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 nutrino has very small mass.
(iv) It interacts very weakly with the matter therefore, it is very difficult to detect it.
(v) In every second trillion of neutrino coming from the sun are passing through our body without any interaction.
Initially, it was thought that during beta decay, a neutron in the parent nucleus is converted to the daughter nuclei by emitting only electron as given by
\({ }_{\mathrm{Z}}^{\mathrm{A}} \mathbf{X} \rightarrow \underset{\mathrm{Z}+1}{\mathrm{~A}} \mathbf{Y}+\mathrm{e}^{-}\)
(i) But the kinetic energy of electron coming out of the nucleus did not match with the experimental results.
(ii) In beta decay, the beta particle (i.e, electron) have a continuous range of energies. But the conservation of energy and momentum gives specific single values for electron energy and the recoiling nucleus Y It seems that the conservation of energy, momentum is violated and could not be explained. So beta decay remained as a puzzle for several years.
(iii) But later w. Pauli proposed a third particle which must be present in beta decay to carry away missing energy and momentum.
(iv) Fermi later named this particle the neutrino.
(v) Finally, the neutrino was detected experimentally in 1956 by Fredrick Reines and Clyde Cowan.

(i) In \(\alpha \text { and } \beta \) decay, most of the daughter nucleus is in the excited state.
(ii) The life time of excited state is approximately 10^-11 s.
(iii) This excited state nucleus immediately returns to the ground state or lower energy state by emitting highly energetic photons called rays of energy in order of MeV.

The gamma decay is given by,
\({ }_{\mathrm{Z}}^{\mathrm{A}} \mathrm{X}^{*} \rightarrow{ }_{\mathrm{Z}}^{\mathrm{A}} \mathrm{X}+\gamma-\text { ray }\)
(iv) Here the asterisk(*) means excited state nucleus.
(a) In gamma decay, there is no change in the mass number or atomic number of the nucleus. when \({ }_{5}^{12} \mathrm{~B}\) undergoes beta decay directly into ground state carbon (\(_{ 6 }^{ 12 }{ C})\) by emitting an electron of maximum of energy 13.4 MeV.
(b) If \({ }_{5}^{12} \mathrm{~B}\) undergoes beta decay to an excited state of carbon \(({ _{ 6 }^{ 12 }{ C } }^{ * })\) by emitting an electron of maximum energy 9.0 MeV followed by gamma decay to ground state by emitting a photon of energy 4.4 MeV. It is represented by,
\(_{ 5 }^{ 12 }{ B }\rightarrow _{ 6 }^{ 12 }{ C+ }{ e }^{ - }+\overline { v } \)
\(_{ 6 }^{ 12 }{ C^* }\rightarrow _{ 6 }^{ 12 }{ C }+\gamma -rays\) 

The average binding energy per nucleon is the energy required to separate single nucleon from the particular nucleus.


\(\overline{\mathrm{BE}}=\frac{\left[\mathrm{Zm}_{\mathrm{H}}+\mathrm{Nm} _{\mathrm{n}}-\mathrm{M}_{\mathrm{A}}\right] \mathrm{c}^{2}}{\mathrm{~A}}\)
\(\overline { BE } \) is plotted against A of all known nuclei.
Important inferences from the average binding energy curve:
(i) The value of \(\overline { BE } \) rises as the mass number increases until it reaches a maximum value of 8.8 MeV for A = 56 (iron) and then it slowly decreases.
(ii) The average binding energy per nucleon is about 8.5 MeV for nuclei having mass number between A = 40 and 120. These elements are comparatively more stable and not radioactive.
(iii) For higher mass numbers, the curve reduces slowly and \(\overline { BE } \) for uranium is about 7.6 MeV. They are unstable and radioactive.
(iv) From Figure, If two light nuclei with A<28 combine with a nucleus with A<56, the binding energy per nucleon is more for final nucleus than initial nuclei. Thus, if the lighter elements combine to produce a nucleus of medium value A, a large amount of energy will be released. This is the basis of nuclear fusion and is the principle of the hydrogen bomb.
(v) If a nucleus of heavy element is split (fission) into two or more nuclei of medium value A, the energy released would again be large. The atom bomb is based on this principle and huge energy of atom bombs comes from this fission when it is uncontrolled.

