147722
When a radioactive isotope ${ }_{88} R^{228}$ decays in series by the emission of $3 \alpha$-particles and $\beta$ particles, the mass number of the isotope finally formed is-
1 83
2 72
3 216
4 228
Explanation:
C Decrease in mass number due to emission of 3 - $\alpha$ particles and $\beta$-particles $=3 \times 4=12$ Decreases in change number in the process - $=3 \times 2-1=5$ So, $\quad \mathrm{A}=228-12=216$
BCECE-2014
NUCLEAR PHYSICS
147730
The ratio of molecular masses of two radioactive substances is $3 / 2$ and the ratio of their decay constants is $4 / 3$. Then, the ratio of their initial activities per mole will be-
1 2
2 $4 / 3$
3 $8 / 9$
4 $9 / 8$
Explanation:
B $\operatorname{Activity}(\mathrm{A})=\frac{\mathrm{dN}}{\mathrm{dt}}=-\lambda \mathrm{N}$ As the number of nuclei $(\mathrm{N})$ per mole are equal for both the substances, irrespective of their molecular mass Therefore, $\mathrm{A} \propto \lambda$ $\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=\frac{\lambda_{1}}{\lambda_{2}}=\frac{4}{3}$
BCECE-2010
NUCLEAR PHYSICS
147733
A nuclear transformation is given by $Y(n, \alpha) \rightarrow_{3} \mathrm{Li}^{7}$. The nucleus of element $\mathrm{Y}$ is
1 ${ }_{5} \mathrm{Be}^{11}$
2 ${ }_{5} \mathrm{~B}^{10}$
3 ${ }_{5} \mathrm{~B}^{9}$
4 ${ }_{6} \mathrm{C}^{12}$
Explanation:
B The notation means that neutron is bombarded on the element then breaks into lithium and $\alpha$-particle. ${ }_{\mathrm{A}}^{\mathrm{Z}} \mathrm{X}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{\mathrm{A}-2}^{\mathrm{Z}-3} \mathrm{Y}+{ }_{2}^{4} \mathrm{He}$ ${ }_{\mathrm{A}-2}^{\mathrm{Z}-3} \mathrm{Y}={ }_{3}^{7} \mathrm{Li}$ $\therefore \quad \mathrm{Z}=10, \mathrm{~A}=5$ So, the element is boron and $\mathrm{X}$ is ${ }_{5} \mathrm{~B}^{10}$
VITEEE-2015
NUCLEAR PHYSICS
147738
In a radioactive decay, an element ${ }_{\mathrm{z}} \mathrm{X}^{\mathrm{A}}$ emits four $\alpha$-particles, three $\beta$-particles and eight gamma photons. The atomic number and mass number of the resulting final nucleus are :
1 $\mathrm{Z}-8, \mathrm{~A}-13$
2 $\mathrm{Z}-11, \mathrm{~A}-16$
3 $\mathrm{Z}-5, \mathrm{~A}-13$
4 $\mathrm{Z}-5, \mathrm{~A}-16$
Explanation:
D Alpha and beta particle are ${ }_{2} \mathrm{He}^{4}$ and ${ }_{-1} \mathrm{e}^{0}$ whereas atomic number as well as mass number of gamma photon is zero. So, nuclear reaction $\mathrm{z}^{\mathrm{A}} \rightarrow \mathrm{z}-5 \mathrm{Y}^{\mathrm{A}-16}+4\left({ }_{2} \mathrm{He}^{4}\right)+3\left({ }_{-1} \mathrm{e}^{0}\right)+8 \gamma$ Atomic number of final nucleus is $\mathrm{Z}-5$ Mass number of final nucleus is $\mathrm{A}-16$
Karnataka CET-2012
NUCLEAR PHYSICS
147747
In deuterium, $D$ and tritium, $T$ fusion reaction $\mathrm{D}+\mathrm{T} \rightarrow \mathrm{X}+\mathrm{n}, \mathrm{X}$ should be
1 ${ }_{1} \mathrm{H}^{3}$
2 ${ }_{1} \mathrm{H}^{2}$
3 ${ }_{2} \mathrm{He}^{4}$
4 ${ }_{1} \mathrm{H}^{1}$
Explanation:
C Considering deuterium (D) and tritium (T) fusion. ${ }_{1} \mathrm{H}^{2}+{ }_{1} \mathrm{H}^{3} \rightarrow{ }_{2} \mathrm{He}^{4}+\mathrm{n}$ So, $\mathrm{X}$ should be ${ }_{2} \mathrm{He}^{4}$.
