147855
A radioactive material decays by simultaneous emission of two particles with respective halflives 1620 and 810 years. The time, in years, after which one-fourth of the material remains is
1 1080
2 2430
3 3240
4 4860
Explanation:
A According to the law of Rutherford-Soddy. $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{n}=\frac{\text { time of decay }(\mathrm{t})}{\text { effective half }- \text { life }}$ The relation between effective disintegration constant $(\lambda)$ and $(T)$ $\lambda=\frac{\ln (2)}{\mathrm{T}}$ Hence, $\lambda_{1}+\lambda_{2}=\frac{\ln (2)}{\mathrm{T}_{1}}+\frac{\ln (2)}{\mathrm{T}_{2}}$ And effective half-life, $\frac{1}{\mathrm{~T}}=\frac{1}{\mathrm{~T}_{1}}+\frac{1}{\mathrm{~T}_{2}}$ $\frac{1}{\mathrm{~T}}=\frac{1}{1620}+\frac{1}{810}$ $\mathrm{T}=540 \text { years }$ So, $\mathrm{n}=\frac{\mathrm{t}}{540}$ From equation (i), we get - $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{4}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{540}}$ $\left(\frac{1}{2}\right)^{2}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{540}}$ On comparing both side we get - $\frac{t}{540}=2$ $t=1080 \text { years }$
AIIMS-2008
NUCLEAR PHYSICS
147856
A radioactive material has half - life of 10 days. What fraction of the material would remain after 30days ?
1 0.5
2 0.25
3 0.125
4 0.33
Explanation:
C Given that, Half-life $\left(\mathrm{T}_{1 / 2}\right)=10$ day Remain time $(\mathrm{t})=30$ day We know that, $\mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}$ $\mathrm{n}=\frac{30}{10}=3$ We know that, $\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{3}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\frac{1}{8}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=0.125$
AIIMS-2005
NUCLEAR PHYSICS
147857
Half life of a substance is 20 minutes, then the time between $33 \%$ decay and $67 \%$ decay will be
1 20 minute
2 40 minute
3 50 minute
4 10 minute
Explanation:
A Let, the initial number of nuclei $=\mathrm{N}_{0}$ Let the time required for $33 \%$ decay $=t_{1}$ So, $0.67 \mathrm{~N}_{0}=\mathrm{N}_{0} \mathrm{e}^{-\lambda t_{1}}$ $\mathrm{e}^{-\lambda \mathrm{t}_{1}}=0.67$ And time required for $67 \%$ decay $=\mathrm{t}_{2}$ Similarly, $\mathrm{e}^{-\lambda \mathrm{t}_{2}}=0.33$ On dividing (ii) by (i) we get- $\mathrm{e}^{-\lambda\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right)}=\frac{0.33}{0.67}$ $\mathrm{e}^{-\lambda\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right)} = \frac{1}{2}$ Taking natural logarithm both the side, we get- $-\lambda\left(t_{2}-t_{1}\right)=\ln \frac{1}{2}$ $\mathrm{t}_{2}-\mathrm{t}_{1}=-\frac{\ln \left(\frac{1}{2}\right)}{\lambda}$ $\mathrm{t}_{2}-\mathrm{t}_{1}=-\frac{\ln \left(\frac{1}{2}\right)}{0.693} \times \mathrm{T}_{1 / 2} \quad\left(\because \lambda=\frac{0.693}{\mathrm{~T}_{1 / 2}}\right)$ $\text { Given, } T_{1 / 2}=20 \mathrm{~min}$ $\mathrm{t}_{2}-\mathrm{t}_{1}=-\frac{\ln \left(\frac{1}{2}\right) \times 20}{-\ln \left(\frac{1}{2}\right)}$ $\mathrm{t}_{2}-\mathrm{t}_{1}=20 \mathrm{~min}$
