147415
If $M_{O}$ is the mass of an oxygen isotope ${ }_{8} \mathrm{O}^{17}, M_{p}$ and $M_{n}$ are the masses of a proton and a neutron, respectively, the nuclear binding energy of the isotope is
D Let for a nucleus of atomic mass A having $Z$ number of protons and $(\mathrm{A}-\mathrm{Z})$ number of neutrons then mass defect $(\Delta \mathrm{m})=\mathrm{ZM}_{\mathrm{P}}+(\mathrm{A}-\mathrm{Z}) \mathrm{M}_{\mathrm{n}}-\mathrm{M}_{\text {nucleus }}$ Number of proton in ${ }_{8} \mathrm{O}^{17}$ is $\mathrm{Z}=8$ Number of neutrons in ${ }_{8} \mathrm{O}^{17}$ is $\mathrm{A}-\mathrm{Z}=17-9=8$ $\Delta \mathrm{m}=\left(8 \mathrm{M}_{\mathrm{P}}+9 \mathrm{M}_{\mathrm{n}}-\mathrm{M}_{0}\right)$ Then binding energy $\mathrm{BE}=(\Delta \mathrm{m}) \mathrm{C}^{2}$ $=\left(8 M_{P}+9 M_{n}-M_{0}\right) C^{2}$
JIPMEER-2015
NUCLEAR PHYSICS
147418
The ratio of the radius of the nucleus of mass number 216 to the radius of the nucleus of mass number 64 is approximately
1 1.0
2 1.2
3 1.5
4 1.8
Explanation:
C Given, $A_{1}=216, A_{2}=64$ We know that, $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{\frac{1}{3}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{216}{64}\right)^{\frac{1}{3}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{6}{4}\right)^{3 \times \frac{1}{3}}=\frac{6}{4}=\frac{3}{2}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=1.5$
AMU-2008
NUCLEAR PHYSICS
147419
A beam of Beryllium nucleus $(z=4)$ of kinetic energy $5.3 \mathrm{MeV}$ is headed towards the nucleus of Gold atom $(z=79)$. What is the distance of closest approach?
1 $10.32 \times 10^{-14} \mathrm{~m}$
2 $8.58 \times 10^{-14} \mathrm{~m}$
3 $3.56 \times 10^{-14} \mathrm{~m}$
4 $1.25 \times 10^{-14} \mathrm{~m}$
Explanation:
B Given, Atomic number of gold $(Z)=79$ Atomic number of beryllium $(\mathrm{z})=4$ Kinetic energy $(\mathrm{K})=5.3 \mathrm{MeV}$ Distance of closest approach $(\mathrm{d})=\frac{1}{4 \pi \varepsilon_{0}} \frac{\text { Ze.ze }}{\mathrm{K}}$ $=\frac{9 \times 10^{9} \times 4 \times 1.6 \times 10^{-19} \times 79 \times 1.6 \times 10^{-19}}{5.3 \times 1.6 \times 10^{-19} \times 10^{6}}$ $\mathrm{~d}=8.58 \times 10^{-14} \mathrm{~m}$
AMU-2007
NUCLEAR PHYSICS
147420
The ratio of the radii of the nuclei ${ }_{13} \mathrm{Al}^{27}$ and ${ }_{52} \mathrm{Te}^{125}$ is
1 $13: 52$
2 $27: 125$
3 $3: 5$
4 $14: 73$
Explanation:
