147498
The ratio of mass defect of the nucleus to its mass number is maximum for
1 $\mathrm{U}^{238}$
2 $\mathrm{N}^{14}$
3 $\mathrm{Si}^{28}$
4 $\mathrm{Fe}^{56}$
Explanation:
D Binding energy per nucleon is maximum for $\mathrm{Fe}^{56}$. Binding energy is directly related with mass defect. Therefore ratio of mass defect to mass number is maximum for $\mathrm{Fe}^{56}$.
J and K CET-2015
NUCLEAR PHYSICS
147502
For the stability of any nucleus
1 binding energy per nucleon will be more
2 binding energy per nucleon will be less
3 number of electrons will be more
4 none of the above
Explanation:
A The binding energy of a nucleus is the energy required to take its nucleons away from one another. It is generally expressed as binding energy per nucleon. It is a measure of the stability of the nucleus. Higher is the binding energy per nucleon, more will be the stability of the nucleus.
UP CPMT-2005
NUCLEAR PHYSICS
147503
$M_{P}$ and $M_{N}$ are masses of proton and neutron respectively, at rest. If they combine to form deuterium nucleus the mass of the nucleus will be
1 Less than $\mathrm{M}_{\mathrm{P}}$
2 Less than $\left(\mathrm{M}_{\mathrm{P}}+\mathrm{M}_{\mathrm{N}}\right)$
3 Less than $\left(\mathrm{M}_{\mathrm{P}}+2 \mathrm{M}_{\mathrm{N}}\right)$
4 Greater than $\left(\mathrm{M}_{\mathrm{P}}+2 \mathrm{M}_{\mathrm{N}}\right)$
Explanation:
B When a nucleus is formed then the mass of nucleus is slightly less than the sum of the mass of proton and neutron. $\mathrm{M} \lt \left(\mathrm{M}_{\mathrm{P}}+\mathrm{M}_{\mathrm{N}}\right)$
UP CPMT-2003
NUCLEAR PHYSICS
147504
If $F_{p p}, F_{n n}$ and $F_{p n}$ represent nuclear forces between proton-proton, neutron-neutron and proton-neutron respectively, then the correct relation is
B In the nucleus the uncharged neutrons and the positive charge protons are held together by the nuclear force in an extremely small space in spite of strong electrostatic repulsion between proton. Since nuclear forces are charge independent. Therefore the force between proton-neutrons, proton-protons and neutron neutron are the same in magnitude. So, $F_{p p}=f_{n n}=f_{p n}$
UP CPMT-2002
NUCLEAR PHYSICS
147506
The mass number of a nucleus is equal to number of
1 neutrons in nucleus
2 protons in nucleus
3 electrons in nucleus
4 nucleons in nucleus
Explanation:
D The mass number of the atom is equal to the sum of the number of protons and neutrons in the nucleus. The mass number of the atom is equal to the number of nucleons present in the nucleus of the atom.
147498
The ratio of mass defect of the nucleus to its mass number is maximum for
1 $\mathrm{U}^{238}$
2 $\mathrm{N}^{14}$
3 $\mathrm{Si}^{28}$
4 $\mathrm{Fe}^{56}$
Explanation:
D Binding energy per nucleon is maximum for $\mathrm{Fe}^{56}$. Binding energy is directly related with mass defect. Therefore ratio of mass defect to mass number is maximum for $\mathrm{Fe}^{56}$.
J and K CET-2015
NUCLEAR PHYSICS
147502
For the stability of any nucleus
1 binding energy per nucleon will be more
2 binding energy per nucleon will be less
3 number of electrons will be more
4 none of the above
Explanation:
A The binding energy of a nucleus is the energy required to take its nucleons away from one another. It is generally expressed as binding energy per nucleon. It is a measure of the stability of the nucleus. Higher is the binding energy per nucleon, more will be the stability of the nucleus.
UP CPMT-2005
NUCLEAR PHYSICS
147503
$M_{P}$ and $M_{N}$ are masses of proton and neutron respectively, at rest. If they combine to form deuterium nucleus the mass of the nucleus will be
1 Less than $\mathrm{M}_{\mathrm{P}}$
2 Less than $\left(\mathrm{M}_{\mathrm{P}}+\mathrm{M}_{\mathrm{N}}\right)$
3 Less than $\left(\mathrm{M}_{\mathrm{P}}+2 \mathrm{M}_{\mathrm{N}}\right)$
4 Greater than $\left(\mathrm{M}_{\mathrm{P}}+2 \mathrm{M}_{\mathrm{N}}\right)$
Explanation:
B When a nucleus is formed then the mass of nucleus is slightly less than the sum of the mass of proton and neutron. $\mathrm{M} \lt \left(\mathrm{M}_{\mathrm{P}}+\mathrm{M}_{\mathrm{N}}\right)$
UP CPMT-2003
NUCLEAR PHYSICS
147504
If $F_{p p}, F_{n n}$ and $F_{p n}$ represent nuclear forces between proton-proton, neutron-neutron and proton-neutron respectively, then the correct relation is
B In the nucleus the uncharged neutrons and the positive charge protons are held together by the nuclear force in an extremely small space in spite of strong electrostatic repulsion between proton. Since nuclear forces are charge independent. Therefore the force between proton-neutrons, proton-protons and neutron neutron are the same in magnitude. So, $F_{p p}=f_{n n}=f_{p n}$
UP CPMT-2002
NUCLEAR PHYSICS
147506
The mass number of a nucleus is equal to number of
1 neutrons in nucleus
2 protons in nucleus
3 electrons in nucleus
4 nucleons in nucleus
Explanation:
D The mass number of the atom is equal to the sum of the number of protons and neutrons in the nucleus. The mass number of the atom is equal to the number of nucleons present in the nucleus of the atom.
