B = ∆m x c²
∆m = \(\frac{B}{c^2}\)
N M_n + Z M_p - M(N,Z) = \(\frac{B}{c^2}\)
N (N,Z) = N M_n + ZM_p - \(\frac{B}{c^2}\)

\(\frac{1}{\lambda}=\mathrm{RZ}^2\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right]  \)
\(\frac{1}{\lambda}=\mathrm{RZ} ^2\left[\frac{1}{1}-\frac{1}{4}\right]=\frac{3}{4} \mathrm{RZ}^2  \)
\(\lambda \propto \frac{1}{Z^2}  \)
\(\lambda_{\mathrm{Li}}: \lambda_{\mathrm{H} c}: \lambda_{\mathrm{H}}=\frac{1}{9}: \frac{1}{4}: \frac{1}{1}=4: 9: 36\)

Cathode rays are stream of negatively charged electron.

When the hydrogen is burnt out, the sun will enter into a new phase called the red giant where helium will fuse to become carbon. During this stage, the sun will expand greatly in size and all its planets will be engulfed in it

Neutrons are classified according to their kinetic energy as
(i) slow neutrons (0 to 1000 eV)
(ii) fast neutrons (0.5 MeV to 10 MeV).

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.

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)) 

General Notation -\(_{ Z }^{ A }{ X }\)
A - Mass number of nucleus - the total number of neutrons and protons in the nucleus
Z - atomic number of nucleus - the total number of protons in the nucleus
X - chemical symbol of the element.

(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.

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

(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.

A = 108, Z  =47, N = 108 - 47 = 61
m_p = 1.007825 u, m_n = 1.008665 u, M = 107.905949 u
(a) \(\Delta \mathrm{m}=Z \mathrm{~m}_{\mathrm{P}}+\mathrm{Nm}_{\mathrm{n}}-\mathrm{M} \)
\(\Delta \mathrm{m}\) = (47 x 1 .007825 + 61 x 1 .008665 - 107 .905949)
\(\Delta \mathrm{m}\) = 47.367775 + 61.528565 -107.905949
\(\Delta \mathrm{m}\) = 108.89634 - 107.905949
\(\Delta \mathrm{m}\) = 0.990391 u
\(\mathrm{BE}=\Delta \mathrm{m} \times 931 \mathrm{MeV} \)
BE = 0.990391 x 931 MeV = 922.054 MeV
(c) \(\overline{\mathbf{B E}}=\frac{\mathbf{B E}}{\mathbf{A}} \)
\(\overline{\mathrm{BE}}=\frac{922.054}{108}=8.537 \mathrm{MeV}=8.5 MeV\)

Density \(\rho=2.3 \times 10^{17} \mathrm{kgm}^{-3}\), Mass M = 5.97 x 10²⁴ kg
\(\rho=\frac{M}{V}=\frac{M}{\frac{4}{3}\pi R^3}\)
\(R=\left[\frac{M}{\frac{4}{3} \pi \rho}\right]^{\frac{1}{3}}=\left[\frac{3 \mathrm{M}}{4 \pi \rho}\right]^{\frac{1}{3}}=\left[\frac{3 \times 5.97 \times 10^{24}}{4 \times 3.14 \times 2.3 \times 10^{17}}\right]^{\frac{1}{3}}=\left[0.62 \times 10^{7}\right]^{\frac{1}{3}}\)
R = 183.7 m
R ≈180 m

(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 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.

The initial number of molecules, N_o = 150
The final number of molecules, N = 150
\(\frac { N }{ { N }_{ 0 } } ={ \left( \frac { 1 }{ 2 } \right) }^{ n } \ or \ \frac { 150 }{ 600 } ={ \left( \frac { 1 }{ 2 } \right) }^{ n }\)
\({ \left( \frac { 1 }{ 2 } \right) }^{ 2 }={ \left( \frac { 1 }{ 2 } \right) }^{ n } \ or \ n=2=\frac { t }{ { T }_{ 1/2 } } \)
t = 2 x 600 s = 1200 s
Best of disintegration,
\(R=\frac { dN }{ dt } =-\lambda N\)
\(=\frac { 0.693 }{ { T }_{ 1/2 } } \times 150\)
\(=\frac { 0.693 }{ { 600} } \times 150\) = 0.173
disintegration/second at the instant when 150 molecules were remaining.

Let the initial number of nuclei X = N_o
Number of nuclei of X left at any time t = N_x
Number of nuclei of Y at any time t = N_o-N_x = N_y
Now, N_X = N₀e^-λt
N_Y = N₀ - N_X = N_oe^-λt
It is given that \(\frac { { N }_{ X } }{ { N }_{ Y } } =\frac { 1 }{ 15 } \)
\(\frac { { N }_{ o }{ e }^{ -\lambda t } }{ { N }_{ o }({ 1-e }^{ -\lambda t }) } =\frac { 1 }{ 15 } \ or \ { e }^{ -\lambda t }=1\)
e^-λt = 16
\(\lambda t={ log }_{ e }2^{ 4 }=4\times { log }_{ e }2=4\times 0.693\)
\(t=4\times \frac { 0.693 }{ \lambda } \)
\(t=4\times { T }_{ 1/2 }=4\times 50=200\) years