C Given, ${ }_{13} \mathrm{~A}^{27} { }_{52} \mathrm{Te}^{125}$ $\mathrm{~A}_1=27 \mathrm{~A}_2=125$ We know that, $\frac{\mathrm{R}_1}{\mathrm{R}_2}=\left(\frac{\mathrm{A}_1}{\mathrm{~A}_2}\right)^{\frac{1}{3}}=\left(\frac{27}{125}\right)^{\frac{1}{3}}=\left(\frac{3}{5}\right)^{3 \times \frac{1}{3}}$ So, \(\quad \mathrm{R}_1: \mathrm{R}_2=3: 5\)
AMU-2013
NUCLEAR PHYSICS
147425
In the nucleus of ${ }_{11} \mathrm{Na}^{23}$, the number of protons, neutrons and electrons are
1 $11,12,0$
2 $23,12,11$
3 $12,11,0$
4 $23,11,12$
Explanation:
A Given, Nucleus is of ${ }_{11} \mathrm{Na}^{23}$ So, Number of protons $=$ atomic number $(\mathrm{Z})=11$ Number of neutrons $=$ Mass number (A) - Atomic number $(\mathrm{Z})$ $=23-11=12$ Since electrons are not present in the nucleus therefore, number of electrons $=0$
147415
If $M_{O}$ is the mass of an oxygen isotope ${ }_{8} \mathrm{O}^{17}, M_{p}$ and $M_{n}$ are the masses of a proton and a neutron, respectively, the nuclear binding energy of the isotope is
D Let for a nucleus of atomic mass A having $Z$ number of protons and $(\mathrm{A}-\mathrm{Z})$ number of neutrons then mass defect $(\Delta \mathrm{m})=\mathrm{ZM}_{\mathrm{P}}+(\mathrm{A}-\mathrm{Z}) \mathrm{M}_{\mathrm{n}}-\mathrm{M}_{\text {nucleus }}$ Number of proton in ${ }_{8} \mathrm{O}^{17}$ is $\mathrm{Z}=8$ Number of neutrons in ${ }_{8} \mathrm{O}^{17}$ is $\mathrm{A}-\mathrm{Z}=17-9=8$ $\Delta \mathrm{m}=\left(8 \mathrm{M}_{\mathrm{P}}+9 \mathrm{M}_{\mathrm{n}}-\mathrm{M}_{0}\right)$ Then binding energy $\mathrm{BE}=(\Delta \mathrm{m}) \mathrm{C}^{2}$ $=\left(8 M_{P}+9 M_{n}-M_{0}\right) C^{2}$
JIPMEER-2015
NUCLEAR PHYSICS
147418
The ratio of the radius of the nucleus of mass number 216 to the radius of the nucleus of mass number 64 is approximately
1 1.0
2 1.2
3 1.5
4 1.8
Explanation:
C Given, $A_{1}=216, A_{2}=64$ We know that, $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{\frac{1}{3}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{216}{64}\right)^{\frac{1}{3}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{6}{4}\right)^{3 \times \frac{1}{3}}=\frac{6}{4}=\frac{3}{2}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=1.5$
AMU-2008
NUCLEAR PHYSICS
147419
A beam of Beryllium nucleus $(z=4)$ of kinetic energy $5.3 \mathrm{MeV}$ is headed towards the nucleus of Gold atom $(z=79)$. What is the distance of closest approach?