AIIMS-2000
NUCLEAR PHYSICS
147858
A nuclear decay is expressed as ${ }_{6} \mathrm{C}^{11} \longrightarrow \mathrm{B}^{11}+\boldsymbol{\beta}^{+}+\mathrm{X}$ Then the unknown particle $X$ is
1 Neutron
2 Antineutrino
3 Proton
4 Neutrino
Explanation:
D ${ }_{6} \mathrm{C}^{11} \rightarrow{ }_{5} \mathrm{~B}^{11}+{ }_{1} \beta^{0}+\mathrm{X}$ For atomic number $6=5+1+Z$ $\mathrm{Z}=0$ For mass number $11=11+0+\mathrm{A}$ $\mathrm{A}=0$ So, the unknown particle $\mathrm{X}$ is neutrino
AIPMT- 2000
NUCLEAR PHYSICS
147860
The activity of a radioactive sample is 1.6 curie and its half life is 2.5 days. The activity after 10 days will be :
1 0.16 curie
2 0.8 curie
3 0.1 curie
4 0.4 curie
Explanation:
C Given that, Time of decay $(\mathrm{t})=10$ days Half-life $\left(\mathrm{T}_{1 / 2}\right)=2.5$ day Initial radioactive nuclei $\left(\mathrm{N}_{\mathrm{o}}\right)=1.6$ curie We know that, $\mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}=\frac{10}{2.5}=4$ We know that, \(\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{n}}\) \(\mathrm{N}=1.6 \times\left(\frac{1}{2}\right)^4\) \(\mathrm{~N}=1.6 \times \frac{1}{16}\) \(\mathrm{~N}=0.1 \text { curie }\)
147855
A radioactive material decays by simultaneous emission of two particles with respective halflives 1620 and 810 years. The time, in years, after which one-fourth of the material remains is
1 1080
2 2430
3 3240
4 4860
Explanation:
A According to the law of Rutherford-Soddy. $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{n}=\frac{\text { time of decay }(\mathrm{t})}{\text { effective half }- \text { life }}$ The relation between effective disintegration constant $(\lambda)$ and $(T)$ $\lambda=\frac{\ln (2)}{\mathrm{T}}$ Hence, $\lambda_{1}+\lambda_{2}=\frac{\ln (2)}{\mathrm{T}_{1}}+\frac{\ln (2)}{\mathrm{T}_{2}}$ And effective half-life, $\frac{1}{\mathrm{~T}}=\frac{1}{\mathrm{~T}_{1}}+\frac{1}{\mathrm{~T}_{2}}$ $\frac{1}{\mathrm{~T}}=\frac{1}{1620}+\frac{1}{810}$ $\mathrm{T}=540 \text { years }$ So, $\mathrm{n}=\frac{\mathrm{t}}{540}$ From equation (i), we get - $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{4}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{540}}$ $\left(\frac{1}{2}\right)^{2}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{540}}$ On comparing both side we get - $\frac{t}{540}=2$ $t=1080 \text { years }$
AIIMS-2008
NUCLEAR PHYSICS
147856
A radioactive material has half - life of 10 days. What fraction of the material would remain after 30days ?