147498
The ratio of mass defect of the nucleus to its mass number is maximum for
1 $\mathrm{U}^{238}$
2 $\mathrm{N}^{14}$
3 $\mathrm{Si}^{28}$
4 $\mathrm{Fe}^{56}$
Explanation:
D Binding energy per nucleon is maximum for $\mathrm{Fe}^{56}$. Binding energy is directly related with mass defect. Therefore ratio of mass defect to mass number is maximum for $\mathrm{Fe}^{56}$.
J and K CET-2015
NUCLEAR PHYSICS
147502
For the stability of any nucleus
1 binding energy per nucleon will be more
2 binding energy per nucleon will be less
3 number of electrons will be more
4 none of the above
Explanation:
A The binding energy of a nucleus is the energy required to take its nucleons away from one another. It is generally expressed as binding energy per nucleon. It is a measure of the stability of the nucleus. Higher is the binding energy per nucleon, more will be the stability of the nucleus.
UP CPMT-2005
NUCLEAR PHYSICS
147503
$M_{P}$ and $M_{N}$ are masses of proton and neutron respectively, at rest. If they combine to form deuterium nucleus the mass of the nucleus will be
1 Less than $\mathrm{M}_{\mathrm{P}}$
2 Less than $\left(\mathrm{M}_{\mathrm{P}}+\mathrm{M}_{\mathrm{N}}\right)$
3 Less than $\left(\mathrm{M}_{\mathrm{P}}+2 \mathrm{M}_{\mathrm{N}}\right)$
4 Greater than $\left(\mathrm{M}_{\mathrm{P}}+2 \mathrm{M}_{\mathrm{N}}\right)$
Explanation:
B When a nucleus is formed then the mass of nucleus is slightly less than the sum of the mass of proton and neutron. $\mathrm{M} \lt \left(\mathrm{M}_{\mathrm{P}}+\mathrm{M}_{\mathrm{N}}\right)$
UP CPMT-2003
NUCLEAR PHYSICS
147504
If $F_{p p}, F_{n n}$ and $F_{p n}$ represent nuclear forces between proton-proton, neutron-neutron and proton-neutron respectively, then the correct relation is
B In the nucleus the uncharged neutrons and the positive charge protons are held together by the nuclear force in an extremely small space in spite of strong electrostatic repulsion between proton. Since nuclear forces are charge independent. Therefore the force between proton-neutrons, proton-protons and neutron neutron are the same in magnitude. So, $F_{p p}=f_{n n}=f_{p n}$
UP CPMT-2002
NUCLEAR PHYSICS
147506
The mass number of a nucleus is equal to number of
1 neutrons in nucleus
2 protons in nucleus
3 electrons in nucleus
4 nucleons in nucleus
Explanation:
D The mass number of the atom is equal to the sum of the number of protons and neutrons in the nucleus. The mass number of the atom is equal to the number of nucleons present in the nucleus of the atom.
147498
The ratio of mass defect of the nucleus to its mass number is maximum for
1 $\mathrm{U}^{238}$
2 $\mathrm{N}^{14}$
3 $\mathrm{Si}^{28}$
4 $\mathrm{Fe}^{56}$
Explanation:
D Binding energy per nucleon is maximum for $\mathrm{Fe}^{56}$. Binding energy is directly related with mass defect. Therefore ratio of mass defect to mass number is maximum for $\mathrm{Fe}^{56}$.
J and K CET-2015
NUCLEAR PHYSICS
147502
For the stability of any nucleus
1 binding energy per nucleon will be more
2 binding energy per nucleon will be less
3 number of electrons will be more
4 none of the above
Explanation:
A The binding energy of a nucleus is the energy required to take its nucleons away from one another. It is generally expressed as binding energy per nucleon. It is a measure of the stability of the nucleus. Higher is the binding energy per nucleon, more will be the stability of the nucleus.