147722
When a radioactive isotope ${ }_{88} R^{228}$ decays in series by the emission of $3 \alpha$-particles and $\beta$ particles, the mass number of the isotope finally formed is-
1 83
2 72
3 216
4 228
Explanation:
C Decrease in mass number due to emission of 3 - $\alpha$ particles and $\beta$-particles $=3 \times 4=12$ Decreases in change number in the process - $=3 \times 2-1=5$ So, $\quad \mathrm{A}=228-12=216$
BCECE-2014
NUCLEAR PHYSICS
147730
The ratio of molecular masses of two radioactive substances is $3 / 2$ and the ratio of their decay constants is $4 / 3$. Then, the ratio of their initial activities per mole will be-
1 2
2 $4 / 3$
3 $8 / 9$
4 $9 / 8$
Explanation:
B $\operatorname{Activity}(\mathrm{A})=\frac{\mathrm{dN}}{\mathrm{dt}}=-\lambda \mathrm{N}$ As the number of nuclei $(\mathrm{N})$ per mole are equal for both the substances, irrespective of their molecular mass Therefore, $\mathrm{A} \propto \lambda$ $\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=\frac{\lambda_{1}}{\lambda_{2}}=\frac{4}{3}$
BCECE-2010
NUCLEAR PHYSICS
147733
A nuclear transformation is given by $Y(n, \alpha) \rightarrow_{3} \mathrm{Li}^{7}$. The nucleus of element $\mathrm{Y}$ is
1 ${ }_{5} \mathrm{Be}^{11}$
2 ${ }_{5} \mathrm{~B}^{10}$
3 ${ }_{5} \mathrm{~B}^{9}$
4 ${ }_{6} \mathrm{C}^{12}$
Explanation:
B The notation means that neutron is bombarded on the element then breaks into lithium and $\alpha$-particle. ${ }_{\mathrm{A}}^{\mathrm{Z}} \mathrm{X}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{\mathrm{A}-2}^{\mathrm{Z}-3} \mathrm{Y}+{ }_{2}^{4} \mathrm{He}$ ${ }_{\mathrm{A}-2}^{\mathrm{Z}-3} \mathrm{Y}={ }_{3}^{7} \mathrm{Li}$ $\therefore \quad \mathrm{Z}=10, \mathrm{~A}=5$ So, the element is boron and $\mathrm{X}$ is ${ }_{5} \mathrm{~B}^{10}$
VITEEE-2015
NUCLEAR PHYSICS
147738
In a radioactive decay, an element ${ }_{\mathrm{z}} \mathrm{X}^{\mathrm{A}}$ emits four $\alpha$-particles, three $\beta$-particles and eight gamma photons. The atomic number and mass number of the resulting final nucleus are :
1 $\mathrm{Z}-8, \mathrm{~A}-13$
2 $\mathrm{Z}-11, \mathrm{~A}-16$
3 $\mathrm{Z}-5, \mathrm{~A}-13$
4 $\mathrm{Z}-5, \mathrm{~A}-16$
Explanation:
D Alpha and beta particle are ${ }_{2} \mathrm{He}^{4}$ and ${ }_{-1} \mathrm{e}^{0}$ whereas atomic number as well as mass number of gamma photon is zero. So, nuclear reaction $\mathrm{z}^{\mathrm{A}} \rightarrow \mathrm{z}-5 \mathrm{Y}^{\mathrm{A}-16}+4\left({ }_{2} \mathrm{He}^{4}\right)+3\left({ }_{-1} \mathrm{e}^{0}\right)+8 \gamma$ Atomic number of final nucleus is $\mathrm{Z}-5$ Mass number of final nucleus is $\mathrm{A}-16$
Karnataka CET-2012
NUCLEAR PHYSICS
147747
In deuterium, $D$ and tritium, $T$ fusion reaction $\mathrm{D}+\mathrm{T} \rightarrow \mathrm{X}+\mathrm{n}, \mathrm{X}$ should be
1 ${ }_{1} \mathrm{H}^{3}$
2 ${ }_{1} \mathrm{H}^{2}$
3 ${ }_{2} \mathrm{He}^{4}$
4 ${ }_{1} \mathrm{H}^{1}$
Explanation:
C Considering deuterium (D) and tritium (T) fusion. ${ }_{1} \mathrm{H}^{2}+{ }_{1} \mathrm{H}^{3} \rightarrow{ }_{2} \mathrm{He}^{4}+\mathrm{n}$ So, $\mathrm{X}$ should be ${ }_{2} \mathrm{He}^{4}$.
