To calculate the age, we need to know the initial activity (R₀) of the caracol (when the sample was alive).
The activity R of the sample
R = R₀ e^-λt   ...(1)
To find the time t, rewriting the above equation (1),
\({ e }^{ \lambda t }=\frac { { R }_{ 0 } }{ R } \)
By taking the logarithm on both sides, we get \(t=\frac { 1 }{ \lambda } In\left( \frac { { R }_{ 0 } }{ R } \right) \) ..(2)
Here R = 38 decays/s = 38 Bq.
To find decay constant, we use the equation
\(\lambda =\frac { 0.6931 }{ { T }_{ 1/2 } } =\frac { 0.6931 }{ 5730yr\times 3.156\times { 10 }^{ 7 }s/yr } \)
[∴ 1yr = 365.25 x 24 x 60 x 60 s = 3.156 x 10⁷ s]
λ = 3.83 x 10⁻¹² s⁻¹
To find the initial activity R₀, we use the equation R₀ = λN₀. Here N₀ is the number of carbon-14 atoms present in the sample when it was alive. The mass of the charcoal is 200 g. In 12 g of carbon, there are 6.02 x 10²³ carbon atoms. So 200 g contains,
\(\frac { 6.02\times { 10 }^{ 23 }atoms/mol }{ 12g/mol } \times 200\approx 1\times { 10 }^{ 25 }\) atoms
When the tree(sample) was alive, the ratio of \(_{ 6 }^{ 14 }{ C }{ e }\) to \(_{ 6 }^{ 12 }{ C }{ e }\) is 1.3 x 10^-12. So the total number of carbon-14 atoms is given by
N₀ = 1 x 10²⁵ x 1.3 x 10^-12 atoms
The initial activity
R₀ = 3.83 x 10^-12 x 1.3 x 10¹³ ≈ 50 decays / s
= 50 Bq
By substituting the value of R₀ and λ in the equation (2), we get
\(t=\frac { 1 }{ 3.83\times { 10 }^{ -12 } } \times In\left[ \frac { 50 }{ 37 } \right] \)
\(t=\frac { 0.301 }{ 3.83 } \times { 10 }^{ 12 }\approx 7.86\times { 10 }^{ 10 }\)sec
In years
\(t=\frac { 7.86\times { 10 }^{ 10 }s }{ 3.156\times { 10 }^{ 7 }s/yr } \approx 2500\) years
In fact, the excavated materials were to USA sent for carbon dating by the Archeological Department of Tamilnadu and the report confirmed that the age of Keezhadi artifacts lies between 2200 years to 2500 years (Sangam era- 400 BC to 200 BC). The Keezhadi excavations experimentally proved that urban civilization existed in Tamil Nadu even 2000 years ago!

(a) To find N₀, we have to find the number of \(_{ 7 }^{ 13 }{ N }\) atoms in 2.6μg. The atomic mass of nitrogen is 13. Therefore, 13 g of  \(_{ 7 }^{ 13 }{ N }\) contains Avogadro number (6.02 x 10²³) of atoms.
In 1 g, the number of \(_{ 7 }^{ 13 }{ N }\) atoms present is equal to \(\frac { 6.02\times { 10 }^{ 23 } }{ 13 } \). So the number of \(_{ 7 }^{ 13 }{ N }\) atoms present in 2.6μg is
\({ N }_{ 0 }=\frac { 6.02\times { 10 }^{ 23 } }{ 13 } \times 2.6\times { 10 }^{ -6 }=12.04\times { 10 }^{ 16 }\) atoms
(b) To find the initial activity R₀, we have to evaluate decay constant λ
\(\lambda =\frac { 0.6931 }{ { T }_{ 1/2 } } =\frac { 0.6931 }{ 10\times 60 } =1.155\times { 10 }^{ -3 }{ s }^{ -1 }\)
Therefore
R₀ = λN₀ = 1.155 x 10^-3 x 12.04 x 10¹⁶
= 13.90 x 10 ¹³ decays/s
= 13.90 x 10 ¹³ Bq
In terms of a curie,
\({ R }_{ 0 }=\frac { 13.90\times { 10 }^{ 13 } }{ 3.7\times { 10 }^{ 10 } } =3.75\times { 10 }^{ 3 }Ci\)
since 1Ci = 3.7 x 10¹⁰Bq
(c) Activity after 2 hours can be calculated in two different ways:
Method 1: R = R₀ e^–λt
At t = 2 hr = 7200 s
R = 3.75 x 10³ x e^{-7200 x 1.155 x 10–3}
R = 3.75 x 10³ x 2.4 x 10^–4 = 0.9 Ci
Method 2: \(R={ \left( \frac { 1 }{ 2 } \right) }^{ n }{ R }_{ 0 }\)
Here \(n=\frac { 120min }{ 10min } =12\)
\(R={ \left( \frac { 1 }{ 2 } \right) }^{ 12 }\times 3.75\times { 10 }^{ 3 }\) ≈ 0.9 Ci
(d) mean life ፒ = \(\frac{T_{1/2}}{0.6931}=\frac{10\times60}{0.6931}\)
= 865.67 s

