(i) The process of breaking up of the nucleus of a heavier atom into two smaller nuclei with the release of a large amount of energy is called nuclear fission.
Examples :
\(_{ 92 }^{ 235 }{ U+ }_{ 0 }^{ 1 }{ n }\rightarrow _{ 92 }^{ 236 }{ { U }^{ * } }\rightarrow _{ 56 }^{ 141 }{ Ba }+_{ 36 }^{ 92 }{ Kr }+3_{ 0 }^{ 1 }{ n }+Q\)
\(_{ 92 }^{ 235 }{ U+ }_{ 0 }^{ 1 }{ n }\rightarrow _{ 92 }^{ 236 }{ { U }^{ * } }\rightarrow _{ 54 }^{ 140 }Xe+_{ 38 }^{ 94 }{ Sr }+2_{ 0 }^{ 1 }{ n }+Q\)
(ii) When the slow neutron is absorbed by the uranium nuclei, the mass number increases by one and goes to an excited state \(_{ 92 }^{ 235 }{ U }\).
(iii) But this excited state does not last longer than 10^-12s and decay into two daughter nuclei along with 2 or 3 neutrons.
Energy released in fission :
(i) We can calculate the energy (Q) released in each uranium fission reaction. We choose the most observed fission reaction which is given in the equation.

\(_{ 92 }^{ 235 }{ U+ }_{ 0 }^{ 1 }{ n }\rightarrow _{ 92 }^{ 236 }{ { U }^{ * } }\rightarrow _{ 56 }^{ 141 }{ Ba }+_{ 36 }^{ 92 }{ Kr }+3_{ 0 }^{ 1 }{ n }+Q\)
Mass of \({ }_{92}^{235} \mathrm{U}\) = 235.045733 u
Mass of \(_{ 0 }^{ 1 }{ n }\) = 1.008665 u
Total mass of reactants = 236.054398 u
Mass of \(_{ 56 }^{ 141 }{ Ba }\) = 140.9177 u
Mass of \(_{ 92 }^{ 36 }{ Kr }\) = 91.8854u
Mass of 3 neutrons = 3.025995 u
The total mass of products = 235.829095 u
Mass detect m =236.054398 u - 235.829095 u
= 0.225303u
So the energy released in each fission
= 0.225303 x 931 MeV = 200.MeV
(ii) This energy first appears as kinetic energy of daughter nuclei and neutrons. But later, this kinetic appears in the form of heat given to the surrounding.
Properties :
(i) The fission is accompanied by the release of neutrons. From each reaction, on an average, 2.5 neutrons are emitted.
(ii) The energy released in the nuclear fission is many times greater than the energy released in chemical reactions.
(iii) Energy released per fission is 200 MeV.

The motion of oil drop inside the chamber can be controlled by adjusting electric field. The oil drop can be moved up or down or even kept balanced in the field of view for sufficiently long time.
Construction:
(i) The apparatus consists of two horizontal circular metal plates A and B each with diameter around 20 cm and are separated by a small distance 1.5 cm.
(ii) These two parallel plates are enclosed in a chamber with glass walls.
(iii) Plates A and B are given a high potential difference around 10 kV such that electric field acts vertically downward
(iv) A small hole is made at the center of the upper plate A.
(v) Atomizer is kept above the hole to spray the liquid.
Working:
(i) When a fine droplet of highly viscous liquid (like glycerine) is sprayed using atomizer, it falls freely downward through the hole under the influence of gravity alone.
(ii) Few oil drops in the chamber can acquire electric charge (negative charge) because of friction with air or passage of x-rays in between the parallel plates.
(iii) The chamber is illuminated by light and oil drops can be seen clearly using microscope.
(iv) These drops can move either upwards or downward.
(v) Let m be the mass of the oil drop and q be its charge. Then the forces acting on the droplet are
(a) gravitational force F_g = mg
(b) electric force F_e = qE
(c) buoyant force F_b
(d) viscous force F_v
(a) Determination of radius of the droplet:
(i) When the electric field is switched off, the oil drop accelerates downwards. Due to presence of air drag forces, the oil drops attain its terminal velocity and moves with constant velocity.
(ii) This velocity can be measured by finding the time taken by the oil drop to fall through a predetermined distance.