1 0.5
2 0.25
3 0.125
4 0.33
Explanation:
C Given that, Half-life $\left(\mathrm{T}_{1 / 2}\right)=10$ day Remain time $(\mathrm{t})=30$ day We know that, $\mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}$ $\mathrm{n}=\frac{30}{10}=3$ We know that, $\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{3}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\frac{1}{8}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=0.125$
AIIMS-2005
NUCLEAR PHYSICS
147857
Half life of a substance is 20 minutes, then the time between $33 \%$ decay and $67 \%$ decay will be
1 20 minute
2 40 minute
3 50 minute
4 10 minute
Explanation:
A Let, the initial number of nuclei $=\mathrm{N}_{0}$ Let the time required for $33 \%$ decay $=t_{1}$ So, $0.67 \mathrm{~N}_{0}=\mathrm{N}_{0} \mathrm{e}^{-\lambda t_{1}}$ $\mathrm{e}^{-\lambda \mathrm{t}_{1}}=0.67$ And time required for $67 \%$ decay $=\mathrm{t}_{2}$ Similarly, $\mathrm{e}^{-\lambda \mathrm{t}_{2}}=0.33$ On dividing (ii) by (i) we get- $\mathrm{e}^{-\lambda\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right)}=\frac{0.33}{0.67}$ $\mathrm{e}^{-\lambda\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right)} = \frac{1}{2}$ Taking natural logarithm both the side, we get- $-\lambda\left(t_{2}-t_{1}\right)=\ln \frac{1}{2}$ $\mathrm{t}_{2}-\mathrm{t}_{1}=-\frac{\ln \left(\frac{1}{2}\right)}{\lambda}$ $\mathrm{t}_{2}-\mathrm{t}_{1}=-\frac{\ln \left(\frac{1}{2}\right)}{0.693} \times \mathrm{T}_{1 / 2} \quad\left(\because \lambda=\frac{0.693}{\mathrm{~T}_{1 / 2}}\right)$ $\text { Given, } T_{1 / 2}=20 \mathrm{~min}$ $\mathrm{t}_{2}-\mathrm{t}_{1}=-\frac{\ln \left(\frac{1}{2}\right) \times 20}{-\ln \left(\frac{1}{2}\right)}$ $\mathrm{t}_{2}-\mathrm{t}_{1}=20 \mathrm{~min}$
AIIMS-2000
NUCLEAR PHYSICS
147858
A nuclear decay is expressed as ${ }_{6} \mathrm{C}^{11} \longrightarrow \mathrm{B}^{11}+\boldsymbol{\beta}^{+}+\mathrm{X}$ Then the unknown particle $X$ is
1 Neutron
2 Antineutrino
3 Proton
4 Neutrino
Explanation:
D ${ }_{6} \mathrm{C}^{11} \rightarrow{ }_{5} \mathrm{~B}^{11}+{ }_{1} \beta^{0}+\mathrm{X}$ For atomic number $6=5+1+Z$ $\mathrm{Z}=0$ For mass number $11=11+0+\mathrm{A}$ $\mathrm{A}=0$ So, the unknown particle $\mathrm{X}$ is neutrino
AIPMT- 2000
NUCLEAR PHYSICS
147860
The activity of a radioactive sample is 1.6 curie and its half life is 2.5 days. The activity after 10 days will be :
1 0.16 curie
2 0.8 curie
3 0.1 curie
4 0.4 curie
Explanation:
C Given that, Time of decay $(\mathrm{t})=10$ days Half-life $\left(\mathrm{T}_{1 / 2}\right)=2.5$ day Initial radioactive nuclei $\left(\mathrm{N}_{\mathrm{o}}\right)=1.6$ curie We know that, $\mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}=\frac{10}{2.5}=4$ We know that, \(\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{n}}\) \(\mathrm{N}=1.6 \times\left(\frac{1}{2}\right)^4\) \(\mathrm{~N}=1.6 \times \frac{1}{16}\) \(\mathrm{~N}=0.1 \text { curie }\)
147855
A radioactive material decays by simultaneous emission of two particles with respective halflives 1620 and 810 years. The time, in years, after which one-fourth of the material remains is
1 1080
2 2430
3 3240
4 4860
Explanation:
A According to the law of Rutherford-Soddy. $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{n}=\frac{\text { time of decay }(\mathrm{t})}{\text { effective half }- \text { life }}$ The relation between effective disintegration constant $(\lambda)$ and $(T)$ $\lambda=\frac{\ln (2)}{\mathrm{T}}$ Hence, $\lambda_{1}+\lambda_{2}=\frac{\ln (2)}{\mathrm{T}_{1}}+\frac{\ln (2)}{\mathrm{T}_{2}}$ And effective half-life, $\frac{1}{\mathrm{~T}}=\frac{1}{\mathrm{~T}_{1}}+\frac{1}{\mathrm{~T}_{2}}$ $\frac{1}{\mathrm{~T}}=\frac{1}{1620}+\frac{1}{810}$ $\mathrm{T}=540 \text { years }$ So, $\mathrm{n}=\frac{\mathrm{t}}{540}$ From equation (i), we get - $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{4}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{540}}$ $\left(\frac{1}{2}\right)^{2}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{540}}$ On comparing both side we get - $\frac{t}{540}=2$ $t=1080 \text { years }$
AIIMS-2008
NUCLEAR PHYSICS
147856
A radioactive material has half - life of 10 days. What fraction of the material would remain after 30days ?