1 $10.32 \times 10^{-14} \mathrm{~m}$
2 $8.58 \times 10^{-14} \mathrm{~m}$
3 $3.56 \times 10^{-14} \mathrm{~m}$
4 $1.25 \times 10^{-14} \mathrm{~m}$
Explanation:
B Given, Atomic number of gold $(Z)=79$ Atomic number of beryllium $(\mathrm{z})=4$ Kinetic energy $(\mathrm{K})=5.3 \mathrm{MeV}$ Distance of closest approach $(\mathrm{d})=\frac{1}{4 \pi \varepsilon_{0}} \frac{\text { Ze.ze }}{\mathrm{K}}$ $=\frac{9 \times 10^{9} \times 4 \times 1.6 \times 10^{-19} \times 79 \times 1.6 \times 10^{-19}}{5.3 \times 1.6 \times 10^{-19} \times 10^{6}}$ $\mathrm{~d}=8.58 \times 10^{-14} \mathrm{~m}$
AMU-2007
NUCLEAR PHYSICS
147420
The ratio of the radii of the nuclei ${ }_{13} \mathrm{Al}^{27}$ and ${ }_{52} \mathrm{Te}^{125}$ is
1 $13: 52$
2 $27: 125$
3 $3: 5$
4 $14: 73$
Explanation:
C Given, ${ }_{13} \mathrm{~A}^{27} { }_{52} \mathrm{Te}^{125}$ $\mathrm{~A}_1=27 \mathrm{~A}_2=125$ We know that, $\frac{\mathrm{R}_1}{\mathrm{R}_2}=\left(\frac{\mathrm{A}_1}{\mathrm{~A}_2}\right)^{\frac{1}{3}}=\left(\frac{27}{125}\right)^{\frac{1}{3}}=\left(\frac{3}{5}\right)^{3 \times \frac{1}{3}}$ So, \(\quad \mathrm{R}_1: \mathrm{R}_2=3: 5\)
AMU-2013
NUCLEAR PHYSICS
147425
In the nucleus of ${ }_{11} \mathrm{Na}^{23}$, the number of protons, neutrons and electrons are
1 $11,12,0$
2 $23,12,11$
3 $12,11,0$
4 $23,11,12$
Explanation:
A Given, Nucleus is of ${ }_{11} \mathrm{Na}^{23}$ So, Number of protons $=$ atomic number $(\mathrm{Z})=11$ Number of neutrons $=$ Mass number (A) - Atomic number $(\mathrm{Z})$ $=23-11=12$ Since electrons are not present in the nucleus therefore, number of electrons $=0$
147415
If $M_{O}$ is the mass of an oxygen isotope ${ }_{8} \mathrm{O}^{17}, M_{p}$ and $M_{n}$ are the masses of a proton and a neutron, respectively, the nuclear binding energy of the isotope is
D Let for a nucleus of atomic mass A having $Z$ number of protons and $(\mathrm{A}-\mathrm{Z})$ number of neutrons then mass defect $(\Delta \mathrm{m})=\mathrm{ZM}_{\mathrm{P}}+(\mathrm{A}-\mathrm{Z}) \mathrm{M}_{\mathrm{n}}-\mathrm{M}_{\text {nucleus }}$ Number of proton in ${ }_{8} \mathrm{O}^{17}$ is $\mathrm{Z}=8$ Number of neutrons in ${ }_{8} \mathrm{O}^{17}$ is $\mathrm{A}-\mathrm{Z}=17-9=8$ $\Delta \mathrm{m}=\left(8 \mathrm{M}_{\mathrm{P}}+9 \mathrm{M}_{\mathrm{n}}-\mathrm{M}_{0}\right)$ Then binding energy $\mathrm{BE}=(\Delta \mathrm{m}) \mathrm{C}^{2}$ $=\left(8 M_{P}+9 M_{n}-M_{0}\right) C^{2}$
JIPMEER-2015
NUCLEAR PHYSICS
147418
The ratio of the radius of the nucleus of mass number 216 to the radius of the nucleus of mass number 64 is approximately
1 1.0
2 1.2
3 1.5
4 1.8
Explanation:
C Given, $A_{1}=216, A_{2}=64$ We know that, $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{\frac{1}{3}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{216}{64}\right)^{\frac{1}{3}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{6}{4}\right)^{3 \times \frac{1}{3}}=\frac{6}{4}=\frac{3}{2}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=1.5$
AMU-2008
NUCLEAR PHYSICS
147419
A beam of Beryllium nucleus $(z=4)$ of kinetic energy $5.3 \mathrm{MeV}$ is headed towards the nucleus of Gold atom $(z=79)$. What is the distance of closest approach?
