We have the expression for nuclear radius as:
R = R₀A¹/³
Where,
R₀ = Constant.
A = Mass number of the nucleus
Nuclear matter density, \(\rho \ =\frac{Mass\ of\ the\ nucles}{Volume\ of\ the\ nucles}\)
Let m be the average mass of the nucleus.
Hence, mass of the nucleus = mA
\(\therefore \rho=m \frac{A}{\frac{4}{3} \pi R^{3}}=\frac{3 \mathrm{~mA}}{4 \pi\left(R_{0} A^{\frac{1}{3}}\right)^{3}}=\frac{3 m A}{4 \pi R_{0}^{3} A}=\frac{3 \mathrm{~m}}{4 \pi R_{0}^{3}}\)
Hence, the nuclear matter density is independent of A. It is nearly constant.

1u = 1.6605 x 10^–27 kg
To convert it into energy units, we multiply it by c² and find that energy equivalent = \(1.6605 \times 10^{-27} \times\left(2.9979 \times 10^{8}\right)^{2} \mathrm{~kg} \mathrm{~m}^{2} / \mathrm{s}^{2}\)
\(=1.4924 \times 10^{-10} \mathrm{~J}\)
\(=\frac{1.4924 \times 10^{-10}}{1.602 \times 10^{-19}} \mathrm{eV}\)
\(=0.9315 \times 10^{9} \mathrm{eV}\)
\(=931.5 \mathrm{MeV}\)
or, \(1 \mathrm{u}=931.5 \mathrm{MeV} / \mathrm{c}^{2}\)
For, \({ }_{8}^{16} \mathrm{O}, \quad \Delta M=0.13691 \mathrm{u}=0.13691 \times 931.5 \mathrm{MeV} / \mathrm{c}^{2}\)
\(=127.5 \mathrm{MeV} / \mathrm{c}^{2}\)
The energy needed to separate \({ }_{8}^{16} \mathrm{O}\) into its constituents is thus 127.5 MeV/c².

Nuclear Fission The phenomenon of splitting of heavy nuclei (mass number> 120) into smaller nuclei of nearly equal masses is known as nuclear fission.
In nuclear fission, the sum of the masses  Y of the product is less than the sum of masses of the reactants. This difference of mass gets converted into energy E = me! and hence sample amount of energy is released in a nuclear fission.
e.g. \(_{ 235 }^{ 92 }{ U }+_{ 0 }^{ 1 }{ n }\rightarrow _{ 56 }^{ 141 }{ Ba }+_{ 36 }^{ 92 }{ Kr }+6_{ 0 }^{1 }{ n }+Q\)
Masses of reactant
= 235.0439 amu + 1.0087 amu
= 236.0526 amu
Masses of product
= 140.9139 + 91.8973 + 3.0261
= 235.8373 amu
Mass defect = 236.0526 - 235.8373
= 0.2153 amu
\(\because\)  1amu \(\equiv\) 931 MeV
\(\Rightarrow\) Energy released = 0.2153 x 931 = 200 MeV nearly
Thus, energy is liberated in nuclear fission \(_{92}^{235}{U}.\)

(i) Mass defect \((\triangle M)\) ,of any nucleus \(_{ Z }^{ A }{ X }\) is the sum of masses of the nucleus (=M) and the sum of masses of its constituent nucleons (=M')
\(\triangle M={ M }^{ ' }-M\)
\(=\left[ { Zm }_{ p }+(A-Z){ m }_{ n } \right] -M\)
where \({ m }_{ p }\) and \({ m }_{ n }\) denote the mass of the proton and the neutron respectively.
(ii) Binding energy is the energy required to seperate a nucleus into its constituent nucleons. The relation between the two is B.E=(mass defect) c²
(iii) There is a release of energy i.e., the reaction is exothermic. Reason: Increase in B.E/nucleon implies that more mass has been converted into energy.