The difference in masses
Δm = (m_U - m_Th - m_α)
= (232.037156–228.028741 – 4.002603)u
The mass lost in this decay = 0.005812 u
Since 1u = 931MeV, the energy Q released is
Q = (0.005812 u) x (931 MeV / u)
= 5.41 MeV
This disintegration energy Q appears as the kinetic energy of α particle and the daughter nucleus. In any decay, the total linear momentum must be conserved.
Total linear momentum of the parent nucleus = total linear momentum of the daughter nucleus and α particle. Since before decay, the uranium nucleus is at rest, its momentum is zero. By applying conservation of momentum, we get
0 = \({ m }_{ Th }{ \overrightarrow { \upsilon } }_{ Th }+{ m }_{ \alpha }\overrightarrow { \upsilon } _{ \alpha }\)
\({ m }_{ \alpha }\overrightarrow { \upsilon } _{ \alpha }\) = - \({ m }_{ Th }{ \overrightarrow { \upsilon } }_{ Th }\)
It implies that the alpha particle and daughter nucleus move in opposite directions.
In magnitude m_α ሀ_α = m_Th ሀ_Th
The velocity of α particle ሀ_{α =} \(\frac { { m }_{ Th } }{ { m }_{ \alpha } } { \upsilon }_{ Th }\)
Since m_Th > m_α , ሀ_α > ሀ_Th. The ratio of the kinetic energy of α particle to that the daughter nucleus,
\(\frac { K.{ E }_{ \alpha } }{ K.{ E }_{ Th } } =\frac { 1/2{ m }_{ \alpha }{ { \upsilon }_{ \alpha } }^{ 2 } }{ 1/2{ m }_{ Th }{ { \upsilon }_{ Th } }^{ 2 } } \)
By substituting, the value of ሀ_α into the above equation, we get \(\frac { K.{ E }_{ \alpha } }{ K.{ E }_{ Th } } =\frac { { m }_{ Th } }{ { m }_{ \alpha } } =\frac { 228.02871 }{ 4.002603 } =57\)
The kinetic energy of α particle is 57 times greater than the kinetic energy of the daughter nucleus (\(_{ 90 }^{ 228 }{ Th }\))
The disintegration energy Q = total kinetic energy of products
K.E_α + K.E_Th = 5.41 MeV
57K.E_Th + K.E _Th = 5.41 MeV
K.E_Th = \(\frac{5.41}{58}\) MeV = 0.0093 MeV
K.E_α = 57K.E_Th = 57 x 0.093 = 5.301 MeV
In fact, 98% of total kinetic energy is taken by the α particle.

(a) Given that E_n = \(\frac { 54.4 }{ { n }^{ 2 } } eV\)
For n = 1, the ground state energy E₁ = –54.4 eV and for n = 2, E₂ = –13.6 eV. Similarly, E₃ = –6.04 eV, E₄ = –3.4 eV and so on.
For large value of principal quantum number – that is, n = ∞, we get E_∞ = 0 eV.

