2}\right)^{4}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{n}=4$ Also we know that, $\mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}$ $\mathrm{~T}_{1 / 2}=\frac{\mathrm{t}}{\mathrm{n}}=\frac{120}{4}=30 \text { days }$
UP CPMT-2003
NUCLEAR PHYSICS
147754
During a mean life of a radioactive element the fraction that disintegrates is
1 $\mathrm{e}$
2 $\frac{1}{\mathrm{e}}$
3 $\frac{\mathrm{e}-1}{\mathrm{e}}$
4 $\frac{\mathrm{e}}{\mathrm{e}-1}$
Explanation:
C The mean lifetime $(\mathrm{t})=\frac{1}{\lambda}$ Fraction that disintegrates is, $\frac{\mathrm{N}_{0}-\mathrm{N}_{\mathrm{t}}}{\mathrm{N}_{0}}=\frac{\mathrm{N}_{0}-\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}}{\mathrm{N}_{0}}$ $=1-\mathrm{e}^{-\lambda \times \frac{1}{\lambda}}$ $=1-\frac{1}{\mathrm{e}}$ $=\frac{\mathrm{e}-1}{\mathrm{e}}$
UP CPMT-2001
NUCLEAR PHYSICS
147760
If ${ }_{92} U^{238}$ emits $8 \alpha$-particles and $6 \beta$-particle then the resulting nucleus is
1 ${ }_{82} \mathrm{U}^{206}$
2 ${ }_{82} \mathrm{~Pb}^{206}$
3 ${ }_{82} \mathrm{U}^{210}$
4 ${ }_{82} \mathrm{U}^{214}$
Explanation:
B After emitting $1 \alpha$ - particles, the atomic number decreases by 2 and mass number decreases by 4. $1 \beta$-particle, the atomic number increases by 1 and mass number is unaffected. So, in this situation, the mass number of the daughter nucleus is - $238-8 \times 4+0=206$ And the atomic number is $92-8 \times 2+6=82$ Thus, the answer is ${ }_{82} \mathrm{~Pb}^{206}$.
147752
In a sample of radioactive material, what percentage of the initial number of active nuclei will decay during one mean life is
1 $37 \%$
2 $50 \%$
3 $63 \%$
4 $69.3 \%$
Explanation:
C Let number of active nuclei initially be $\mathrm{N}_{0}$ Mean life $=\frac{1}{\lambda}$ Using the law of radioactive decay, $\mathrm{N}=\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-\lambda \mathrm{t}}$ We need to find the number of nuclei that decayed, which can be given by $\mathrm{N}^{\prime}=\mathrm{N}_{\mathrm{o}}-\mathrm{N}$ $\mathrm{N}^{\prime}=\mathrm{N}-\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-\lambda \mathrm{t}}$ \(\mathrm{N}^{\prime}=\mathrm{N}_{\mathrm{o}}\left(1-\mathrm{e}^{-\lambda \mathrm{t}}\right)\) \(\mathrm{N}^{\prime}=\mathrm{N}_{\mathrm{o}}\left(1-\mathrm{e}^{-\lambda(1 / \lambda)}\right)\) \(\mathrm{N}^{\prime}=\mathrm{N}_{\mathrm{o}}\left(1-\mathrm{e}^{-1}\right)\) \(\mathrm{N}^{\prime}=0.63 \mathrm{~N}_{\mathrm{o}}\) \(\therefore\) Percentage decayed \(=63 \%\)
UP CPMT-2005
NUCLEAR PHYSICS
147753
If a sample of $16 \mathrm{~g}$ radioactive substance disintegrate to $1 \mathrm{~g}$ in 120 days, then what will be the half-life of the sample -
1 15 days
2 7.5 days
3 30 days
4 60 days
Explanation:
C Given, $\mathrm{N}=1 \mathrm{~g}, \mathrm{~N}_{0}=16 \mathrm{~g}, \mathrm{t}=120$ days We know that, $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{1}{16}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\left(\frac{1}{2}\right)^{4}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{n}=4$ Also we know that, $\mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}$ $\mathrm{~T}_{1 / 2}=\frac{\mathrm{t}}{\mathrm{n}}=\frac{120}{4}=30 \text { days }$
UP CPMT-2003
NUCLEAR PHYSICS
147754
During a mean life of a radioactive element the fraction that disintegrates is
1 $\mathrm{e}$
2 $\frac{1}{\mathrm{e}}$
3 $\frac{\mathrm{e}-1}{\mathrm{e}}$
4 $\frac{\mathrm{e}}{\mathrm{e}-1}$
Explanation:
C The mean lifetime $(\mathrm{t})=\frac{1}{\lambda}$ Fraction that disintegrates is, $\frac{\mathrm{N}_{0}-\mathrm{N}_{\mathrm{t}}}{\mathrm{N}_{0}}=\frac{\mathrm{N}_{0}-\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}}{\mathrm{N}_{0}}$ $=1-\mathrm{e}^{-\lambda \times \frac{1}{\lambda}}$ $=1-\frac{1}{\mathrm{e}}$ $=\frac{\mathrm{e}-1}{\mathrm{e}}$
UP CPMT-2001
NUCLEAR PHYSICS
147760
If ${ }_{92} U^{238}$ emits $8 \alpha$-particles and $6 \beta$-particle then the resulting nucleus is
1 ${ }_{82} \mathrm{U}^{206}$
2 ${ }_{82} \mathrm{~Pb}^{206}$
3 ${ }_{82} \mathrm{U}^{210}$
4 ${ }_{82} \mathrm{U}^{214}$
Explanation:
B After emitting $1 \alpha$ - particles, the atomic number decreases by 2 and mass number decreases by 4. $1 \beta$-particle, the atomic number increases by 1 and mass number is unaffected. So, in this situation, the mass number of the daughter nucleus is - $238-8 \times 4+0=206$ And the atomic number is $92-8 \times 2+6=82$ Thus, the answer is ${ }_{82} \mathrm{~Pb}^{206}$.