1 $10.32 \times 10^{-14} \mathrm{~m}$
2 $8.58 \times 10^{-14} \mathrm{~m}$
3 $3.56 \times 10^{-14} \mathrm{~m}$
4 $1.25 \times 10^{-14} \mathrm{~m}$
Explanation:
B Given, Atomic number of gold $(Z)=79$ Atomic number of beryllium $(\mathrm{z})=4$ Kinetic energy $(\mathrm{K})=5.3 \mathrm{MeV}$ Distance of closest approach $(\mathrm{d})=\frac{1}{4 \pi \varepsilon_{0}} \frac{\text { Ze.ze }}{\mathrm{K}}$ $=\frac{9 \times 10^{9} \times 4 \times 1.6 \times 10^{-19} \times 79 \times 1.6 \times 10^{-19}}{5.3 \times 1.6 \times 10^{-19} \times 10^{6}}$ $\mathrm{~d}=8.58 \times 10^{-14} \mathrm{~m}$
AMU-2007
NUCLEAR PHYSICS
147420
The ratio of the radii of the nuclei ${ }_{13} \mathrm{Al}^{27}$ and ${ }_{52} \mathrm{Te}^{125}$ is
1 $13: 52$
2 $27: 125$
3 $3: 5$
4 $14: 73$
Explanation:
C Given, ${ }_{13} \mathrm{~A}^{27} { }_{52} \mathrm{Te}^{125}$ $\mathrm{~A}_1=27 \mathrm{~A}_2=125$ We know that, $\frac{\mathrm{R}_1}{\mathrm{R}_2}=\left(\frac{\mathrm{A}_1}{\mathrm{~A}_2}\right)^{\frac{1}{3}}=\left(\frac{27}{125}\right)^{\frac{1}{3}}=\left(\frac{3}{5}\right)^{3 \times \frac{1}{3}}$ So, \(\quad \mathrm{R}_1: \mathrm{R}_2=3: 5\)
AMU-2013
NUCLEAR PHYSICS
147425
In the nucleus of ${ }_{11} \mathrm{Na}^{23}$, the number of protons, neutrons and electrons are
1 $11,12,0$
2 $23,12,11$
3 $12,11,0$
4 $23,11,12$
Explanation:
A Given, Nucleus is of ${ }_{11} \mathrm{Na}^{23}$ So, Number of protons $=$ atomic number $(\mathrm{Z})=11$ Number of neutrons $=$ Mass number (A) - Atomic number $(\mathrm{Z})$ $=23-11=12$ Since electrons are not present in the nucleus therefore, number of electrons $=0$
147415
If $M_{O}$ is the mass of an oxygen isotope ${ }_{8} \mathrm{O}^{17}, M_{p}$ and $M_{n}$ are the masses of a proton and a neutron, respectively, the nuclear binding energy of the isotope is
D Let for a nucleus of atomic mass A having $Z$ number of protons and $(\mathrm{A}-\mathrm{Z})$ number of neutrons then mass defect $(\Delta \mathrm{m})=\mathrm{ZM}_{\mathrm{P}}+(\mathrm{A}-\mathrm{Z}) \mathrm{M}_{\mathrm{n}}-\mathrm{M}_{\text {nucleus }}$ Number of proton in ${ }_{8} \mathrm{O}^{17}$ is $\mathrm{Z}=8$ Number of neutrons in ${ }_{8} \mathrm{O}^{17}$ is $\mathrm{A}-\mathrm{Z}=17-9=8$ $\Delta \mathrm{m}=\left(8 \mathrm{M}_{\mathrm{P}}+9 \mathrm{M}_{\mathrm{n}}-\mathrm{M}_{0}\right)$ Then binding energy $\mathrm{BE}=(\Delta \mathrm{m}) \mathrm{C}^{2}$ $=\left(8 M_{P}+9 M_{n}-M_{0}\right) C^{2}$
JIPMEER-2015
NUCLEAR PHYSICS
147418
The ratio of the radius of the nucleus of mass number 216 to the radius of the nucleus of mass number 64 is approximately
1 1.0
2 1.2
3 1.5
4 1.8
Explanation:
C Given, $A_{1}=216, A_{2}=64$ We know that, $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{\frac{1}{3}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{216}{64}\right)^{\frac{1}{3}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{6}{4}\right)^{3 \times \frac{1}{3}}=\frac{6}{4}=\frac{3}{2}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=1.5$
AMU-2008
NUCLEAR PHYSICS
147419
A beam of Beryllium nucleus $(z=4)$ of kinetic energy $5.3 \mathrm{MeV}$ is headed towards the nucleus of Gold atom $(z=79)$. What is the distance of closest approach?