Total B.E of \(_{ 6 }{ { C }^{ 12 } }\)\(​​​​​​=12\times 7.68=92.16\quad MeV\)
Total B.E of \(_{ 6 }{ { C }^{ 13 } }\)  \(=13\times 7.47=97.11\quad MeV\)
As \(_{ 6 }{ { C }^{ 13 } }\) has one excess neutron than \(_{ 6 }{ { C }^{ 12 } }\),
\(\therefore \) Energy required to remove a neutron
= 97.11-92.16 = 4.95 MeV

\(2.27\times { 10 }^{ -14 }m\)
Here, \(b=?\ KE=5MeV=5\times 1.6\times { 10 }^{ -13 }J\)
\(\theta =90°,\ \quad Z=79\)
\( b=\frac { Z{ e }^{ 2 }cot{ \theta }/{ 2 } }{ 4\pi { \epsilon }_{ 0 }(KE) } \)
\( =\frac { 9\times { 10 }^{ 9 }\times 79{ \left( 1.6\times { 10 }^{ -19 } \right) }^{ 2 }cot45° }{ 5\times 1.6\times { 10 }^{ -13 } }\)
\( b=2.27\times { 10 }^{ -14 }m\)

Since impact parameter is given by:
\(b={Ze^2cot{\theta\over2}\over4\pi\epsilon_0 E}\)
when b = 0, \(cot{\theta\over2}=0\)
or \({\theta\over2}=90^0\)
or  \(\theta =180^0\)
i.e. \(\alpha\) -particle retrace its path as its angle of scattering is 180⁰

Given, nuclear radius of  ²⁷Al, r₁ = 3.6 fm
Nuclear radius of ¹²⁵Fe, r₂ = ?
A₁ = 27,  A₂ = 125
The nuclear radius is given by 
\(R=R_0 A^{\frac{1}{3}} \Rightarrow R \propto A^{\frac{1}{3}} \)
\(\therefore\) \(\frac{R_2}{R_1}=\left(\frac{A_2}{A_1}\right)^{\frac{1}{3}}=\left(\frac{125}{27}\right)^{\frac{1}{3}}=\frac{5}{3}\)
\(\therefore\)\(R_2=\frac{5}{3} R_1=\frac{5}{3}\)x 3.6 = 6 fm
\(\therefore\) R₂ = 6 fm

Energy, \(E=10^{-3} \times\left(3 \times 10^{8}\right)^{2} \mathrm{~J}\)
\(E=10^{-3} \times 9 \times 10^{16}=9 \times 10^{13} \mathrm{~J}\)
Thus, if one gram of matter is converted to energy, there is a release of enormous amount of energy.

m_Fe = 55.85
u = 9.27 × 10^–26 kg
Nuclear density = \(\frac{\text { mass }}{\text { volume }}=\frac{9.27 \times 10^{-26}}{(4 \pi / 3)\left(1.2 \times 10^{-15}\right)^{3}} \times \frac{1}{56}\)
= 2.29 × 10¹⁷ kg m^–3
The density of matter in neutron stars (an astrophysical object) is comparable to this density. This shows that matter in these objects has been compressed to such an extent that they resemble a big nucleus.

\(\text { Isotopes }\left({ }_{15} P^{30}{ }_{, 15} P^{31}\right) \text {, Isotones }\left({ }_6 C^{13}{ }_7 N^{14}\right) \text {. }\)

Nuclear forces are the strongest forces in nature.They are effective only inside the nucleus.

When an electron of mass m is confined to move on a line of length l with velocity v, the de-Broglie wavelength \(\lambda \) associated with electron is  \(\lambda =\frac { h }{ mv } =\frac { h }{ p } \ \ or \ \ p=\frac { h }{ \lambda } =\frac { h }{ { 2l }/{ n } } =\frac { nh }{ 2l } \)
When electron revolves in a circular orbit of radius r; then \(2l=2\pi r\)
\( \ \therefore \ \ p=\frac { nh }{ 2\pi r } \\ \ \ or \ \ p\times r=\frac { nh }{ 2\pi } \)
i.e., angular momentum  \((p\times r)\) of electron is integral multiple of \({ h }/{ 2\pi }\) . This is Bohr's quantization condition of angular momentum.

