147386
A particle having almost zero mass and exactly zero charge is #[Qdiff: N/A, QCat: N/A, examname: 46. When $0.50 \AA \mathrm{X}$-rays strike a material, $=0.01302 \mathrm{AMU}$ , $=0.01302 \times 931 \mathrm{MeV}$ , $=12.13 \mathrm{MeV}$ ]#
1 positron
2 electron
3 neutron
4 neutrino
Explanation:
D We know that the neutrino is the particle which has almost zero mass and exactly zero charge. Positron has the positive charge and mass equal to $1.6 \times 10^{-27}$. Electron is a particle which has the negative charge and mass equal to $9.1 \times 10^{-31} \mathrm{~kg}$. Neutron is particle which has zero charge and mass equal to same as proton. Ans: d : Binding energy of last proton. $\mathrm{M}\left({ }_{8} \mathrm{O}^{16}\right)=\mathrm{M}\left({ }_{7} \mathrm{~N}^{15}\right)+{ }^{1} \mathrm{M}_{\mathrm{P}}$ Binding energy of last proton $=\mathrm{M}\left({ }_{7} \mathrm{~N}^{15}\right)+\mathrm{M}_{\mathrm{P}}-\mathrm{M}\left({ }_{8} \mathrm{O}^{16}\right)$ $=15.00011+1.00783-15.99492$ Ans: b : Given that, Magnetic field $(\mathrm{B})=2 \times 10^{-2} \mathrm{~T}$ Radius $(\mathrm{r})=23 \mathrm{~mm}=23 \times 10^{-3} \mathrm{~m}$ Wavelength $(\lambda)=0.50 \AA=0.50 \times 10^{-10} \mathrm{~m}$ We know that, $\mathrm{qvB}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}$ $\mathrm{v}=\frac{\mathrm{q}}{\mathrm{m}} \mathrm{Br}=\frac{\mathrm{e}}{\mathrm{m}} \mathrm{Br}$ The kinetic energy of the photoelectron $=\frac{1}{2} \mathrm{mv}^{2}$ $=\frac{1}{2} \mathrm{~m}\left(\frac{\mathrm{e}}{\mathrm{m}} \mathrm{Br}\right)^{2}$ $=\frac{1}{2} \mathrm{~m} \frac{\mathrm{e}^{2} \mathrm{~B}^{2} \mathrm{r}^{2}}{\mathrm{~m}^{2}}$ $=\frac{1}{2} \frac{\mathrm{e}^{2} \mathrm{~B}^{2} \mathrm{r}^{2}}{\mathrm{~m}}$ $=\frac{\frac{1}{2} \times\left(1.6 \times 10^{-19}\right)^{2} \times\left(2 \times 10^{-2}\right)^{2} \times\left(23 \times 10^{-3}\right)^{2}}{\mathrm{KE}=2.97 \times 10^{-15} \mathrm{~J}}$ $\mathrm{KE}=\frac{2.97 \times 10^{-31}}{1.6 \times 10^{-19}}=18.6 \mathrm{keV}$ Energy of the incident photon $\mathrm{E}=\frac{\mathrm{hc}}{\lambda}=\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{0.50 \times 10^{-10}}=39.75 \times 10^{-16} \mathrm{~J}$ $\mathrm{E}=\frac{39.75 \times 10^{-16}}{1.6 \times 10^{-19}}=24.8 \times 10^{3} \mathrm{eV}$ $\mathrm{E}=24.8 \mathrm{keV}$ Therefore, binding energy $=24.8-18.6=6.2 \mathrm{keV}$
NUCLEAR PHYSICS
147388
The natural boron of atomic weight 10.81 is found to have two isotopes $B^{10}$ and $B^{11}$. The ratio of abundance of isotopes in natural boron should be
1 $11: 10$
2 $81: 19$
3 $10: 11$
4 $15: 16$
5 $19: 81$
Explanation:
E Let abundance of $\mathrm{B}^{10}$ be $\mathrm{m} \%$ So abundance of $\mathrm{B}^{11}$ be $(100-\mathrm{m}) \%$ $\therefore 10.81=\frac{(10 \times \mathrm{m})+11(100-\mathrm{m})}{100}$ $1081=10 \mathrm{~m}+1100-11 \mathrm{~m}$ $\mathrm{~m}=19$ $\therefore$ Ratio of abundances $=\frac{\mathrm{B}^{10}}{\mathrm{~B}^{11}}=\frac{\mathrm{m}}{100-\mathrm{m}}$ $=\frac{19}{100-19}$ $=\frac{19}{81}$
Kerala CEE 2007
NUCLEAR PHYSICS
147390
The binding energy per nucleon of ${ }_{8} \mathrm{O}^{16}$ is 7.97 $\mathrm{MeV}$ and that ${ }_{8} \mathrm{O}^{\mathrm{I7}}$ of is $7.75 \mathrm{MeV}$. The energy required to remove one neutron from ${ }_{8} \mathrm{O}^{17}$ is $\mathrm{MeV}$.