(b) For a hydrogen-like atom, ground state energy is
E₁ =\(\frac { 13.6 }{ { n }^{ 2 } } { Z }^{ 2 }eV\)
where Z is the atomic number. Hence, comparing this energy with given energy, we get, – 13.6 Z² = – 54.4 ⇒ Z = ±2. Since, atomic number cannot be negative number, Z = 2.
(c) The first excitation energy is
E₁ = E₂ - E₁ = -13.6 eV - (-54.4eV)
= 40.8 eV
Hence, the first excitation potential is
\({ V }_{ 1 }=\frac { 1 }{ e } { E }_{ 1 }=\frac { (40.8eV) }{ e } \)
= 40.8 volt
The first ionization energy is
E_ionization = E_∞ -  E₁ = 0 -(-54.4eV)
= 54.4 eV
Hence, the first ionization potential is
\({ V }_{ ionization }=\frac { 1 }{ e } { E }_{ ionization }=\frac { (54.4eV) }{ e } \)
= 54.4 volt
(d) Consider two photons to be A and B.
Given that photon A with energy 42 eV and photon B with energy 51 eV
From Bohr assumption, difference in energy levels is equal to photon energy, then atom will absorb energy, otherwise, not.
E₂ - E₁ = -13.6eV - (-54.4eV)
= 40.8eV ≈ 41 eV
Similarly,
E₃ - E₁ = -6.04 eV - (-54.4eV)
= 48.36 eV
E₄ - E₁ = -3.4eV - (-54.4eV)
= 51 eV
E₃ - E₂ = -6.04eV - (-13.6eV)
= 7.56 eV
and so on.
But note that E₂ – E₁ ≠ 42 eV, E₃ – E₁ ≠ 42 eV, E₄ – E₁ ≠ 42 eV and E₃ – E₂ ≠ 4₂ eV
For all possibilities, no difference in energy is an integer multiple of photon energy. Hence, photon A is not absorbed by this atom. But for Photon B, E₄ – E₁ = 51 eV, which means, Photon B can be absorbed by this atom
(d) The radius of Bohr orbit is \(r_n=\frac { a_o\times n^2 }{ z }\)
For n = 1, z = 2
\(r_1=\frac { a_o }{ 2 }\)
\(=\frac { 0.529 }{ 2 }\)
= 0.265 Å
(e) Since total energy is equal to negative of kinetic energy in Bohr atom model, we get
\(K{ E }_{ n }=-{ E }_{ n }=-\left( -\frac { 54.4 }{ { n }^{ 2 } } eV \right) \)
\(=-\frac { 54.4 }{ { n }^{ 2 } } eV\)
Since, potential energy is negative of twice the kinetic energy, 
\( U_{ n }=-2K{ E }_{ n }=-2\left( -\frac { 54.4 }{ { n }^{ 2 } } eV \right) \)
\(​​=-\frac { 108.8 }{ { n }^{ 2 } } eV\)
For a ground state, put n = 1
Kinetic energy is KE₁ = 54.4 eV and Potential energy is U₁ = –108.8 eV

Let the mass of the electron be m and mass of the nucleus be M. Since there is no external force acting on the system, the centre of mass of hydrogen atom remains at rest. Hence, both nucleus and electron move about the centre of mass as shown in figure.

Let V be the velocity of the nuclear motion and υ be the velocity of electron motion. Since the total linear momentum of the system is zero,
−mሀ + Mሀ = 0 or
Mሀ = mሀ = p
\(\vec { { p }_{ e } } +\vec { { p }_{ n } } =\vec { 0 } \) or
\(|\vec { { p }_{ e } } |=|\vec { { p }_{ n } } |=p\)
Hence, the kinetic energy of the system is
KE = \(\frac { { { p }^{ 2 } }_{ n } }{ 2M } +\frac { { { p }^{ 2 } }_{ e } }{ 2m } =\frac { { p }^{ 2 } }{ 2 } \left( \frac { 1 }{ M } +\frac { 1 }{ m } \right) \)
Let \(\frac { 1 }{ M } +\frac { 1 }{ m } =\frac { 1 }{ { \mu }_{ m } } \). Here the reduced mass is
\({ \mu }_{ m }=\frac { mM }{ M+m } \)
Therefore, the kinetic energy of the system now is
\(KE=\frac { { p }^{ 2 } }{ { 2\mu }_{ m } } \)
Since the potential energy of the system is same, the total energy of the hydrogen can be expressed by replacing mass by reduced mass, which is
\({ E }_{ n }=-\frac { { \mu }_{ m }{ e }_{ 4 } }{ 8{ \epsilon ^{ 2 } }_{ 0 }{ h }^{ 2 } } \frac { 1 }{ { n }^{ 2 } } \)
Since the nucleus is very heavy compared to the electron, the reduced mass is closer to the mass of the electron.