The electrostatic force is a conservative force, the potential energy for the electron in n^th orbit is
\(U_{n} =\frac{1}{4 \pi \varepsilon_{0}} \frac{(+Z e)(-e)}{r_{n}}=-\frac{1}{4 \pi \varepsilon_{0}} \frac{Z^{2}}{r_{n}} \)      \(\left[ \because r_n=\frac{\varepsilon_{0} h^{2} n^{2}}{\pi m Z e^{2}}\right]\)
\(U_{n} =-\frac{1}{4\varepsilon_{0}} -\frac{Z^{2} \mathrm{me}^{4}}{h^{2} n^{2}} \)
The kinetic energy of electron in n^th orbit is
\(\mathrm{KE}_{\mathrm{n}}=\frac{1}{2} \mathrm{mv}_{\mathrm{n}}^{2}=\frac{\mathrm{Z}^{2} m \mathrm{e}^{4}}{8 \varepsilon_{0}^{2} \mathrm{~h}^{2} \mathrm{n}^{2}}\)
This implies that U_n = -2KE_n
Total energy of electron in the n^th orbit is
\(E_{n}=K E_{n}+U_{n}=K E_{n}-2 K E_{n}=-K E_{n} \)
\(E_{n}=-\frac{Z^{2} m e^{4}}{8 \varepsilon_{0}^{2} h^{2} n^{2}} \)
For Hydrogen atom Z = 1
\(E_{n}=-\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{2} n^{2}} \text { joule }\)
n - principal quantum number
The negative sign indicates that the electron is bound to the nucleus.
Substituting the values of mass and charge of an electron (m and e), permittivity of free space \(\varepsilon^{0}\)  and Planck's constant h and expressing in terms of (+(eV)), we get
\(E_{n}=-13.6\left(\frac{1}{n^{2}}\right) e V\)
(i) For the first orbit (ground state), the total energy of electron is E₁ = - 13.6 eV.
(ii) For the second orbit (first excited state), the total energy of electron is E₂ = -3.4 eV.
(iii) For the third orbit (second excited state), the total energy of electron is E₃ = -1.51 eV and so on.

(i) The electron in an atom moves around nucleus in circular orbits under the influence of Coulomb electrostatic force of attraction. This Coulomb force gives necessary centripetal force for the electron to undergo circular motion.
(ii) Electrons in an atom revolve around the nucleus only in certain discrete orbits called stationary orbits where it does not radiate electromagnetic energy. Only those discrete orbits allowed are stable orbits.
(iii) The angular momentum of the electron in these stationary orbits are quantized (ie) L = \(\frac{nh}{2\pi}\) This is known as Bohr quantization condition.
(iv) The energy of the orbits are not continuous but only discrete. This is called quantization of energy.
(v) An electron can jump from one orbit to another orbit by absorbing or emitting a photon whose energy is equal to the difference in energy between the two orbital levels.

(i) Cathode rays possess energy and momentum and travel in a straight line with high speed of the order of 10⁷m s^-1or \({ \left( \frac { 1 }{ 10 } \right) }^{ th }\) of the speed of light.
(ii) It can be deflected by application of electric and magnetic fields. The direction of deflection indicates that they are negatively charged particles.
(ii) When the cathode rays are allowed to fall on matter, they produce heat. They affect the photographic plates and also produce fluorescence when they fall on certain crystals and minerals.
(iii) When the cathode rays fall on a material of high atomic weight, x-rays are produced.
(iv) Cathode rays ionize the gas through which they pass.

235 g of \(_{ 92 }^{ 235 }{ U }\) has 6.02 x 10²³ atoms. In one gram of \(_{ 92 }^{ 235 }{ U }\), the number of atoms is equal to \(\frac { 6.02\times { 10 }^{ 23 } }{ 235 } =2.56\times { 10 }^{ 21 }\)
So the number of atoms in 1 kg of \(_{ 92 }^{ 235 }{ U }\) = 2.56 x 10²¹ x 1000 = 2.56 x 10²⁴
Each \(_{ 92 }^{ 235 }{ U }\) nucleus releases 200 MeV of energy during the fission. The total energy released by 1kg of \(_{ 92 }^{ 235 }{ U }\) is
Q = 2.56 x 10²⁴ x 200MeV = 5.12 x 10²⁶ MeV
In terms of joules,
Q = 5.12 x  10²⁶ x 1.6 x 10^-13 J = 8.192 x 10¹³ J
In terms of Kilowatt hour,
Q = \(\frac { 8.192\times { 10 }^{ 13 } }{ 3.6\times { 10 }^{ 6 } } =2.27\times { 10 }^{ 7 }\) kWh