1 0.5
2 0.25
3 0.125
4 0.33
Explanation:
C Given that, Half-life $\left(\mathrm{T}_{1 / 2}\right)=10$ day Remain time $(\mathrm{t})=30$ day We know that, $\mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}$ $\mathrm{n}=\frac{30}{10}=3$ We know that, $\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{3}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\frac{1}{8}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=0.125$
AIIMS-2005
NUCLEAR PHYSICS
147857
Half life of a substance is 20 minutes, then the time between $33 \%$ decay and $67 \%$ decay will be
1 20 minute
2 40 minute
3 50 minute
4 10 minute
Explanation:
A Let, the initial number of nuclei $=\mathrm{N}_{0}$ Let the time required for $33 \%$ decay $=t_{1}$ So, $0.67 \mathrm{~N}_{0}=\mathrm{N}_{0} \mathrm{e}^{-\lambda t_{1}}$ $\mathrm{e}^{-\lambda \mathrm{t}_{1}}=0.67$ And time required for $67 \%$ decay $=\mathrm{t}_{2}$ Similarly, $\mathrm{e}^{-\lambda \mathrm{t}_{2}}=0.33$ On dividing (ii) by (i) we get- $\mathrm{e}^{-\lambda\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right)}=\frac{0.33}{0.67}$ $\mathrm{e}^{-\lambda\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right)} = \frac{1}{2}$ Taking natural logarithm both the side, we get- $-\lambda\left(t_{2}-t_{1}\right)=\ln \frac{1}{2}$ $\mathrm{t}_{2}-\mathrm{t}_{1}=-\frac{\ln \left(\frac{1}{2}\right)}{\lambda}$ $\mathrm{t}_{2}-\mathrm{t}_{1}=-\frac{\ln \left(\frac{1}{2}\right)}{0.693} \times \mathrm{T}_{1 / 2} \quad\left(\because \lambda=\frac{0.693}{\mathrm{~T}_{1 / 2}}\right)$ $\text { Given, } T_{1 / 2}=20 \mathrm{~min}$ $\mathrm{t}_{2}-\mathrm{t}_{1}=-\frac{\ln \left(\frac{1}{2}\right) \times 20}{-\ln \left(\frac{1}{2}\right)}$ $\mathrm{t}_{2}-\mathrm{t}_{1}=20 \mathrm{~min}$
AIIMS-2000
NUCLEAR PHYSICS
147858
A nuclear decay is expressed as ${ }_{6} \mathrm{C}^{11} \longrightarrow \mathrm{B}^{11}+\boldsymbol{\beta}^{+}+\mathrm{X}$ Then the unknown particle $X$ is
1 Neutron
2 Antineutrino
3 Proton
4 Neutrino
Explanation:
D ${ }_{6} \mathrm{C}^{11} \rightarrow{ }_{5} \mathrm{~B}^{11}+{ }_{1} \beta^{0}+\mathrm{X}$ For atomic number $6=5+1+Z$ $\mathrm{Z}=0$ For mass number $11=11+0+\mathrm{A}$ $\mathrm{A}=0$ So, the unknown particle $\mathrm{X}$ is neutrino
AIPMT- 2000
NUCLEAR PHYSICS
147860
The activity of a radioactive sample is 1.6 curie and its half life is 2.5 days. The activity after 10 days will be :
1 0.16 curie
2 0.8 curie
3 0.1 curie
4 0.4 curie
Explanation:
C Given that, Time of decay $(\mathrm{t})=10$ days Half-life $\left(\mathrm{T}_{1 / 2}\right)=2.5$ day Initial radioactive nuclei $\left(\mathrm{N}_{\mathrm{o}}\right)=1.6$ curie We know that, $\mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}=\frac{10}{2.5}=4$ We know that, \(\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{n}}\) \(\mathrm{N}=1.6 \times\left(\frac{1}{2}\right)^4\) \(\mathrm{~N}=1.6 \times \frac{1}{16}\) \(\mathrm{~N}=0.1 \text { curie }\)