1 $10.32 \times 10^{-14} \mathrm{~m}$
2 $8.58 \times 10^{-14} \mathrm{~m}$
3 $3.56 \times 10^{-14} \mathrm{~m}$
4 $1.25 \times 10^{-14} \mathrm{~m}$
Explanation:
B Given, Atomic number of gold $(Z)=79$ Atomic number of beryllium $(\mathrm{z})=4$ Kinetic energy $(\mathrm{K})=5.3 \mathrm{MeV}$ Distance of closest approach $(\mathrm{d})=\frac{1}{4 \pi \varepsilon_{0}} \frac{\text { Ze.ze }}{\mathrm{K}}$ $=\frac{9 \times 10^{9} \times 4 \times 1.6 \times 10^{-19} \times 79 \times 1.6 \times 10^{-19}}{5.3 \times 1.6 \times 10^{-19} \times 10^{6}}$ $\mathrm{~d}=8.58 \times 10^{-14} \mathrm{~m}$
AMU-2007
NUCLEAR PHYSICS
147420
The ratio of the radii of the nuclei ${ }_{13} \mathrm{Al}^{27}$ and ${ }_{52} \mathrm{Te}^{125}$ is
1 $13: 52$
2 $27: 125$
3 $3: 5$
4 $14: 73$
Explanation:
C Given, ${ }_{13} \mathrm{~A}^{27} { }_{52} \mathrm{Te}^{125}$ $\mathrm{~A}_1=27 \mathrm{~A}_2=125$ We know that, $\frac{\mathrm{R}_1}{\mathrm{R}_2}=\left(\frac{\mathrm{A}_1}{\mathrm{~A}_2}\right)^{\frac{1}{3}}=\left(\frac{27}{125}\right)^{\frac{1}{3}}=\left(\frac{3}{5}\right)^{3 \times \frac{1}{3}}$ So, \(\quad \mathrm{R}_1: \mathrm{R}_2=3: 5\)
AMU-2013
NUCLEAR PHYSICS
147425
In the nucleus of ${ }_{11} \mathrm{Na}^{23}$, the number of protons, neutrons and electrons are
1 $11,12,0$
2 $23,12,11$
3 $12,11,0$
4 $23,11,12$
Explanation:
A Given, Nucleus is of ${ }_{11} \mathrm{Na}^{23}$ So, Number of protons $=$ atomic number $(\mathrm{Z})=11$ Number of neutrons $=$ Mass number (A) - Atomic number $(\mathrm{Z})$ $=23-11=12$ Since electrons are not present in the nucleus therefore, number of electrons $=0$
147415
If $M_{O}$ is the mass of an oxygen isotope ${ }_{8} \mathrm{O}^{17}, M_{p}$ and $M_{n}$ are the masses of a proton and a neutron, respectively, the nuclear binding energy of the isotope is
D Let for a nucleus of atomic mass A having $Z$ number of protons and $(\mathrm{A}-\mathrm{Z})$ number of neutrons then mass defect $(\Delta \mathrm{m})=\mathrm{ZM}_{\mathrm{P}}+(\mathrm{A}-\mathrm{Z}) \mathrm{M}_{\mathrm{n}}-\mathrm{M}_{\text {nucleus }}$ Number of proton in ${ }_{8} \mathrm{O}^{17}$ is $\mathrm{Z}=8$ Number of neutrons in ${ }_{8} \mathrm{O}^{17}$ is $\mathrm{A}-\mathrm{Z}=17-9=8$ $\Delta \mathrm{m}=\left(8 \mathrm{M}_{\mathrm{P}}+9 \mathrm{M}_{\mathrm{n}}-\mathrm{M}_{0}\right)$ Then binding energy $\mathrm{BE}=(\Delta \mathrm{m}) \mathrm{C}^{2}$ $=\left(8 M_{P}+9 M_{n}-M_{0}\right) C^{2}$
