A nucleus of mass number $$A$$ and atomic number $$Z$$ contains $$Z$$ protons and $$\left( {A - Z} \right)$$  neutrons. As an atom is electrically neutral, therefore number of peripheral electrons must be equal to $$Z,$$ the number of protons inside the nucleus. In $$_{11}N{a^{23}},Z = 11$$    i.e. number of protons = 11,
Mass number $$A = 23$$
Number of neutrons $$ = A - Z = 23 - 11 = 12$$
There is no electron in the nucleus. So, number of protons, neutrons and electrons are 11, 12, 0.

Let the percentage of $${B^{10}}$$ atoms be $$x,$$ then average atomic weight
$$\eqalign{ & = \frac{{10x + 11\left( {100 - x} \right)}}{{100}} = 10.81 \cr & \Rightarrow x = 19 \cr & \therefore \frac{{{N_B}10}}{{{N_B}11}} = \frac{{19}}{{81}} \cr} $$

$$4_1^1{H^ + } \to _2^4H{e^{2 + }} + 2{e^ - } + 26MeV$$
represent a fusion reaction.

In fission process, when a parent nucleus breaks into daughter products, then some mass is lost in the form of energy. Thus, mass of fission products < mass of parent nucleus
$$ \Rightarrow \frac{{{\text{Mass of fission products}}}}{{{\text{Mass of parent nucleus}}}} < 1$$

We know that energy is released when heavy nuclei undergo fission or light nuclei undergo fusion. Therefore statement (1) is correct.
The second statement is false because for heavy nuclei the binding energy per nucleon decreases with increasing $$Z$$ and for light nuclei, $$B.E$$ /nucleon increases with increasing $$Z.$$

Given, $$\Delta m = 0.02866\,u$$
∴ Energy liberated $$ = \Delta m{c^2}$$
⇒ Energy liberated per $$u = \frac{{0.02866 \times 931}}{4} = \frac{{26.7}}{4}MeV$$
$$\eqalign{ & = \frac{{{E_b}}}{A} \cr & = 6.675\,MeV \cr} $$

Nuclear forces are the short range 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 are charge independent forces. They are the strongest forces in nature. The magnitude of nuclear forces is 100 times that of electrostatic forces and $${10^{38}}$$  times that of gravitational forces between nucleons. They are operative upto distances of the order of a few fermi. They are non-central forces.

Binding energy for light nuclei $$\left( {A < 20} \right)$$  is much smaller than the binding energy for heavier nuclei. This suggests a process that is reverse of fission. When two light nuclei combine to form a heavier nucleus, the process is called nuclear fusion. The union of two light nuclei into heavier nuclei also lead to a transfer of mass and a consequent liberation of large amount energy.

$$_Z{X^A}{ + _0}{n^1}{ \to _3}L{i^{\text{7}}}{ + _2}H{e^4}$$
On comparison,
$$A = 7 + 4 - 1 = 10,z = 3 + 2 - 0 = 5$$
It is boron $$_5{B^{10}}$$

We use the formula,
$$R = {R_0}{A^{\frac{1}{3}}}$$
This represents relation between atomic mass and radius of the nucleus.
For berillium, $${R_1} = {R_0}{\left( 9 \right)^{\frac{1}{3}}}$$
For germanium, $${R_2} = {R_0}{A^{\frac{1}{3}}}$$
$$\eqalign{ & \frac{{{R_1}}}{{{R_2}}} = \frac{{{{\left( 9 \right)}^{\frac{1}{3}}}}}{{{{\left( A \right)}^{\frac{1}{3}}}}} \cr & \Rightarrow \frac{1}{2} = \frac{{{{\left( 9 \right)}^{\frac{1}{3}}}}}{{{{\left( A \right)}^{\frac{1}{3}}}}} \cr & \Rightarrow \frac{1}{8} = \frac{9}{A} \cr & \Rightarrow A = 8 \times 9 = 72. \cr} $$