147855
A radioactive material decays by simultaneous emission of two particles with respective halflives 1620 and 810 years. The time, in years, after which one-fourth of the material remains is
1 1080
2 2430
3 3240
4 4860
Explanation:
A According to the law of Rutherford-Soddy. $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{n}=\frac{\text { time of decay }(\mathrm{t})}{\text { effective half }- \text { life }}$ The relation between effective disintegration constant $(\lambda)$ and $(T)$ $\lambda=\frac{\ln (2)}{\mathrm{T}}$ Hence, $\lambda_{1}+\lambda_{2}=\frac{\ln (2)}{\mathrm{T}_{1}}+\frac{\ln (2)}{\mathrm{T}_{2}}$ And effective half-life, $\frac{1}{\mathrm{~T}}=\frac{1}{\mathrm{~T}_{1}}+\frac{1}{\mathrm{~T}_{2}}$ $\frac{1}{\mathrm{~T}}=\frac{1}{1620}+\frac{1}{810}$ $\mathrm{T}=540 \text { years }$ So, $\mathrm{n}=\frac{\mathrm{t}}{540}$ From equation (i), we get - $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{4}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{540}}$ $\left(\frac{1}{2}\right)^{2}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{540}}$ On comparing both side we get - $\frac{t}{540}=2$ $t=1080 \text { years }$
AIIMS-2008
NUCLEAR PHYSICS
147856
A radioactive material has half - life of 10 days. What fraction of the material would remain after 30days ?
1 0.5
2 0.25
3 0.125
4 0.33
Explanation:
C Given that, Half-life $\left(\mathrm{T}_{1 / 2}\right)=10$ day Remain time $(\mathrm{t})=30$ day We know that, $\mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}$ $\mathrm{n}=\frac{30}{10}=3$ We know that, $\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{3}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\frac{1}{8}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=0.125$
AIIMS-2005
NUCLEAR PHYSICS
147857
Half life of a substance is 20 minutes, then the time between $33 \%$ decay and $67 \%$ decay will be
1 20 minute
2 40 minute
3 50 minute
4 10 minute
Explanation:
A Let, the initial number of nuclei $=\mathrm{N}_{0}$ Let the time required for $33 \%$ decay $=t_{1}$ So, $0.67 \mathrm{~N}_{0}=\mathrm{N}_{0} \mathrm{e}^{-\lambda t_{1}}$ $\mathrm{e}^{-\lambda \mathrm{t}_{1}}=0.67$ And time required for $67 \%$ decay $=\mathrm{t}_{2}$ Similarly, $\mathrm{e}^{-\lambda \mathrm{t}_{2}}=0.33$ On dividing (ii) by (i) we get- $\mathrm{e}^{-\lambda\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right)}=\frac{0.33}{0.67}$ $\mathrm{e}^{-\lambda\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right)} = \frac{1}{2}$ Taking natural logarithm both the side, we get- $-\lambda\left(t_{2}-t_{1}\right)=\ln \frac{1}{2}$ $\mathrm{t}_{2}-\mathrm{t}_{1}=-\frac{\ln \left(\frac{1}{2}\right)}{\lambda}$ $\mathrm{t}_{2}-\mathrm{t}_{1}=-\frac{\ln \left(\frac{1}{2}\right)}{0.693} \times \mathrm{T}_{1 / 2} \quad\left(\because \lambda=\frac{0.693}{\mathrm{~T}_{1 / 2}}\right)$ $\text { Given, } T_{1 / 2}=20 \mathrm{~min}$ $\mathrm{t}_{2}-\mathrm{t}_{1}=-\frac{\ln \left(\frac{1}{2}\right) \times 20}{-\ln \left(\frac{1}{2}\right)}$ $\mathrm{t}_{2}-\mathrm{t}_{1}=20 \mathrm{~min}$