(iii) From the free body diagram, we note that viscous force and buoyant force (upward) balance the gravitational force (downward).
(iv) Let us assume that oil drop to be spherical in shape.
(v) Let p be the density of the oil drop, and r be the radius of the oil drop, then the mass of the oil drop, the oil drop can be expressed in terms of its density as, \( \rho=m/v \Rightarrow m=\left(\frac{4}{3} \pi r^{3}\right) \rho\)
∵ (Volume of the sphere, V = \(\frac{4}{3} \pi r^{3})\)
Then gravitational force \(\mathrm{F}_{\mathrm{g}}=\mathrm{mg}=\left(\frac{4}{3} \pi \mathrm{r}^{3}\right) \rho g\)
(vi) Let σ be the density of air, the upthrust force experienced by the oil drop due to displaced air is \(F_{b}=\left(\frac{4}{3} \pi r^{3}\right) σ g\)
(vii) Once the oil drop attains a terminal velocity v, the net downward force acting on the oil drop is equal to the viscous force acting opposite to the direction of motion of the oil drop. From Stokes law, the viscous force on the oil drop is
\(\mathrm{F}_{\mathrm{v}}=6 \pi \eta \mathrm{rv}\)
(ix) From free body diagram, the force balancing equation is,
\(F_{g}=F_{b}+F_{v} \)
\(\left(\frac{4}{3} \pi r^{3}\right) \rho g=\left(\frac{4}{3} \pi r^{3}\right) \sigma g+6 \pi \eta r v \)
\(\frac{4}{3} \pi r^{3}(\rho-\sigma) g=6 \pi \eta r v \)
\(\frac{2}{3} r^{2}(\rho-\sigma) g=3 \eta v \)
Hence radius of the oil drop is \(r=\left[\frac{9 \eta v}{2(\rho-\sigma) g}\right]^{\frac{1}{2}} \ldots\)               ........(1)
(b) Determination of electric charge:
(i) Now switch on the electric field.
(ii) Upward electric force on charged oil drops is qE.
(iii) Choose any one drop in the field of view of microscope.
(iv) Strength of the electric field is adjusted to make that particular drop to be stationary.
(v) Being oil drop is at rest, the viscous force acting on the oil drop is zero.
(vi) Then, from the free body diagram.
\(F_{g}=F_{b}+F_{v} \)
\(\left(\frac{4}{3} \pi r^{3}\right) \rho g=\left(\frac{4}{3} \pi r^{3}\right) \sigma g+q E \)
\(\frac{4}{3} \pi r^{3}(\rho-\sigma) g=q E \)
\(q=\frac{4}{3 E} \pi r^{3}(\rho-\sigma) g \ldots \ldots \ldots \ldots \ldots \) (2)
Substituting (1) in (2)
\(q=\frac{18 \pi}{E}\left(\frac{\eta^{3} v^{3}}{2(\rho-\sigma) g}\right)^{\frac{1}{2}}\)
(vii) Millikan repeated this experiment several times and computed the charges on oil drops. He found that the charge of any oil drop can be written as integral multiple of a basic value, -1.6 x 10^-19 C which is nothing but the charge of an electron.

Principle:
Cathode rays are deflected in electric and magnetic fields.
By the variation of electric and magnetic fields, mass normalized charge or the specific charge (charge per unit mass) of the cathode rays is measured.

Construction and Working:
(i) Cathode rays (electron beam) produced at cathode of a highly evacuated discharge tube. Cathode rays are attracted towards anode disc A.
(ii) Pin hole in the anode disc allows only a narrow beam of cathode rays.
(iii) These cathode rays are now allowed to pass through the parallel metal plates, maintained at high voltage.
(iv) Further, discharge tube is kept in between pole pieces of magnet.
(v) Electric and magnetic fields are perpendicular to each other.
(vi) When the cathode rays strike the zinc sulphide coated screen (O), produces scintillation and hence bright spot is observed.
(i) Determination of velocity of cathode rays:
(a) For a fixed electric field between the plates, the magnetic field is adjusted such that the cathode rays (electron beam) strike at the original position O.
(b) This means that the magnitude of electric force is balanced by the magnitude of force due to magnetic field.
(ie) Ee = Bev
\(\Rightarrow v=\frac { E }{ B } \)  ...(1)

(ii) Determination of specific charge:
(a) Since accelerated from cathode to anode, the potential energy of the electron beam at the cathode is converted into kinetic energy of the electron beam at the anode.
(b) Let V be the potential difference between anode and cathode, then the potential energy is eV.
Then from law of conservation of energy,
\(eV=\frac { 1 }{ 2 } { mv }^{ 2 } \)
\(\frac { e }{ m } =\frac { { v }^{ 2 } }{ 2V } \)   ....(2)
Substituting (1) in (2),
\(\frac { e }{ m } =\frac { 1 }{ 2V } \frac { { E }^{ 2 } }{ { B }^{ 2 } } \)  
By substituting known values, we get
\(\frac { e }{ m } =1.7\times { 10 }^{ 11 }{ CKg }^{ -1 }\)
The specific charge \(\frac{\mathrm{e}}{\mathrm{m}}\) is independent of (a) gas used (b) nature of the electrodes

(i) The phenomenon of spontaneous emission of highly penetrating radiations such as α, β, γ rays by heavy elements having atomic number greater than 82 is called radioactivity and the substances which emit these radiations are called radioactive elements.
(ii) The radioactive phenomenon spontaneous and is unaffected by any external agent like temperature, pressure, and magnetic fields.

(i) Fusion requires high temperature controlled reaction is not yet obtained. Highly sophisticated technology will be required.
(ii) However, the fuel is easily available, cheap and causes very less pollution.
(iii) There is no problem of waste management.
(iv) Fission controlled chain reaction is possible. The technology is well developed and established.
(v) There also exists the problem of waste management.