147722
When a radioactive isotope ${ }_{88} R^{228}$ decays in series by the emission of $3 \alpha$-particles and $\beta$ particles, the mass number of the isotope finally formed is-
1 83
2 72
3 216
4 228
Explanation:
C Decrease in mass number due to emission of 3 - $\alpha$ particles and $\beta$-particles $=3 \times 4=12$ Decreases in change number in the process - $=3 \times 2-1=5$ So, $\quad \mathrm{A}=228-12=216$
BCECE-2014
NUCLEAR PHYSICS
147730
The ratio of molecular masses of two radioactive substances is $3 / 2$ and the ratio of their decay constants is $4 / 3$. Then, the ratio of their initial activities per mole will be-
1 2
2 $4 / 3$
3 $8 / 9$
4 $9 / 8$
Explanation:
B $\operatorname{Activity}(\mathrm{A})=\frac{\mathrm{dN}}{\mathrm{dt}}=-\lambda \mathrm{N}$ As the number of nuclei $(\mathrm{N})$ per mole are equal for both the substances, irrespective of their molecular mass Therefore, $\mathrm{A} \propto \lambda$ $\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=\frac{\lambda_{1}}{\lambda_{2}}=\frac{4}{3}$
BCECE-2010
NUCLEAR PHYSICS
147733
A nuclear transformation is given by $Y(n, \alpha) \rightarrow_{3} \mathrm{Li}^{7}$. The nucleus of element $\mathrm{Y}$ is
1 ${ }_{5} \mathrm{Be}^{11}$
2 ${ }_{5} \mathrm{~B}^{10}$
3 ${ }_{5} \mathrm{~B}^{9}$
4 ${ }_{6} \mathrm{C}^{12}$
Explanation:
B The notation means that neutron is bombarded on the element then breaks into lithium and $\alpha$-particle. ${ }_{\mathrm{A}}^{\mathrm{Z}} \mathrm{X}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{\mathrm{A}-2}^{\mathrm{Z}-3} \mathrm{Y}+{ }_{2}^{4} \mathrm{He}$ ${ }_{\mathrm{A}-2}^{\mathrm{Z}-3} \mathrm{Y}={ }_{3}^{7} \mathrm{Li}$ $\therefore \quad \mathrm{Z}=10, \mathrm{~A}=5$ So, the element is boron and $\mathrm{X}$ is ${ }_{5} \mathrm{~B}^{10}$
VITEEE-2015
NUCLEAR PHYSICS
147738
In a radioactive decay, an element ${ }_{\mathrm{z}} \mathrm{X}^{\mathrm{A}}$ emits four $\alpha$-particles, three $\beta$-particles and eight gamma photons. The atomic number and mass number of the resulting final nucleus are :
1 $\mathrm{Z}-8, \mathrm{~A}-13$
2 $\mathrm{Z}-11, \mathrm{~A}-16$
3 $\mathrm{Z}-5, \mathrm{~A}-13$
4 $\mathrm{Z}-5, \mathrm{~A}-16$
Explanation:
D Alpha and beta particle are ${ }_{2} \mathrm{He}^{4}$ and ${ }_{-1} \mathrm{e}^{0}$ whereas atomic number as well as mass number of gamma photon is zero. So, nuclear reaction $\mathrm{z}^{\mathrm{A}} \rightarrow \mathrm{z}-5 \mathrm{Y}^{\mathrm{A}-16}+4\left({ }_{2} \mathrm{He}^{4}\right)+3\left({ }_{-1} \mathrm{e}^{0}\right)+8 \gamma$ Atomic number of final nucleus is $\mathrm{Z}-5$ Mass number of final nucleus is $\mathrm{A}-16$
Karnataka CET-2012
NUCLEAR PHYSICS
147747
In deuterium, $D$ and tritium, $T$ fusion reaction $\mathrm{D}+\mathrm{T} \rightarrow \mathrm{X}+\mathrm{n}, \mathrm{X}$ should be
1 ${ }_{1} \mathrm{H}^{3}$
2 ${ }_{1} \mathrm{H}^{2}$
3 ${ }_{2} \mathrm{He}^{4}$
4 ${ }_{1} \mathrm{H}^{1}$
Explanation:
C Considering deuterium (D) and tritium (T) fusion. ${ }_{1} \mathrm{H}^{2}+{ }_{1} \mathrm{H}^{3} \rightarrow{ }_{2} \mathrm{He}^{4}+\mathrm{n}$ So, $\mathrm{X}$ should be ${ }_{2} \mathrm{He}^{4}$.