AIIMS-2000
NUCLEAR PHYSICS
147858
A nuclear decay is expressed as ${ }_{6} \mathrm{C}^{11} \longrightarrow \mathrm{B}^{11}+\boldsymbol{\beta}^{+}+\mathrm{X}$ Then the unknown particle $X$ is
1 Neutron
2 Antineutrino
3 Proton
4 Neutrino
Explanation:
D ${ }_{6} \mathrm{C}^{11} \rightarrow{ }_{5} \mathrm{~B}^{11}+{ }_{1} \beta^{0}+\mathrm{X}$ For atomic number $6=5+1+Z$ $\mathrm{Z}=0$ For mass number $11=11+0+\mathrm{A}$ $\mathrm{A}=0$ So, the unknown particle $\mathrm{X}$ is neutrino
AIPMT- 2000
NUCLEAR PHYSICS
147860
The activity of a radioactive sample is 1.6 curie and its half life is 2.5 days. The activity after 10 days will be :
1 0.16 curie
2 0.8 curie
3 0.1 curie
4 0.4 curie
Explanation:
C Given that, Time of decay $(\mathrm{t})=10$ days Half-life $\left(\mathrm{T}_{1 / 2}\right)=2.5$ day Initial radioactive nuclei $\left(\mathrm{N}_{\mathrm{o}}\right)=1.6$ curie We know that, $\mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}=\frac{10}{2.5}=4$ We know that, \(\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{n}}\) \(\mathrm{N}=1.6 \times\left(\frac{1}{2}\right)^4\) \(\mathrm{~N}=1.6 \times \frac{1}{16}\) \(\mathrm{~N}=0.1 \text { curie }\)
147855
A radioactive material decays by simultaneous emission of two particles with respective halflives 1620 and 810 years. The time, in years, after which one-fourth of the material remains is
1 1080
2 2430
3 3240
4 4860
Explanation:
A According to the law of Rutherford-Soddy. $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{n}=\frac{\text { time of decay }(\mathrm{t})}{\text { effective half }- \text { life }}$ The relation between effective disintegration constant $(\lambda)$ and $(T)$ $\lambda=\frac{\ln (2)}{\mathrm{T}}$ Hence, $\lambda_{1}+\lambda_{2}=\frac{\ln (2)}{\mathrm{T}_{1}}+\frac{\ln (2)}{\mathrm{T}_{2}}$ And effective half-life, $\frac{1}{\mathrm{~T}}=\frac{1}{\mathrm{~T}_{1}}+\frac{1}{\mathrm{~T}_{2}}$ $\frac{1}{\mathrm{~T}}=\frac{1}{1620}+\frac{1}{810}$ $\mathrm{T}=540 \text { years }$ So, $\mathrm{n}=\frac{\mathrm{t}}{540}$ From equation (i), we get - $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{4}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{540}}$ $\left(\frac{1}{2}\right)^{2}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{540}}$ On comparing both side we get - $\frac{t}{540}=2$ $t=1080 \text { years }$
AIIMS-2008
NUCLEAR PHYSICS
147856
A radioactive material has half - life of 10 days. What fraction of the material would remain after 30days ?