On the ascending part of the curve, B.E./A increases with the mass number (A).
So, B.E. of the resultant will be greater than that of the nuclei which are fused together. Hence the fusion of the nuclei on the ascending part results in an increase in binding energy, therefore, more stable nucleus
On the descending part of the curve, the B.E. of the heavier nuclei is lower. So the fission of the heavier nuclei, on the descending part, will cause an increase in B.E. and, therefore, more stable nucleus

The given fusion reaction is:
\({ }_{1}^{2} \mathrm{H}+{ }_{1}^{2} \mathrm{H} \rightarrow{ }_{2}^{3} \mathrm{He}+\mathrm{n}+3.27 \mathrm{MeV}\)
Amount of deuterium, m = 2 kg
1 mole, i.e., 2 g of deuterium contains 6.023 x 10²³ atoms.
∴ 2.0 kg of deuterium contains \(=\frac{6.023 \times 10^{23}}{2} \times 2000=6.023 \times 10^{26} \text { atoms }\)
It can be inferred from the given reaction that when two atoms of deuterium fuse, 3.27 MeV energy is released.
∴Total energy per nucleus released in the fusion reaction:
\(E=\frac{3.27}{2} \times 6.023 \times 10^{26} \mathrm{MeV}\)
\(=\frac{3.27}{2} \times 6.023 \times 10^{26} \times 1.6 \times 10^{-19} \times 10^{6}\)
\(=1.576 \times 10^{14} J\)
Power of the electric lamp, P = 100 W = 100 J/s
Hence, the energy consumed by the lamp per second = 100 J
The total time for which the electric lamp will glow is calculated as:
\(\frac{1.576 \times 10^{14}}{100} s\)
\(\frac{1.576 \times 10^{14}}{100 \times 60 \times 60 \times 24 \times 365} \approx 4.9 \times 10^{4} \text { year }\)

Atomic mass of \(\begin{array}{l} 26 \\ 56 \end{array} \text { Fe }\),m₁ = 55.934939 u
\(\begin{array}{l} 26 \\ 56 \end{array} \text { Fe }\) nucleus has 26 protons and (56 − 26) = 30 neutrons
Hence, the mass defect of the nucleus,Δm = 26 x m_H + 30 x m_n − m₁
Where,
Mass of a proton, m_H = 1.007825 u
Mass of a neutron, m_n = 1.008665 u
∴Δm = 26 x 1.007825 + 30 x 1.008665 − 55.934939
= 26.20345 + 30.25995 − 55.934939
= 0.528461 u
But 1 u = 931.5 MeV/c²
∴Δm = 0.528461 x 931.5 MeV/c²
The binding energy of this nucleus is given as:
E_b1 = Δmc²
Where,
c = Speed of light
∴E_b1 = 0.528461 x 931.5 \(\left(\frac{M e V}{c^{2}}\right) \times c^{2}\)
= 492.26 MeV
Average binding energy per nucleon =\(\frac{492.26}{56}=8.79 \mathrm{MeV}\)
Atomic mass of \(\begin{array}{l} 209 \\ 83 \end{array} \text { Bi }\), m₂ = 208.980388 u
\(\begin{array}{l} 209 \\ 83 \end{array} \text { Bi }\) nucleus has 83 protons and (209 − 83) 126 neutrons.
Hence, the mass defect of this nucleus is given as:
Δm' = 83 x m_H + 126 x m_n − m₂
Where,
Mass of a proton, m_H = 1.007825 u
Mass of a neutron, m_n = 1.008665 u
∴Δm' = 83 x 1.007825 + 126 x 1.008665 − 208.980388
= 83.649475 + 127.091790 − 208.980388
= 1.760877 u
But 1 u = 931.5 MeV/c²
∴Δm' = 1.760877 x 931.5 MeV/c²
Hence, the binding energy of this nucleus is given as:
E_b2 = Δm'c²
= 1.760877 x 931.5 \(\left(\frac{M e V}{c^{2}}\right) \times c^{2}\)
= 1640.26 MeV
Average bindingenergy per nucleon \(=\frac{1640.26}{209}=7.848 \mathrm{MeV}\)