147722
When a radioactive isotope ${ }_{88} R^{228}$ decays in series by the emission of $3 \alpha$-particles and $\beta$ particles, the mass number of the isotope finally formed is-
1 83
2 72
3 216
4 228
Explanation:
C Decrease in mass number due to emission of 3 - $\alpha$ particles and $\beta$-particles $=3 \times 4=12$ Decreases in change number in the process - $=3 \times 2-1=5$ So, $\quad \mathrm{A}=228-12=216$
BCECE-2014
NUCLEAR PHYSICS
147730
The ratio of molecular masses of two radioactive substances is $3 / 2$ and the ratio of their decay constants is $4 / 3$. Then, the ratio of their initial activities per mole will be-
1 2
2 $4 / 3$
3 $8 / 9$
4 $9 / 8$
Explanation:
B $\operatorname{Activity}(\mathrm{A})=\frac{\mathrm{dN}}{\mathrm{dt}}=-\lambda \mathrm{N}$ As the number of nuclei $(\mathrm{N})$ per mole are equal for both the substances, irrespective of their molecular mass Therefore, $\mathrm{A} \propto \lambda$ $\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=\frac{\lambda_{1}}{\lambda_{2}}=\frac{4}{3}$
BCECE-2010
NUCLEAR PHYSICS
147733
A nuclear transformation is given by $Y(n, \alpha) \rightarrow_{3} \mathrm{Li}^{7}$. The nucleus of element $\mathrm{Y}$ is
1 ${ }_{5} \mathrm{Be}^{11}$
2 ${ }_{5} \mathrm{~B}^{10}$
3 ${ }_{5} \mathrm{~B}^{9}$
4 ${ }_{6} \mathrm{C}^{12}$
Explanation:
B The notation means that neutron is bombarded on the element then breaks into lithium and $\alpha$-particle. ${ }_{\mathrm{A}}^{\mathrm{Z}} \mathrm{X}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{\mathrm{A}-2}^{\mathrm{Z}-3} \mathrm{Y}+{ }_{2}^{4} \mathrm{He}$ ${ }_{\mathrm{A}-2}^{\mathrm{Z}-3} \mathrm{Y}={ }_{3}^{7} \mathrm{Li}$ $\therefore \quad \mathrm{Z}=10, \mathrm{~A}=5$ So, the element is boron and $\mathrm{X}$ is ${ }_{5} \mathrm{~B}^{10}$
VITEEE-2015
NUCLEAR PHYSICS
147738
In a radioactive decay, an element ${ }_{\mathrm{z}} \mathrm{X}^{\mathrm{A}}$ emits four $\alpha$-particles, three $\beta$-particles and eight gamma photons. The atomic number and mass number of the resulting final nucleus are :
1 $\mathrm{Z}-8, \mathrm{~A}-13$
2 $\mathrm{Z}-11, \mathrm{~A}-16$
3 $\mathrm{Z}-5, \mathrm{~A}-13$
4 $\mathrm{Z}-5, \mathrm{~A}-16$
Explanation:
D Alpha and beta particle are ${ }_{2} \mathrm{He}^{4}$ and ${ }_{-1} \mathrm{e}^{0}$ whereas atomic number as well as mass number of gamma photon is zero. So, nuclear reaction $\mathrm{z}^{\mathrm{A}} \rightarrow \mathrm{z}-5 \mathrm{Y}^{\mathrm{A}-16}+4\left({ }_{2} \mathrm{He}^{4}\right)+3\left({ }_{-1} \mathrm{e}^{0}\right)+8 \gamma$ Atomic number of final nucleus is $\mathrm{Z}-5$ Mass number of final nucleus is $\mathrm{A}-16$
Karnataka CET-2012
NUCLEAR PHYSICS
147747
In deuterium, $D$ and tritium, $T$ fusion reaction $\mathrm{D}+\mathrm{T} \rightarrow \mathrm{X}+\mathrm{n}, \mathrm{X}$ should be
1 ${ }_{1} \mathrm{H}^{3}$
2 ${ }_{1} \mathrm{H}^{2}$
3 ${ }_{2} \mathrm{He}^{4}$
4 ${ }_{1} \mathrm{H}^{1}$
Explanation:
C Considering deuterium (D) and tritium (T) fusion. ${ }_{1} \mathrm{H}^{2}+{ }_{1} \mathrm{H}^{3} \rightarrow{ }_{2} \mathrm{He}^{4}+\mathrm{n}$ So, $\mathrm{X}$ should be ${ }_{2} \mathrm{He}^{4}$.