1 3.52
2 3.62
3 4.23
4 7.86
Explanation:
C Binding energy per nucleon of ${ }_{8} \mathrm{O}^{16}=7.97$ $\mathrm{MeV}$ Binding energy per nucleon of ${ }_{8} \mathrm{O}^{17}=7.75 \mathrm{MeV}$ When one neutron is remove ${ }_{8} \mathrm{O}^{17} \rightarrow{ }_{8} \mathrm{O}^{16}+{ }_{0} \mathrm{n}^{1}$ Energy required $=$ Binding energy ${ }_{8} \mathrm{O}^{17}-$ Binding energy of ${ }_{8} \mathrm{O}^{16}$ $=17 \times 7.75-16 \times 7.97$ $=131.75-127.52$ $=4.23 \mathrm{MeV}$
GUJCET-2015
NUCLEAR PHYSICS
147392
Nuclear radius of ${ }^{27} \mathrm{Al}$ is $3 / 5$ times that of ${ }^{\mathrm{A}} \mathrm{X}$. Value of the mass number $A$ is
1 54
2 81
3 125
4 186
Explanation:
C Given that, atomic mass of $\mathrm{Al}$ and $\mathrm{X}$ is 27 and $\mathrm{A}$ $\therefore \quad \frac{\mathrm{R}_{\mathrm{Al}}}{\mathrm{R}_{\mathrm{X}}}=\frac{3}{5}$ Radius of nucleus is given by - $R=R_{0} A^{1 / 3}$ $\text { So, }R \propto A^{1 / 3}$ Ratio of radius of $\mathrm{Al}$ and $\mathrm{X}-$ $\therefore \quad \frac{\mathrm{R}_{\mathrm{Al}}}{\mathrm{R}_{\mathrm{X}}}=\left(\frac{{ }^{\mathrm{A}} \mathrm{Al}}{{ }^{\mathrm{A}} \mathrm{X}}\right)^{1 / 3}$ $\frac{3}{5}=\left(\frac{27}{{ }^{\mathrm{A}} \mathrm{X}}\right)^{1 / 3}$ $\left(\frac{3}{5}\right)^{3}=\frac{27}{\mathrm{~A}_{\mathrm{X}}}$ ${ }^{\mathrm{A}} \mathrm{X}=125$
NEET Test Series from KOTA - 10 Papers In MS WORD
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NUCLEAR PHYSICS
147386
A particle having almost zero mass and exactly zero charge is #[Qdiff: N/A, QCat: N/A, examname: 46. When $0.50 \AA \mathrm{X}$-rays strike a material, $=0.01302 \mathrm{AMU}$ , $=0.01302 \times 931 \mathrm{MeV}$ , $=12.13 \mathrm{MeV}$ ]#
1 positron
2 electron
3 neutron
4 neutrino
Explanation:
D We know that the neutrino is the particle which has almost zero mass and exactly zero charge. Positron has the positive charge and mass equal to $1.6 \times 10^{-27}$. Electron is a particle which has the negative charge and mass equal to $9.1 \times 10^{-31} \mathrm{~kg}$. Neutron is particle which has zero charge and mass equal to same as proton. Ans: d : Binding energy of last proton. $\mathrm{M}\left({ }_{8} \mathrm{O}^{16}\right)=\mathrm{M}\left({ }_{7} \mathrm{~N}^{15}\right)+{ }^{1} \mathrm{M}_{\mathrm{P}}$ Binding energy of last proton $=\mathrm{M}\left({ }_{7} \mathrm{~N}^{15}\right)+\mathrm{M}_{\mathrm{P}}-\mathrm{M}\left({ }_{8} \mathrm{O}^{16}\right)$ $=15.00011+1.00783-15.99492$ Ans: b : Given that, Magnetic field $(\mathrm{B})=2 \times 10^{-2} \mathrm{~T}$ Radius $(\mathrm{r})=23 \mathrm{~mm}=23 \times 10^{-3} \mathrm{~m}$ Wavelength $(\lambda)=0.50 \AA=0.50 \times 10^{-10} \mathrm{~m}$ We know that, $\mathrm{qvB}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}$ $\mathrm{v}=\frac{\mathrm{q}}{\mathrm{m}} \mathrm{Br}=\frac{\mathrm{e}}{\mathrm{m}} \mathrm{Br}$ The kinetic energy of the photoelectron $=\frac{1}{2} \mathrm{mv}^{2}$ $=\frac{1}{2} \mathrm{~m}\left(\frac{\mathrm{e}}{\mathrm{m}} \mathrm{Br}\right)^{2}$ $=\frac{1}{2} \mathrm{~m} \frac{\mathrm{e}^{2} \mathrm{~B}^{2} \mathrm{r}^{2}}{\mathrm{~m}^{2}}$ $=\frac{1}{2} \frac{\mathrm{e}^{2} \mathrm{~B}^{2} \mathrm{r}^{2}}{\mathrm{~m}}$ $=\frac{\frac{1}{2} \times\left(1.6 \times 10^{-19}\right)^{2} \times\left(2 \times 10^{-2}\right)^{2} \times\left(23 \times 10^{-3}\right)^{2}}{\mathrm{KE}=2.97 \times 10^{-15} \mathrm{~J}}$ $\mathrm{KE}=\frac{2.97 \times 10^{-31}}{1.6 \times 10^{-19}}=18.6 \mathrm{keV}$ Energy of the incident photon $\mathrm{E}=\frac{\mathrm{hc}}{\lambda}=\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{0.50 \times 10^{-10}}=39.75 \times 10^{-16} \mathrm{~J}$ $\mathrm{E}=\frac{39.75 \times 10^{-16}}{1.6 \times 10^{-19}}=24.8 \times 10^{3} \mathrm{eV}$ $\mathrm{E}=24.8 \mathrm{keV}$ Therefore, binding energy $=24.8-18.6=6.2 \mathrm{keV}$
NUCLEAR PHYSICS
147388
The natural boron of atomic weight 10.81 is found to have two isotopes $B^{10}$ and $B^{11}$. The ratio of abundance of isotopes in natural boron should be
1 $11: 10$
2 $81: 19$
3 $10: 11$
4 $15: 16$
5 $19: 81$
Explanation:
E Let abundance of $\mathrm{B}^{10}$ be $\mathrm{m} \%$ So abundance of $\mathrm{B}^{11}$ be $(100-\mathrm{m}) \%$ $\therefore 10.81=\frac{(10 \times \mathrm{m})+11(100-\mathrm{m})}{100}$ $1081=10 \mathrm{~m}+1100-11 \mathrm{~m}$ $\mathrm{~m}=19$ $\therefore$ Ratio of abundances $=\frac{\mathrm{B}^{10}}{\mathrm{~B}^{11}}=\frac{\mathrm{m}}{100-\mathrm{m}}$ $=\frac{19}{100-19}$ $=\frac{19}{81}$
Kerala CEE 2007
NUCLEAR PHYSICS
147390
The binding energy per nucleon of ${ }_{8} \mathrm{O}^{16}$ is 7.97 $\mathrm{MeV}$ and that ${ }_{8} \mathrm{O}^{\mathrm{I7}}$ of is $7.75 \mathrm{MeV}$. The energy required to remove one neutron from ${ }_{8} \mathrm{O}^{17}$ is $\mathrm{MeV}$.