JIPMEER-2015
NUCLEAR PHYSICS
147418
The ratio of the radius of the nucleus of mass number 216 to the radius of the nucleus of mass number 64 is approximately
1 1.0
2 1.2
3 1.5
4 1.8
Explanation:
C Given, $A_{1}=216, A_{2}=64$ We know that, $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{\frac{1}{3}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{216}{64}\right)^{\frac{1}{3}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{6}{4}\right)^{3 \times \frac{1}{3}}=\frac{6}{4}=\frac{3}{2}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=1.5$
AMU-2008
NUCLEAR PHYSICS
147419
A beam of Beryllium nucleus $(z=4)$ of kinetic energy $5.3 \mathrm{MeV}$ is headed towards the nucleus of Gold atom $(z=79)$. What is the distance of closest approach?
1 $10.32 \times 10^{-14} \mathrm{~m}$
2 $8.58 \times 10^{-14} \mathrm{~m}$
3 $3.56 \times 10^{-14} \mathrm{~m}$
4 $1.25 \times 10^{-14} \mathrm{~m}$
Explanation:
B Given, Atomic number of gold $(Z)=79$ Atomic number of beryllium $(\mathrm{z})=4$ Kinetic energy $(\mathrm{K})=5.3 \mathrm{MeV}$ Distance of closest approach $(\mathrm{d})=\frac{1}{4 \pi \varepsilon_{0}} \frac{\text { Ze.ze }}{\mathrm{K}}$ $=\frac{9 \times 10^{9} \times 4 \times 1.6 \times 10^{-19} \times 79 \times 1.6 \times 10^{-19}}{5.3 \times 1.6 \times 10^{-19} \times 10^{6}}$ $\mathrm{~d}=8.58 \times 10^{-14} \mathrm{~m}$
AMU-2007
NUCLEAR PHYSICS
147420
The ratio of the radii of the nuclei ${ }_{13} \mathrm{Al}^{27}$ and ${ }_{52} \mathrm{Te}^{125}$ is
1 $13: 52$
2 $27: 125$
3 $3: 5$
4 $14: 73$
Explanation:
C Given, ${ }_{13} \mathrm{~A}^{27} { }_{52} \mathrm{Te}^{125}$ $\mathrm{~A}_1=27 \mathrm{~A}_2=125$ We know that, $\frac{\mathrm{R}_1}{\mathrm{R}_2}=\left(\frac{\mathrm{A}_1}{\mathrm{~A}_2}\right)^{\frac{1}{3}}=\left(\frac{27}{125}\right)^{\frac{1}{3}}=\left(\frac{3}{5}\right)^{3 \times \frac{1}{3}}$ So, \(\quad \mathrm{R}_1: \mathrm{R}_2=3: 5\)
AMU-2013
NUCLEAR PHYSICS
147425
In the nucleus of ${ }_{11} \mathrm{Na}^{23}$, the number of protons, neutrons and electrons are
1 $11,12,0$
2 $23,12,11$
3 $12,11,0$
4 $23,11,12$
Explanation:
A Given, Nucleus is of ${ }_{11} \mathrm{Na}^{23}$ So, Number of protons $=$ atomic number $(\mathrm{Z})=11$ Number of neutrons $=$ Mass number (A) - Atomic number $(\mathrm{Z})$ $=23-11=12$ Since electrons are not present in the nucleus therefore, number of electrons $=0$