(i) Here,     \(\Delta m=1\quad a.m.u.=1.66\times { 10 }^{ -27 }kg\)
\(E=\left( \Delta m \right) { c }^{ 2 }=1.66\times { 10 }^{ -27 }{ \left( 3\times { 10 }^{ 8 } \right) }^{ 2 }=1.49\times { 10 }^{ -10 }J\)
(ii) Yes, in the phenomenon of pair production. Under suitable conditions, a photon materialises into an electron and a position :   \(\gamma ={ e }^{ -1 }+{ e }^{ +1 }\)
(iii) Einstein's relation, \(E=\left( \Delta m \right) { c }^{ 2 }\) emphasis that when certain mass disappears, an equivalent amount of energy appears. The reverse is also true. It Implies that to gain something, you have to lose another in equivalent amount. No one can have all gains together or all losses together. It also implies that nothing come for free. You have to pay the price in one form and acquire something in the desired form.

\(In \ _{ 26 }{ { Fe }^{ 56 } },\ no.of \ protons \ 26;\)
no. of neutrons = 56 - 26 = 30
\( \therefore \ Mass\ defect=26{ m }_{ p }+30{ m }_{ n }-{ M }_{ Fe }\)
\(=26\times 1.007825+30\times 1.008665-55.934939\)
\( =26.20345+30.25995-55.934939\)
\( =0.528461\ u\)
\( B.E/nucleon=\frac { 0.528461\times 931 }{ 56 } =8.79MeV/N\)

(i) (a) : As nearly 99.9% mass of atom is in nucleus.
\(\therefore \quad \frac{\text { Mass of nucleus }}{\text { Mass of atom }}=\frac{99.9}{100}=0.99 \approx 1\)
(ii) (a): Since, the nuclei of deuterium and tritium are isotopes of hydrogen, they must contain only one proton each. But the masses of the nuclei of hydrogen, deuterium and tritium are in the ration of 1 : 2 : 3,because of presence of neutral matter in deuterium and tritium nuclei.
(iii) (c)
(iv) (a) : \(R=R_{0} A^{1 / 3}\)
\(\log R=\log R_{0}+\frac{1}{3} \log A\)
On comparing the above equation of straight line; y = mx + c. So, the graph between log A and log R is a straight line also.
(v) (a): Here, A₁ = 197 and A₂ = 107
\(\therefore \quad \frac{R_{1}}{R_{2}}=\left(\frac{A_{1}}{A_{2}}\right)^{1 / 3}=\left(\frac{197}{107}\right)^{1 / 3}=1.225 \simeq 1.23\)

(i) (b)
(ii) (b): Fast neutrons are slowed down by elastic scattering with light nuclei as each collision takes away nearly 50% of energy.
(iii) (d): Reactions I and II represent fission of uranium isotope \({ }_{92}^{235} \mathrm{U}\), when bombarded with neutrons that breaks it into two intermediate mass nuclear fragments. However, reaction III represents two deuterons fuses together to from the light isotope of helium.
(iv) (c): On an average 2.5 neutrons are released per fission of the uranium atom.
The energy of the neutron released per fission of the uranium atom is 2 MeV.
(v) (a): In fission process, when a parent nucleus breaks. into daughter products, the some mass is lost in the form of energy. Thus,mass of fission products < mass of parent nucleus.
\(\Rightarrow \frac{\text { Mass of fission products }}{\text { Mass of parent nucleus }}<1\)

(b): Percentage of mass converted to energy in a fission reaction is 0.1% whereas in a fusion reaction it is 0.4%. Consequently the amount of energy released is more in a fusion than in a fission reaction. It is not easy to control a fusion reaction.

(a): According to the planetary model of the atom the mass of the atom is concentrate at the centre, the nucleus. The electrons orbit around the nucleus in circular paths of different radii. The radius of the outermost orbit .gives the size of atom. The size of the atom is \(\sim\) 10^-10m as compared to the size of the nucleus 10-¹⁵ m. Electrons being extremely light (me = 5.49 x 10-4 u = 9.11 x 10^-31kg) as compared to protons and neutrons have a negligible contribution to the weight of the atom. They mainly increase the volume of the atom. Therefore, the nuclear density is much higher that the atomic density (10³ kg/m³).

(c): Isotopes have same number of electrons and protons but different neutron number. That's why the mass number of isotopes are different and can be separated by using a mass spectrometer.