1 3.52
2 3.62
3 4.23
4 7.86
Explanation:
C Binding energy per nucleon of ${ }_{8} \mathrm{O}^{16}=7.97$ $\mathrm{MeV}$ Binding energy per nucleon of ${ }_{8} \mathrm{O}^{17}=7.75 \mathrm{MeV}$ When one neutron is remove ${ }_{8} \mathrm{O}^{17} \rightarrow{ }_{8} \mathrm{O}^{16}+{ }_{0} \mathrm{n}^{1}$ Energy required $=$ Binding energy ${ }_{8} \mathrm{O}^{17}-$ Binding energy of ${ }_{8} \mathrm{O}^{16}$ $=17 \times 7.75-16 \times 7.97$ $=131.75-127.52$ $=4.23 \mathrm{MeV}$
GUJCET-2015
NUCLEAR PHYSICS
147392
Nuclear radius of ${ }^{27} \mathrm{Al}$ is $3 / 5$ times that of ${ }^{\mathrm{A}} \mathrm{X}$. Value of the mass number $A$ is
1 54
2 81
3 125
4 186
Explanation:
C Given that, atomic mass of $\mathrm{Al}$ and $\mathrm{X}$ is 27 and $\mathrm{A}$ $\therefore \quad \frac{\mathrm{R}_{\mathrm{Al}}}{\mathrm{R}_{\mathrm{X}}}=\frac{3}{5}$ Radius of nucleus is given by - $R=R_{0} A^{1 / 3}$ $\text { So, }R \propto A^{1 / 3}$ Ratio of radius of $\mathrm{Al}$ and $\mathrm{X}-$ $\therefore \quad \frac{\mathrm{R}_{\mathrm{Al}}}{\mathrm{R}_{\mathrm{X}}}=\left(\frac{{ }^{\mathrm{A}} \mathrm{Al}}{{ }^{\mathrm{A}} \mathrm{X}}\right)^{1 / 3}$ $\frac{3}{5}=\left(\frac{27}{{ }^{\mathrm{A}} \mathrm{X}}\right)^{1 / 3}$ $\left(\frac{3}{5}\right)^{3}=\frac{27}{\mathrm{~A}_{\mathrm{X}}}$ ${ }^{\mathrm{A}} \mathrm{X}=125$
147386
A particle having almost zero mass and exactly zero charge is #[Qdiff: N/A, QCat: N/A, examname: 46. When $0.50 \AA \mathrm{X}$-rays strike a material, $=0.01302 \mathrm{AMU}$ , $=0.01302 \times 931 \mathrm{MeV}$ , $=12.13 \mathrm{MeV}$ ]#
1 positron
2 electron
3 neutron
4 neutrino
Explanation:
D We know that the neutrino is the particle which has almost zero mass and exactly zero charge. Positron has the positive charge and mass equal to $1.6 \times 10^{-27}$. Electron is a particle which has the negative charge and mass equal to $9.1 \times 10^{-31} \mathrm{~kg}$. Neutron is particle which has zero charge and mass equal to same as proton. Ans: d : Binding energy of last proton. $\mathrm{M}\left({ }_{8} \mathrm{O}^{16}\right)=\mathrm{M}\left({ }_{7} \mathrm{~N}^{15}\right)+{ }^{1} \mathrm{M}_{\mathrm{P}}$ Binding energy of last proton $=\mathrm{M}\left({ }_{7} \mathrm{~N}^{15}\right)+\mathrm{M}_{\mathrm{P}}-\mathrm{M}\left({ }_{8} \mathrm{O}^{16}\right)$ $=15.00011+1.00783-15.99492$ Ans: b : Given that, Magnetic field $(\mathrm{B})=2 \times 10^{-2} \mathrm{~T}$ Radius $(\mathrm{r})=23 \mathrm{~mm}=23 \times 10^{-3} \mathrm{~m}$ Wavelength $(\lambda)=0.50 \AA=0.50 \times 10^{-10} \mathrm{~m}$ We know that, $\mathrm{qvB}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}$ $\mathrm{v}=\frac{\mathrm{q}}{\mathrm{m}} \mathrm{Br}=\frac{\mathrm{e}}{\mathrm{m}} \mathrm{Br}$ The kinetic energy of the photoelectron $=\frac{1}{2} \mathrm{mv}^{2}$ $=\frac{1}{2} \mathrm{~m}\left(\frac{\mathrm{e}}{\mathrm{m}} \mathrm{Br}\right)^{2}$ $=\frac{1}{2} \mathrm{~m} \frac{\mathrm{e}^{2} \mathrm{~B}^{2} \mathrm{r}^{2}}{\mathrm{~m}^{2}}$ $=\frac{1}{2} \frac{\mathrm{e}^{2} \mathrm{~B}^{2} \mathrm{r}^{2}}{\mathrm{~m}}$ $=\frac{\frac{1}{2} \times\left(1.6 \times 10^{-19}\right)^{2} \times\left(2 \times 10^{-2}\right)^{2} \times\left(23 \times 10^{-3}\right)^{2}}{\mathrm{KE}=2.97 \times 10^{-15} \mathrm{~J}}$ $\mathrm{KE}=\frac{2.97 \times 10^{-31}}{1.6 \times 10^{-19}}=18.6 \mathrm{keV}$ Energy of the incident photon $\mathrm{E}=\frac{\mathrm{hc}}{\lambda}=\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{0.50 \times 10^{-10}}=39.75 \times 10^{-16} \mathrm{~J}$ $\mathrm{E}=\frac{39.75 \times 10^{-16}}{1.6 \times 10^{-19}}=24.8 \times 10^{3} \mathrm{eV}$ $\mathrm{E}=24.8 \mathrm{keV}$ Therefore, binding energy $=24.8-18.6=6.2 \mathrm{keV}$
NUCLEAR PHYSICS
147388
The natural boron of atomic weight 10.81 is found to have two isotopes $B^{10}$ and $B^{11}$. The ratio of abundance of isotopes in natural boron should be
1 $11: 10$
2 $81: 19$
3 $10: 11$
4 $15: 16$
5 $19: 81$
Explanation:
E Let abundance of $\mathrm{B}^{10}$ be $\mathrm{m} \%$ So abundance of $\mathrm{B}^{11}$ be $(100-\mathrm{m}) \%$ $\therefore 10.81=\frac{(10 \times \mathrm{m})+11(100-\mathrm{m})}{100}$ $1081=10 \mathrm{~m}+1100-11 \mathrm{~m}$ $\mathrm{~m}=19$ $\therefore$ Ratio of abundances $=\frac{\mathrm{B}^{10}}{\mathrm{~B}^{11}}=\frac{\mathrm{m}}{100-\mathrm{m}}$ $=\frac{19}{100-19}$ $=\frac{19}{81}$
Kerala CEE 2007
NUCLEAR PHYSICS
147390
The binding energy per nucleon of ${ }_{8} \mathrm{O}^{16}$ is 7.97 $\mathrm{MeV}$ and that ${ }_{8} \mathrm{O}^{\mathrm{I7}}$ of is $7.75 \mathrm{MeV}$. The energy required to remove one neutron from ${ }_{8} \mathrm{O}^{17}$ is $\mathrm{MeV}$.
