Nuclear force - class-XII

Energy released = (Total binding energy of products) - (Total binding energy of reactants). Reactants: 238 nucleons * 7.5 MeV = 1785 MeV. Products: 2 * (119 nucleons * 8.5 MeV) = 2023 MeV. Energy released = 2023 - 1785 = 238 MeV.

Answer is (B)

Nuclear forces properties :
1) They will act upto nearly one ferms. Beyond that they become negligible.
2) Charge independent : $F _{pp} = F _{pn} = F _{nn}$
3) They do not act along the line joining the centres of the nuclears, so they are non-central forces.
4) Nuclear forces are strong in nature.
So, correct choice is option A.

At separation less than one fermi,  the nuclear force of attraction is strongly active and Nuclear force is charge independent force.

Nuclear force is short range, charge independent and spin independent force.

Neutrons have no charge on them. So, there is no electrostatic force. But we have gravitational and nuclear forces on two neutrons because neutrons have masses and gravitational force depends on mass.
So, correct choice is option C.

Since, masses of electrons, protons and neutrons are negligible. So, we often neglect gravitational interaction in nuclear physics.

Two protons attracts each other when distance between them is less than $10^{-15}m.$

At separation greater than fermi, nuclear force is negligible. So, electromagnetic force is much powerful than nuclear force.
So, correct chioce is option B.

The forces that bind fermions (electron, proton, neutron) together are termed as weak forces. Whereas the forces that bind the constituents of a nucleus are strong nuclear force.

Electromagnetic force.

As we know that nucleus force is a strong force , they are strongest in magnitude. They are short range because the distance between the nucleon $0.7$ fermi which is very small.
They are charged independent. They result for interaction of every nucleon with the nearest limited number of nucleons.

Q = (BE of products) - (BE of reactants). Reactants: 2 * (2 * 1.15) = 4.6 MeV. Products: 4 * 7.1 = 28.4 MeV. Q = 28.4 - 4.6 = 23.8 MeV.

Nuclear force
The nuclear force (or nucleon interaction or residual strong force) is the force of protons and neutrons, subatomic particles that are collectively called nucleons. The nuclear force is responsible for binding protons and neutrons into atomic nuclei. Neutrons and protons are affected by the nuclear force almost identically. Since protons have charge +1 e, they experience a Coulomb repulsion that tends to push them apart, but at short range, the nuclear force is sufficiently attractive as to overcome the electromagnetic repulsive force. The mass of a nucleus is less than the sum total of the individual masses of the protons and neutrons which form it.
Gravitational force
This force acts b.w two particles having non zero mass
Electromagnetic force
Electromagnetism is the study of the electromagnetic force which is a type of physical interaction that occurs between electrically charged particles. 

Nuclear force
The nuclear force (or nucleon interaction or residual strong force) is the force of protons and neutrons, subatomic particles that are collectively called nucleons. The nuclear force is responsible for binding protons and neutrons into atomic nuclei. Neutrons and protons are affected by the nuclear force almost identically. Since protons have charge +1 e, they experience a Coulomb repulsion that tends to push them apart, but at short range the nuclear force is sufficiently attractive as to overcome the electromagnetic repulsive force. The mass of a nucleus is less than the sum total of the individual masses of the protons and neutrons which form it.
Gravitational force
This force acts b.w two particles having non zero mass
Electromagnetic force
Electromagnetism is the study of the electromagnetic force which is a type of physical interaction that occurs between electrically charged particles.

Electrons, protons and helium ions are deflected in the magnetic field.

Answer is (B).

Nuclear force is independent of charge of the nucleons.

All of the above.

Nuclear forces are the strongest forces in nature.
Nuclear forces are
the strong forces of attraction which hold together the nucleons
(neutrons and protons) in the tiny nucleus of an atom, inspite of strong
electrostatic forces of repulsion between protons. Nuclear forces act
between a pair of neutrons, a pair of protons and also between a
neutron, proton pair with the same strength. This shows that nuclear
forces are independent of charge.
The attractive nuclear force is the
same for any pair of nucleons. Thus, $F _{1} = F _{3}$ when there are
no electrostatic forces, but
$F _{2} =$ attractive nuclear force $-$ repulsive electrostatic force
Hence, $F _{1} = F _{3} > F _{2}$

A radioactive nucleus always decays into more stable nucleus and high binding energy corresponds to more stable nucleus. Therefore, after emitting an alpha particle energy $E _2$ will be more than the initial energy $E _1$.

Since neutron is neutral in nature, the electrostatic force between neutron and proton is zero. They have attractive nuclear force between them which causes the nucleus to bind together. 

The range of the weak nuclear force is extremely short, approximately 10^-18 meters.

Energy is released when the product has a higher specific binding energy than the reactants. (1) D(7 MeV) -> 2B(8.5 MeV): 8.5 > 7, so energy is released. (2) C(8 MeV) -> B(8.5 MeV) + A(5 MeV): Average BE is (8.5+5)/2 = 6.75 < 8, so energy is absorbed. (3) B(8.5 MeV) -> 2A(5 MeV): 5 < 8.5, so energy is absorbed.

At separation less than one fermi, hence nuclear force of attraction is strongly active.
Nuclear force is charge independent force.
So, $F _{pp} = F _{pn} = F _{nn}$

The expression for the radius of the nucleus is as shown below.
$r _{nucleus} =1.3\times 10^{-13}(A)^{1/3}$; where $A$ is mass number
Radius of  $ _{92}^{238}\textrm{U}=1.3\times10^{-13}\times (238)^{1/3}$
                           $= 8.06\times 10^{-13}cm$
Radius of $ _{2}^{4}\textrm{He}=1.3\times10^{-13}\times (4)^{1/3}$
                         $=2.06\times10^{-13}cm$
Total distance between uranium and helium nuclei is equal to the sum of their radii. 

Density of nucleus is: $\rho=\dfrac{A}{\dfrac{4}{3}\pi R^3}$
The radius of a nucleus, $R=r _0A^{1/3},$ so density of nucleus is independent of A and $R\propto {^3\sqrt{A}}$
The nuclear force is a short-range force because the distance between the nucleon is less than $0.7$ fermi (then the force is repulsive) and if greater than $10.7$ fermi (the force is attractive).