(i) Energy-produced per fusion is less but the number of nuclei per unit mass is larger on account of smaller mass number.
(ii) So, energy released per unit mass is greater. On the other hand, the fissionable materials have high atomic weight.
(iii) The number of fission nuclei per unit mass is less. So, energy released is less.

(i) Since proton number and neutron number are conserved in a nuclear reaction, the total rest mass of neutrons and protons is the same on either side of a reaction.
(ii) But the total binding energy of nudei on the left side need not be the same as that on the right hand side.
(iii) The difference in these binding energies appears as energy released or obsorbed in nuclear reaction.

Atomic mass unit is defined as \(\frac{1}{12}\) ^th of the mass of one \(_{ 6 }^{ 12 }{ C }\) atom.
E = mc²
= 1.66 x 10-27 x 3 X 108 x 3 x 108 J
= 1.66 x 9 x 10-11 J
= \(\frac { 1.66\times 9\times { 10 }^{ -11 } }{ 1.6\times { 10 }^{ -13 } } \) MeV = 931 MeV.

Rate of disintegration of a radioactive sample,
R = \(\frac{dN}{dt}\)
According to radioactive decay law
- \(\frac{dN}{dt}\) ∝ N
- \(\frac{dN}{dt}\) = λN or R = λN

The radius of the nth orbit 
\({ r }_{ n }=\frac { { n }^{ 2 }{ h }^{ 2 }{ \epsilon }_{ 0 } }{ \pi m{ Ze }^{ 2 } } \)
r₁ = 0.53Å
∴ The diameter of H₂ = 2 x r₁
= 0.53 x 2 = 1.06Å

The neutrino has the following properties
(i) It has zero charge
(ii) It has an antiparticle called anti-neutrino.
(iii) Recent experiments showed that the neutrino has very tiny mass
(iv) It interacts very weakly with the matter. Therefore, it is very difficult to detect In fact, in every second, trillions of neutrinos coming from the sun are passing through our body without any interaction.

(i) Moderators: The moderator is a material used to convert fast neutrons into slow neutrons. Usually the moderators are chosen in such a way that it must be very light nucleus having mass comparable to that of neutrons.
(ii) Hence, these light nuclei undergo collision with fast neutrons and the speed of the neutron is reduced.
(iii) A billiard ball striking a stationary billiard ball of equal mass would itself be stopped but the same billiard ball bounces off almost with same speed when it strikes a heavier mass.
(iv) This is the reason for using lighter nuclei as moderators.

(i) Angular momentum is given by
l = nħ = \(\frac { nh }{ 2\pi } \)
= \(\frac { 5\times 6.6\times { 10 }^{ -34 } }{ 2\times 3.14 } \) = 5.25 x 10^-34 kgm²s^-1
(ii) Velocity is given by
Velocity \(\nu =\frac { l }{ mr } \)

v = 4.4 x 10⁵ms⁻¹

(i) The application of beta decay is radioactive dating or carbon dating.
(ii) Using this technique, the age of an ancient object can be calculated.
(iii) All living organisms absorb carbon dioxide (CO₂) from air to synthesize organic molecules.
(iv) In this absorbed CO₂, the major part is \(_{ 6 }^{ 12 }{ C }\) and very small fraction (1.3 x 10^-12) is radioactive \(_{ 6 }^{ 14 }{ C }\)whose half-life is 5730 years
(v) Carbon-14 in the atmosphere is always decaying but at the same time, cosmic rays from outer space are continuously bombarding the atoms in the atmosphere which produces \(_{ 6 }^{ 14 }{ C }\)
(v) So the Continuous production and decay of \(_{ 6 }^{ 14 }{ C }\) in the atmosphere keep the ratio of \(_{ 6 }^{ 14 }{ C }\) always constant.
(vi) Since our human body, tree or any living organism continuously absorbs CO₂ from the atmosphere, the ratio of \(_{ 6 }^{ 14 }{ C }\) to \(_{ 6 }^{ 12 }{ C }\) in the living organism is also nearly constant.
(vii) But when the organism get dies, it stops absorbing CO₂ 
(viii) Now \(_{ 6 }^{ 14 }{ C }\) starts to decay, the ratio of \(_{ 6 }^{ 14 }{ C }\) to \(_{ 6 }^{ 12 }{ C }\) in a dead organism or specimen decreases over the years.
(ix) Suppose the ratio of \(_{ 6 }^{ 14 }{ C }\) to \(_{ 6 }^{ 12 }{ C }\) in the ancient tree pieces excavated is known, then the age of the tree pieces can be calculated.

The element chlorine is a mixture of 75.77% of  \(_{ 17 }^{ 35 }{ Cl }\) and 24.23% of  \(_{ 17 }^{ 37 }{ Cl }\). So the average atomic mass will be
\(\frac { 75.77 }{ 100 } \times 34.96885u+\frac { 24.23 }{ 100 } \times 36.96593u\)
= 35.453 u
In fact, the chemist uses the average atomic mass or simply called chemical atomic weight (35.453 u for chlorine) of an element. So it must be remembered that the atomic mass which is mentioned in the periodic table is basically averaged atomic mass.