1 3.52
2 3.62
3 4.23
4 7.86
Explanation:
C Binding energy per nucleon of ${ }_{8} \mathrm{O}^{16}=7.97$ $\mathrm{MeV}$ Binding energy per nucleon of ${ }_{8} \mathrm{O}^{17}=7.75 \mathrm{MeV}$ When one neutron is remove ${ }_{8} \mathrm{O}^{17} \rightarrow{ }_{8} \mathrm{O}^{16}+{ }_{0} \mathrm{n}^{1}$ Energy required $=$ Binding energy ${ }_{8} \mathrm{O}^{17}-$ Binding energy of ${ }_{8} \mathrm{O}^{16}$ $=17 \times 7.75-16 \times 7.97$ $=131.75-127.52$ $=4.23 \mathrm{MeV}$
GUJCET-2015
NUCLEAR PHYSICS
147392
Nuclear radius of ${ }^{27} \mathrm{Al}$ is $3 / 5$ times that of ${ }^{\mathrm{A}} \mathrm{X}$. Value of the mass number $A$ is
1 54
2 81
3 125
4 186
Explanation:
C Given that, atomic mass of $\mathrm{Al}$ and $\mathrm{X}$ is 27 and $\mathrm{A}$ $\therefore \quad \frac{\mathrm{R}_{\mathrm{Al}}}{\mathrm{R}_{\mathrm{X}}}=\frac{3}{5}$ Radius of nucleus is given by - $R=R_{0} A^{1 / 3}$ $\text { So, }R \propto A^{1 / 3}$ Ratio of radius of $\mathrm{Al}$ and $\mathrm{X}-$ $\therefore \quad \frac{\mathrm{R}_{\mathrm{Al}}}{\mathrm{R}_{\mathrm{X}}}=\left(\frac{{ }^{\mathrm{A}} \mathrm{Al}}{{ }^{\mathrm{A}} \mathrm{X}}\right)^{1 / 3}$ $\frac{3}{5}=\left(\frac{27}{{ }^{\mathrm{A}} \mathrm{X}}\right)^{1 / 3}$ $\left(\frac{3}{5}\right)^{3}=\frac{27}{\mathrm{~A}_{\mathrm{X}}}$ ${ }^{\mathrm{A}} \mathrm{X}=125$
147386
A particle having almost zero mass and exactly zero charge is #[Qdiff: N/A, QCat: N/A, examname: 46. When $0.50 \AA \mathrm{X}$-rays strike a material, $=0.01302 \mathrm{AMU}$ , $=0.01302 \times 931 \mathrm{MeV}$ , $=12.13 \mathrm{MeV}$ ]#
1 positron
2 electron
3 neutron
4 neutrino
Explanation:
D We know that the neutrino is the particle which has almost zero mass and exactly zero charge. Positron has the positive charge and mass equal to $1.6 \times 10^{-27}$. Electron is a particle which has the negative charge and mass equal to $9.1 \times 10^{-31} \mathrm{~kg}$. Neutron is particle which has zero charge and mass equal to same as proton. Ans: d : Binding energy of last proton. $\mathrm{M}\left({ }_{8} \mathrm{O}^{16}\right)=\mathrm{M}\left({ }_{7} \mathrm{~N}^{15}\right)+{ }^{1} \mathrm{M}_{\mathrm{P}}$ Binding energy of last proton $=\mathrm{M}\left({ }_{7} \mathrm{~N}^{15}\right)+\mathrm{M}_{\mathrm{P}}-\mathrm{M}\left({ }_{8} \mathrm{O}^{16}\right)$ $=15.00011+1.00783-15.99492$ Ans: b : Given that, Magnetic field $(\mathrm{B})=2 \times 10^{-2} \mathrm{~T}$ Radius $(\mathrm{r})=23 \mathrm{~mm}=23 \times 10^{-3} \mathrm{~m}$ Wavelength $(\lambda)=0.50 \AA=0.50 \times 10^{-10} \mathrm{~m}$ We know that, $\mathrm{qvB}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}$ $\mathrm{v}=\frac{\mathrm{q}}{\mathrm{m}} \mathrm{Br}=\frac{\mathrm{e}}{\mathrm{m}} \mathrm{Br}$ The kinetic energy of the photoelectron $=\frac{1}{2} \mathrm{mv}^{2}$ $=\frac{1}{2} \mathrm{~m}\left(\frac{\mathrm{e}}{\mathrm{m}} \mathrm{Br}\right)^{2}$ $=\frac{1}{2} \mathrm{~m} \frac{\mathrm{e}^{2} \mathrm{~B}^{2} \mathrm{r}^{2}}{\mathrm{~m}^{2}}$ $=\frac{1}{2} \frac{\mathrm{e}^{2} \mathrm{~B}^{2} \mathrm{r}^{2}}{\mathrm{~m}}$ $=\frac{\frac{1}{2} \times\left(1.6 \times 10^{-19}\right)^{2} \times\left(2 \times 10^{-2}\right)^{2} \times\left(23 \times 10^{-3}\right)^{2}}{\mathrm{KE}=2.97 \times 10^{-15} \mathrm{~J}}$ $\mathrm{KE}=\frac{2.97 \times 10^{-31}}{1.6 \times 10^{-19}}=18.6 \mathrm{keV}$ Energy of the incident photon $\mathrm{E}=\frac{\mathrm{hc}}{\lambda}=\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{0.50 \times 10^{-10}}=39.75 \times 10^{-16} \mathrm{~J}$ $\mathrm{E}=\frac{39.75 \times 10^{-16}}{1.6 \times 10^{-19}}=24.8 \times 10^{3} \mathrm{eV}$ $\mathrm{E}=24.8 \mathrm{keV}$ Therefore, binding energy $=24.8-18.6=6.2 \mathrm{keV}$
NUCLEAR PHYSICS
147388
The natural boron of atomic weight 10.81 is found to have two isotopes $B^{10}$ and $B^{11}$. The ratio of abundance of isotopes in natural boron should be
1 $11: 10$
2 $81: 19$
3 $10: 11$
4 $15: 16$
5 $19: 81$
Explanation:
E Let abundance of $\mathrm{B}^{10}$ be $\mathrm{m} \%$ So abundance of $\mathrm{B}^{11}$ be $(100-\mathrm{m}) \%$ $\therefore 10.81=\frac{(10 \times \mathrm{m})+11(100-\mathrm{m})}{100}$ $1081=10 \mathrm{~m}+1100-11 \mathrm{~m}$ $\mathrm{~m}=19$ $\therefore$ Ratio of abundances $=\frac{\mathrm{B}^{10}}{\mathrm{~B}^{11}}=\frac{\mathrm{m}}{100-\mathrm{m}}$ $=\frac{19}{100-19}$ $=\frac{19}{81}$
Kerala CEE 2007
NUCLEAR PHYSICS
147390
The binding energy per nucleon of ${ }_{8} \mathrm{O}^{16}$ is 7.97 $\mathrm{MeV}$ and that ${ }_{8} \mathrm{O}^{\mathrm{I7}}$ of is $7.75 \mathrm{MeV}$. The energy required to remove one neutron from ${ }_{8} \mathrm{O}^{17}$ is $\mathrm{MeV}$.
1 3.52
2 3.62
3 4.23
4 7.86
Explanation:
C Binding energy per nucleon of ${ }_{8} \mathrm{O}^{16}=7.97$ $\mathrm{MeV}$ Binding energy per nucleon of ${ }_{8} \mathrm{O}^{17}=7.75 \mathrm{MeV}$ When one neutron is remove ${ }_{8} \mathrm{O}^{17} \rightarrow{ }_{8} \mathrm{O}^{16}+{ }_{0} \mathrm{n}^{1}$ Energy required $=$ Binding energy ${ }_{8} \mathrm{O}^{17}-$ Binding energy of ${ }_{8} \mathrm{O}^{16}$ $=17 \times 7.75-16 \times 7.97$ $=131.75-127.52$ $=4.23 \mathrm{MeV}$
GUJCET-2015
NUCLEAR PHYSICS
147392
Nuclear radius of ${ }^{27} \mathrm{Al}$ is $3 / 5$ times that of ${ }^{\mathrm{A}} \mathrm{X}$. Value of the mass number $A$ is
1 54
2 81
3 125
4 186
Explanation:
C Given that, atomic mass of $\mathrm{Al}$ and $\mathrm{X}$ is 27 and $\mathrm{A}$ $\therefore \quad \frac{\mathrm{R}_{\mathrm{Al}}}{\mathrm{R}_{\mathrm{X}}}=\frac{3}{5}$ Radius of nucleus is given by - $R=R_{0} A^{1 / 3}$ $\text { So, }R \propto A^{1 / 3}$ Ratio of radius of $\mathrm{Al}$ and $\mathrm{X}-$ $\therefore \quad \frac{\mathrm{R}_{\mathrm{Al}}}{\mathrm{R}_{\mathrm{X}}}=\left(\frac{{ }^{\mathrm{A}} \mathrm{Al}}{{ }^{\mathrm{A}} \mathrm{X}}\right)^{1 / 3}$ $\frac{3}{5}=\left(\frac{27}{{ }^{\mathrm{A}} \mathrm{X}}\right)^{1 / 3}$ $\left(\frac{3}{5}\right)^{3}=\frac{27}{\mathrm{~A}_{\mathrm{X}}}$ ${ }^{\mathrm{A}} \mathrm{X}=125$