UP CPMT-2005
NUCLEAR PHYSICS
147503
$M_{P}$ and $M_{N}$ are masses of proton and neutron respectively, at rest. If they combine to form deuterium nucleus the mass of the nucleus will be
1 Less than $\mathrm{M}_{\mathrm{P}}$
2 Less than $\left(\mathrm{M}_{\mathrm{P}}+\mathrm{M}_{\mathrm{N}}\right)$
3 Less than $\left(\mathrm{M}_{\mathrm{P}}+2 \mathrm{M}_{\mathrm{N}}\right)$
4 Greater than $\left(\mathrm{M}_{\mathrm{P}}+2 \mathrm{M}_{\mathrm{N}}\right)$
Explanation:
B When a nucleus is formed then the mass of nucleus is slightly less than the sum of the mass of proton and neutron. $\mathrm{M} \lt \left(\mathrm{M}_{\mathrm{P}}+\mathrm{M}_{\mathrm{N}}\right)$
UP CPMT-2003
NUCLEAR PHYSICS
147504
If $F_{p p}, F_{n n}$ and $F_{p n}$ represent nuclear forces between proton-proton, neutron-neutron and proton-neutron respectively, then the correct relation is
B In the nucleus the uncharged neutrons and the positive charge protons are held together by the nuclear force in an extremely small space in spite of strong electrostatic repulsion between proton. Since nuclear forces are charge independent. Therefore the force between proton-neutrons, proton-protons and neutron neutron are the same in magnitude. So, $F_{p p}=f_{n n}=f_{p n}$
UP CPMT-2002
NUCLEAR PHYSICS
147506
The mass number of a nucleus is equal to number of
1 neutrons in nucleus
2 protons in nucleus
3 electrons in nucleus
4 nucleons in nucleus
Explanation:
D The mass number of the atom is equal to the sum of the number of protons and neutrons in the nucleus. The mass number of the atom is equal to the number of nucleons present in the nucleus of the atom.
147498
The ratio of mass defect of the nucleus to its mass number is maximum for
1 $\mathrm{U}^{238}$
2 $\mathrm{N}^{14}$
3 $\mathrm{Si}^{28}$
4 $\mathrm{Fe}^{56}$
Explanation:
D Binding energy per nucleon is maximum for $\mathrm{Fe}^{56}$. Binding energy is directly related with mass defect. Therefore ratio of mass defect to mass number is maximum for $\mathrm{Fe}^{56}$.
J and K CET-2015
NUCLEAR PHYSICS
147502
For the stability of any nucleus
1 binding energy per nucleon will be more
2 binding energy per nucleon will be less
3 number of electrons will be more
4 none of the above
Explanation:
A The binding energy of a nucleus is the energy required to take its nucleons away from one another. It is generally expressed as binding energy per nucleon. It is a measure of the stability of the nucleus. Higher is the binding energy per nucleon, more will be the stability of the nucleus.
UP CPMT-2005
NUCLEAR PHYSICS
147503
$M_{P}$ and $M_{N}$ are masses of proton and neutron respectively, at rest. If they combine to form deuterium nucleus the mass of the nucleus will be
1 Less than $\mathrm{M}_{\mathrm{P}}$
2 Less than $\left(\mathrm{M}_{\mathrm{P}}+\mathrm{M}_{\mathrm{N}}\right)$
3 Less than $\left(\mathrm{M}_{\mathrm{P}}+2 \mathrm{M}_{\mathrm{N}}\right)$
4 Greater than $\left(\mathrm{M}_{\mathrm{P}}+2 \mathrm{M}_{\mathrm{N}}\right)$
Explanation:
B When a nucleus is formed then the mass of nucleus is slightly less than the sum of the mass of proton and neutron. $\mathrm{M} \lt \left(\mathrm{M}_{\mathrm{P}}+\mathrm{M}_{\mathrm{N}}\right)$
UP CPMT-2003
NUCLEAR PHYSICS
147504
If $F_{p p}, F_{n n}$ and $F_{p n}$ represent nuclear forces between proton-proton, neutron-neutron and proton-neutron respectively, then the correct relation is
B In the nucleus the uncharged neutrons and the positive charge protons are held together by the nuclear force in an extremely small space in spite of strong electrostatic repulsion between proton. Since nuclear forces are charge independent. Therefore the force between proton-neutrons, proton-protons and neutron neutron are the same in magnitude. So, $F_{p p}=f_{n n}=f_{p n}$
UP CPMT-2002
NUCLEAR PHYSICS
147506
The mass number of a nucleus is equal to number of
1 neutrons in nucleus
2 protons in nucleus
3 electrons in nucleus
4 nucleons in nucleus
Explanation:
D The mass number of the atom is equal to the sum of the number of protons and neutrons in the nucleus. The mass number of the atom is equal to the number of nucleons present in the nucleus of the atom.
