Radius $$R$$ as a nucleus changes with the nucleon number $$A$$ of the nucleus as
$$R = 1.3 \times {10^{ - 15}} \times {A^{\frac{1}{3}}}m$$
Hence, $$\frac{{{R_2}}}{{{R_1}}} = {\left( {\frac{{{A_2}}}{{{A_1}}}} \right)^{\frac{1}{3}}} = {\left( {\frac{{128}}{{16}}} \right)^{\frac{1}{3}}} = {\left( 8 \right)^{\frac{1}{3}}} = 2$$
$$\therefore {R_2} = 2{R_1} = 2\left( {3 \times {{10}^{ - 15}}} \right)m = 6 \times {10^{ - 15}}m$$

The relation between nuclei radius $$\left( R \right)$$  and mass number $$\left( A \right)$$  is given by
$$\eqalign{ & R \propto {A^{\frac{1}{3}}}\,......\left( {\text{i}} \right) \cr & {\text{or}}\,\,A \propto {R^3} \cr & {\text{or}}\,\,\frac{{{A_1}}}{{{A_2}}} = {\left( {\frac{{{R_1}}}{{{R_2}}}} \right)^3} \cr & {\text{Given,}}\,\,{R_1} = R,{R_2} = \frac{R}{2},A = 56 \cr & \therefore \frac{{56}}{{{A_2}}} = {\left( {\frac{R}{{\frac{R}{2}}}} \right)^3} = {2^3} = 8 \cr & {\text{or}}\,\,{A_2} = \frac{{56}}{8} = 7 \cr} $$
Thus, required stable nucleus will be $$L{i^7}.$$

Density of nuclear matter is the ratio of mass of nucleus and its volume.
If $$m$$ is average mass of a nucleon and $$R$$ is the nuclear radius, then mass of nucleus $$= mA,$$  where $$A$$ is the mass number of the element.
Volume of nucleus $$ = \frac{4}{3}\pi {R^3}$$
$$\eqalign{ & = \frac{4}{3}\pi {\left( {{R_0}{A^{\frac{1}{3}}}} \right)^3} \cr & = \frac{4}{3}\pi R_0^3A \cr} $$
As density of nuclear matter $$ = \frac{{{\text{mass of nucleus}}}}{{{\text{volume of nucleus}}}}$$
$$\eqalign{ & \rho = \frac{{mA}}{{\frac{4}{3}\pi R_0^3A}} \cr & \therefore \rho = \frac{{3m}}{{4\pi R_0^3}} \cr} $$
As $$m$$ and $${{R_0}}$$ are constants, therefore density $$\rho $$ of nuclear matter is constant.

$$\eqalign{ & P = n\left( {\frac{E}{t}} \right) \cr & \Rightarrow 1000 = \frac{{n \times 200 \times {{10}^6} \times 1.6 \times {{10}^{ - 19}}}}{t} \cr & \Rightarrow \frac{n}{t} = 3.125 \times {10^{13}}. \cr} $$

From law of conservation of momentum
$$\eqalign{ & 0 = {m_1}{v_1} + {m_2}{v_2} \cr & \therefore {m_1}{v_1} = - {m_2}{v_2} \cr & {\text{or}}\,\,\frac{{{v_1}}}{{{v_2}}} = - \frac{{{m_1}}}{{{m_2}}} = \frac{2}{1} \cr} $$
One sign indicates that velocity is in opposite direction
As nucleus is assumed to be spherical of radius $$r,$$ density $$\rho .$$
$$\therefore m = \frac{4}{3}\pi {r^3}\rho \Rightarrow m \propto {r^3}$$
So for two different parts of nuclei,
$$\eqalign{ & \frac{{{m_2}}}{{{m_1}}} = \frac{{r_2^3}}{{r_1^3}} \cr & \therefore \frac{{{v_1}}}{{{v_2}}} = \frac{{r_2^3}}{{r_1^3}}\,\,{\text{or}}\,\,\frac{{{r_1}}}{{{r_2}}} = {\left( {\frac{{{v_2}}}{{{v_1}}}} \right)^{\frac{1}{3}}} = {\left( {\frac{1}{2}} \right)^{\frac{1}{3}}} \cr & \Rightarrow {r_1}:{r_2} = 1:{2^{\frac{1}{3}}} \cr} $$

Average kinetic energy per molecule $$ = \frac{3}{2}kT$$
This kinetic energy should be able to provide the repulsive potential energy
$$\eqalign{ & \therefore \frac{3}{2}kT = 7.7 \times {10^{ - 14}} \cr & \Rightarrow T = \frac{{2 \times 7.7 \times {{10}^{ - 14}}}}{{3 \times 1.38 \times {{10}^{ - 23}}}} = 3.27 \times {10^9}K \cr} $$

The function of a moderator is to slow down the fast moving secondary neutrons produced during the fission as fission reaction can only be initiated by slow moving neutrons.
The material of moderator should be light and it should not absorb neutrons. Usually, heavy water, graphite, deuterium, paraffin etc. can act as moderators. These moderators are rich in protons.

When a nucleus is formed, then the mass of nucleus is slightly less than the sum of masses of $$Z$$ protons and $$N$$ neutrons.
i.e, $$m < \left( {Z{m_p} + N{m_n}} \right)$$

Nuclear radius, $$r \propto {A^{\frac{1}{3}}}$$
where $$A$$ is mass number
$$\eqalign{ & \frac{{{r_{Cu}}}}{{{r_{Al}}}} = {\left( {\frac{{{A_{Cu}}}}{{{A_{Al}}}}} \right)^{\frac{1}{3}}} = {\left( {\frac{{64}}{{27}}} \right)^{\frac{1}{3}}} \cr & \Rightarrow {r_{Cu}} = 3.6\left( {\frac{4}{3}} \right) = 4.8\,Fermi \cr} $$

It has been known that a nucleus of mass number $$A$$ has radius
$$R = {R_0}{A^{\frac{1}{3}}},$$
where $${R_0} = 1.2 \times {10^{ - 15}}m$$     and $$A=$$  mass number
In case of $$_{13}^{27}A\ell ,$$  let nuclear radius be $${R_1}$$ and for $$_{32}^{125}Te,$$  nuclear radius be $${R_2}$$
$$\eqalign{ & {\text{For}}\,\,_{13}^{27}Al,\,{R_1} = {R_0}{\left( {27} \right)^{\frac{1}{3}}} = 3{R_0} \cr & {\text{For}}\,\,_{32}^{125}Te,\,{R_2} = {R_0}{\left( {125} \right)^{\frac{1}{3}}} = 5{R_0} \cr & \frac{{{R_2}}}{{{R_1}}} = \frac{{5{R_0}}}{{3{R_0}}} = \frac{5}{3}{R_1} = \frac{5}{3} \times 3.6 = 6\,fm. \cr} $$