147722
When a radioactive isotope ${ }_{88} R^{228}$ decays in series by the emission of $3 \alpha$-particles and $\beta$ particles, the mass number of the isotope finally formed is-
1 83
2 72
3 216
4 228
Explanation:
C Decrease in mass number due to emission of 3 - $\alpha$ particles and $\beta$-particles $=3 \times 4=12$ Decreases in change number in the process - $=3 \times 2-1=5$ So, $\quad \mathrm{A}=228-12=216$
BCECE-2014
NUCLEAR PHYSICS
147730
The ratio of molecular masses of two radioactive substances is $3 / 2$ and the ratio of their decay constants is $4 / 3$. Then, the ratio of their initial activities per mole will be-
1 2
2 $4 / 3$
3 $8 / 9$
4 $9 / 8$
Explanation:
B $\operatorname{Activity}(\mathrm{A})=\frac{\mathrm{dN}}{\mathrm{dt}}=-\lambda \mathrm{N}$ As the number of nuclei $(\mathrm{N})$ per mole are equal for both the substances, irrespective of their molecular mass Therefore, $\mathrm{A} \propto \lambda$ $\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=\frac{\lambda_{1}}{\lambda_{2}}=\frac{4}{3}$
BCECE-2010
NUCLEAR PHYSICS
147733
A nuclear transformation is given by $Y(n, \alpha) \rightarrow_{3} \mathrm{Li}^{7}$. The nucleus of element $\mathrm{Y}$ is
1 ${ }_{5} \mathrm{Be}^{11}$
2 ${ }_{5} \mathrm{~B}^{10}$
3 ${ }_{5} \mathrm{~B}^{9}$
4 ${ }_{6} \mathrm{C}^{12}$
Explanation:
B The notation means that neutron is bombarded on the element then breaks into lithium and $\alpha$-particle. ${ }_{\mathrm{A}}^{\mathrm{Z}} \mathrm{X}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{\mathrm{A}-2}^{\mathrm{Z}-3} \mathrm{Y}+{ }_{2}^{4} \mathrm{He}$ ${ }_{\mathrm{A}-2}^{\mathrm{Z}-3} \mathrm{Y}={ }_{3}^{7} \mathrm{Li}$ $\therefore \quad \mathrm{Z}=10, \mathrm{~A}=5$ So, the element is boron and $\mathrm{X}$ is ${ }_{5} \mathrm{~B}^{10}$
VITEEE-2015
NUCLEAR PHYSICS
147738
In a radioactive decay, an element ${ }_{\mathrm{z}} \mathrm{X}^{\mathrm{A}}$ emits four $\alpha$-particles, three $\beta$-particles and eight gamma photons. The atomic number and mass number of the resulting final nucleus are :
1 $\mathrm{Z}-8, \mathrm{~A}-13$
2 $\mathrm{Z}-11, \mathrm{~A}-16$
3 $\mathrm{Z}-5, \mathrm{~A}-13$
4 $\mathrm{Z}-5, \mathrm{~A}-16$
Explanation:
D Alpha and beta particle are ${ }_{2} \mathrm{He}^{4}$ and ${ }_{-1} \mathrm{e}^{0}$ whereas atomic number as well as mass number of gamma photon is zero. So, nuclear reaction $\mathrm{z}^{\mathrm{A}} \rightarrow \mathrm{z}-5 \mathrm{Y}^{\mathrm{A}-16}+4\left({ }_{2} \mathrm{He}^{4}\right)+3\left({ }_{-1} \mathrm{e}^{0}\right)+8 \gamma$ Atomic number of final nucleus is $\mathrm{Z}-5$ Mass number of final nucleus is $\mathrm{A}-16$
Karnataka CET-2012
NUCLEAR PHYSICS
147747
In deuterium, $D$ and tritium, $T$ fusion reaction $\mathrm{D}+\mathrm{T} \rightarrow \mathrm{X}+\mathrm{n}, \mathrm{X}$ should be
1 ${ }_{1} \mathrm{H}^{3}$
2 ${ }_{1} \mathrm{H}^{2}$
3 ${ }_{2} \mathrm{He}^{4}$
4 ${ }_{1} \mathrm{H}^{1}$
Explanation:
C Considering deuterium (D) and tritium (T) fusion. ${ }_{1} \mathrm{H}^{2}+{ }_{1} \mathrm{H}^{3} \rightarrow{ }_{2} \mathrm{He}^{4}+\mathrm{n}$ So, $\mathrm{X}$ should be ${ }_{2} \mathrm{He}^{4}$.