1 0.5
2 0.25
3 0.125
4 0.33
Explanation:
C Given that, Half-life $\left(\mathrm{T}_{1 / 2}\right)=10$ day Remain time $(\mathrm{t})=30$ day We know that, $\mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}$ $\mathrm{n}=\frac{30}{10}=3$ We know that, $\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{3}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\frac{1}{8}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=0.125$
AIIMS-2005
NUCLEAR PHYSICS
147857
Half life of a substance is 20 minutes, then the time between $33 \%$ decay and $67 \%$ decay will be
1 20 minute
2 40 minute
3 50 minute
4 10 minute
Explanation:
A Let, the initial number of nuclei $=\mathrm{N}_{0}$ Let the time required for $33 \%$ decay $=t_{1}$ So, $0.67 \mathrm{~N}_{0}=\mathrm{N}_{0} \mathrm{e}^{-\lambda t_{1}}$ $\mathrm{e}^{-\lambda \mathrm{t}_{1}}=0.67$ And time required for $67 \%$ decay $=\mathrm{t}_{2}$ Similarly, $\mathrm{e}^{-\lambda \mathrm{t}_{2}}=0.33$ On dividing (ii) by (i) we get- $\mathrm{e}^{-\lambda\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right)}=\frac{0.33}{0.67}$ $\mathrm{e}^{-\lambda\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right)} = \frac{1}{2}$ Taking natural logarithm both the side, we get- $-\lambda\left(t_{2}-t_{1}\right)=\ln \frac{1}{2}$ $\mathrm{t}_{2}-\mathrm{t}_{1}=-\frac{\ln \left(\frac{1}{2}\right)}{\lambda}$ $\mathrm{t}_{2}-\mathrm{t}_{1}=-\frac{\ln \left(\frac{1}{2}\right)}{0.693} \times \mathrm{T}_{1 / 2} \quad\left(\because \lambda=\frac{0.693}{\mathrm{~T}_{1 / 2}}\right)$ $\text { Given, } T_{1 / 2}=20 \mathrm{~min}$ $\mathrm{t}_{2}-\mathrm{t}_{1}=-\frac{\ln \left(\frac{1}{2}\right) \times 20}{-\ln \left(\frac{1}{2}\right)}$ $\mathrm{t}_{2}-\mathrm{t}_{1}=20 \mathrm{~min}$
AIIMS-2000
NUCLEAR PHYSICS
147858
A nuclear decay is expressed as ${ }_{6} \mathrm{C}^{11} \longrightarrow \mathrm{B}^{11}+\boldsymbol{\beta}^{+}+\mathrm{X}$ Then the unknown particle $X$ is
1 Neutron
2 Antineutrino
3 Proton
4 Neutrino
Explanation:
D ${ }_{6} \mathrm{C}^{11} \rightarrow{ }_{5} \mathrm{~B}^{11}+{ }_{1} \beta^{0}+\mathrm{X}$ For atomic number $6=5+1+Z$ $\mathrm{Z}=0$ For mass number $11=11+0+\mathrm{A}$ $\mathrm{A}=0$ So, the unknown particle $\mathrm{X}$ is neutrino
AIPMT- 2000
NUCLEAR PHYSICS
147860
The activity of a radioactive sample is 1.6 curie and its half life is 2.5 days. The activity after 10 days will be :
1 0.16 curie
2 0.8 curie
3 0.1 curie
4 0.4 curie
Explanation:
C Given that, Time of decay $(\mathrm{t})=10$ days Half-life $\left(\mathrm{T}_{1 / 2}\right)=2.5$ day Initial radioactive nuclei $\left(\mathrm{N}_{\mathrm{o}}\right)=1.6$ curie We know that, $\mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}=\frac{10}{2.5}=4$ We know that, \(\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{n}}\) \(\mathrm{N}=1.6 \times\left(\frac{1}{2}\right)^4\) \(\mathrm{~N}=1.6 \times \frac{1}{16}\) \(\mathrm{~N}=0.1 \text { curie }\)