To get the time interval in terms of half life, \(n=\frac { t }{ { T }_{ 1/2 } } =\frac { 22,920 \ yr }{ 5730 \ yr } =4\)
The number of nuclei remaining undecayed after 22,920 years,
\(N={ \left( \frac { 1 }{ 2 } \right) }^{ n }{ N }_{ 0 }={ \left( \frac { 1 }{ 2 } \right) }^{ 4 }\times 10,000\)
N = 625

From equation (9.19), the radius of the nuclecus, R = R₀ \(A^\frac13\). Then the volume of the nucleus
\(V=\frac { 4 }{ 3 } \pi { R }^{ 3 }=\frac { 4 }{ 3 } { \pi { R }_{ 0 } }^{ 3 }A\)
By ignoring the mass difference between the proton and neutron, the total mass of the nucleus having mass number A is equal to A.m where m is mass of the proton and is equal to 1.6726 x 10^-27 kg.
Nuclear density.

R = R₀A^\(\frac13\)
R = 1.2 x 10^{−15 x} (197)^\(\frac13\) = 6.97 x 10⁻¹⁵ m
Or R = 6.97 F

\(T_{1 / 2}=1600 \text { years }=1600 \times 365 \times 24 \times 60 \times 60 s\)
R = 1 curie = 3.7 x 10¹⁰ Bq, Show that m = 1g
R = λN
Number of atoms Present, N = \(\frac{\mathrm{R}}{\lambda}=\frac{\mathrm{R}}{0.6931} \mathrm{~T}_{1 / 2}\)
Mass of 6.023 x 10²³ atoms of  \({ }_{88}^{226} R a=226 g\)
Mass of 1 atom of  \({ }_{88}^{226} \mathrm{Ra}=\frac{226}{6.023 \times 10^{23}} \mathrm{~g}\)
Mass of N atoms of \({ }_{88}^{{ }{266}} \mathrm{Ra}=\frac{226}{6.023 \times 10^{23}} \times \mathrm{Ng}\)
Mass of N atoms of \({ }_{88}^{226} \mathrm{Ra}(\mathrm{m})=\frac{226}{6.023 \times 10^{23}} \times \frac{\mathrm{R}}{0.6931} \mathrm{~T}_{1 / 2} \mathrm{~g}\)
\(\mathrm{m}=\frac{226}{6.023 \times 10^{23}} \times \frac{3.7 \times 10^{10}}{0.6931} \times 1600 \times 365 \times 24 \times 60 \times 60 \mathrm{~g}\)
m = 1.01 g

According to quark model,
(i) Proton is made up of two up quarks and one down quark.
(ii) Neutron is made up of one up quark and two down quarks.

One curie was defined as number of decays per second in 1 gram of radium. 1 1 curie = 3.7 X 10¹⁰ decays/s.

Half-life T_1/2 of nucleus is the time required for the number of atoms initially present to reduce to one half of the initial amount.
\(\mathrm{T}_{1 / 2}=\frac{0.6931}{\lambda}\)
\(\lambda\) is the decay constant.

The mean life time of the nucleus is the ratio of sum or integration of life times of all nuclei to the total number nuclei present initially.
\(\tau =\frac { 1 }{ \lambda } \)

The phenomenon of spontaneous emission of highly penetrating radiations such as α, β and ⋎ rays by an element is called radioactivity and the substances which emit these radiations are called radioactive elements.

While forming a nucleus, the mass disappear is converted into energy. This energy is called binding energy of a nucleus.
\(\therefore \mathrm{BE}=\left(\mathrm{Zm}_{\mathrm{P}}+\mathrm{Nm}_{\mathrm{n}}-\mathrm{M}\right) \times \mathrm{c}^{2}\)
where, c - velocity of light
 

One atomic mass unit (u) is defined as the (1/12)^th of the mass of the isotope of carbon \(_{ 6 }^{ 12 }{ C }\)  
\(\mathrm{lu}=\frac{\text { mass of }_{6}^{12} \mathrm{C} \text { atom }}{12}=\frac{1.9926 \times 10^{-26}}{12}=1.660 \times 10^{-27} \mathrm{~kg}\)

Isotopes are atoms of the same element having same atomic number Z, but different mass number A.
(Ex: Hydrogen, \(_{ 1 }^{ 1 }{ H }\) ((hydrogen), \(_{ 1 }^{ 2 }{ H }\) (deuterium),and \(_{ 1 }^{ 3 }{ H }\) (tritium)) 

The impact parameter is defined as the perpendicular distance between the centre of the gold nucleus and the direction of velocity vector of alpha particle when it is at a large distance in the Rutherford's alpha particles scattering experiment.

(i) Minimum energy required to remove an electron from an atom in the ground state is known as binding energy or ionization energy.
(ii) Ionization potential is defined as ionization energy per unit charge.

(i) At any instant t, the number of decays per unit time, called rate of decay \(\left( \frac { dN }{ dt } \right) \) is proportional to the number of nuclei (N) at the same instant.
\(\left( \frac { dN }{ dt } \right) \propto N\)
\(\frac { dN }{ dt } =-\lambda N\) ...(1)
(ii) Here, proportionality constant \( \lambda\) is called decay constant which is different for different radioactive sample and the negative sign in the equation implies that the N is decreasing with time. From (1)
\(\frac{\mathrm{dN}}{\mathrm{N}}=-\lambda \mathrm{dt}\)   ...(2)
(iii) Here, dN represents number of nuclei decaying in the time interval dt.
(iv) Let us assume that at time t = 0 s, the number of nuclei present in the radioactive sample is N_o.
(v) By integrating the equation (2), we can calculate the number of undecayed nuclei N at any time t.
\(\int_{N_{0}}^{N} \frac{d N}{N}=-\int_{0}^{t} \lambda d t \)
\({[\ln N]_{N_{0}}^{N}=-\lambda t} \)
\(\ln \left[\frac{N}{N_{0}}\right]=-\lambda t \)
Taking exponential on both sides, we get
\(\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda t}\)         .....(3)
(vi) Equation (3) is called the law of radioactive decay.
(vii) Here N denotes the number of undecayed nuclei present at any time t and N_o denotes the number of nuclei present initially time t = 0.
(viii) From equation (3) the number of atoms is decreasing exponentially over the time. This implies that the time taken for all the radioactive nuclei to decay will be infinite.

When electron jumps from higher energy stationary orbit (m) to lower energy stationary orbit (n), emit radiation.
Wave number (or) Wave length of the emitted radiation is given by,
\(\frac{1}{\lambda}=R\left[\frac{1}{n^{2}}-\frac{1}{m^{2}}\right]=\bar{v}\)
V - Ware number; R- Rydberg constant  = 1.09737 x 10⁷ m^-1
m > n (where m, n are integers) 
(a) Lyman series:
When electron jumps from any outer orbit to first orbit, (n = 1 and m = 2,3,4.......) the wave number or wavelength of spectral lines lies in ultra-violet region is
\(\bar{v}=\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{1^{2}}-\frac{1}{\mathrm{~m}^{2}}\right]\)
(b) Balmer series:
Wher electron jumps from any outer orbit to second orbit, (n = 2 and m = 3, 4, 5.......) the wave number or wavelength of spectral lines lies in visible region is,
\(\bar{v}=\frac{1}{\lambda}=R\left[\frac{1}{2^{2}}-\frac{1}{m^{2}}\right]\)
(c) Paschen series:
When electron jumps from any outer orbit to third orbit, (n = 3 and m = 4, 5, 6.......) the wave number or wavelength of spectral lines lies in infra red (Near IR) region is
\(\bar{v}=\frac{1}{\lambda}=R\left[\frac{1}{3^{2}}-\frac{1}{m^{2}}\right]\)
(d) Bracket series:
When electron jumps from any outer orbit to fourth orbit, (n = 4 and m = 5, 6, 7.......) the wave number or wavelength of spectral lines lies in infra red (Middle IR) region is
\(\bar{v}=\frac{1}{\lambda}=R\left[\frac{1}{4^{2}}-\frac{1}{m^{2}}\right]\)
(e) Pfund series:
When electron jumps from any outer orbit to fifth orbit, (n - 5 and m = 6, 7, 8.......) the wave number or wavelength of spectral lines lies in infra red (far IR) region is
\(\bar{v}=\frac{1}{\lambda}=R\left[\frac{1}{5^{2}}-\frac{1}{m^{2}}\right]\)