(a) The velocity of an electron in n^th orbit is
\(\upsilon _{ n }=\frac { h }{ 2\pi m{ a }_{ 0 } } \frac { Z }{ n } \)
Where \({ a }_{ 0 }=\frac { { \epsilon }_{ 0 }{ h }^{ 2 } }{ \pi { me }^{ 2 } } \) = Bohr radius. Substituting for a₀ in ሀ_n,
\({ \upsilon }_{ n }=\frac { { e }^{ 2 } }{ 2{ \epsilon }_{ 0 }h } \frac { Z }{ n } =c\left( \frac { { e }^{ 2 } }{ 2{ \epsilon }_{ 0 }hc } \right) \frac { Z }{ n } =\frac { \alpha cZ }{ n } \)
where c is the speed of light in free space or vacuum and its value is c = 3 x 10⁸ m s^–1 and α is called fine structure constant.
For a hydrogen atom, Z = 1 and for the first orbit, n = 1, the ratio of velocity of electron in first orbit to the speed of light in vacuum or free space is
\(\frac { { \upsilon }_{ 1 } }{ c } =\alpha =\frac { { e }^{ 2 } }{ 2{ \epsilon }_{ 0 }hc } \)
\(\alpha =\frac { { (1.6\times { 10 }^{ -19 }C })^{ 2 } }{ 2\times (8.854\times { 10 }^{ -12 }{ C }^{ 2 }{ N }^{ -1 }{ m }^{ -2 }) } \) x \(\frac { 1 }{ (6.6\times { 10 }^{ -34 }{ Nms)\times (3\times { 10 }^{ 8 } }{ ms }^{ -1 }) } \)
≈ \(\frac{1}{136.9}=\frac{1}{137}\) which is a dimensionless number
⇒ α = \(\frac{1}{137}\)
(b) Using fine structure constant, the velocity of electron can be written as v_n = \(\frac{αcZ}{n}\)
For hydrogen atom (Z = 1) the velocity of electron in n^th orbit is v_n = \(\frac{c}{137}\frac{1}{n}=(2.19\times10^6)\frac{1}{n}ms^{-1}\)
For the first orbit (ground state), the velocity of electron is v₁ = 2.19 x 10⁶ms⁻¹
For the second orbit (first excited state), the velocity of electron is v₂ = 1.095 x 10⁶ms⁻¹
For the third orbit (second excited state), the velocity of electron is v₃ = 0.73 x 10⁶ms⁻¹
Here, v₁ > v₂ > v₃

Decayed = 60 %
Left undecayed = 40 %(ie) \(\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{40}{100} \)
\(\mathrm{~T}_{\frac{1}{2}}=3.8 \text { days } \)
\(\mathbf{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda t} \)
\(\frac{\mathrm{N}}{\mathrm{N}_{0}}=\mathrm{e}^{-\lambda t} \)
\(\frac{40}{100}=\mathrm{e}^{-\lambda t} \Rightarrow \frac{100}{40}=2.5=\mathrm{e}^{\lambda t} \)
\(\therefore \mathrm{e}^{\lambda t} \) = 2.5
Taking log on both sides
\(\lambda t=\ln [2.5]=2.3026 \times \log (2.5)=2.3026 \times 0.3974 \)
\(t=\frac{0.9163}{\lambda}=\frac{0.9163}{0.6931} \times T_{1 / 2} \)
\(t=1.322 \times 3.8=5.022 \text { days } \)

\(\mathrm{T}_{\frac{1}{2}}=5.01 \text { days} \), \(\mathrm{R}_{o}=1 μ \mathrm{Ci}=3.7 \times 10^{10} \times 10^{-6} \)  decays per second
\(if\ \mathrm{t}=1\ year\ \mathrm{R}= ? \), \(\tau=? \), \(\lambda=? \), \(\mathrm{~N}_{o}=? \)
a) \(\mathrm{R}=\mathrm{R}_{o} \mathrm{e}^{-\lambda t} \)
\(\mathrm{R}=\mathrm{R}_{o} \mathrm{e}^{-\frac{0.6931}{\mathrm{~T}_{1 / 2} }t} \)
\(\mathrm{R}=\mathrm{R}_{o} \mathrm{e}^{-\frac{0.6931}{5.01} \times 365} \)
\(\mathrm{R}=\mathrm{R}_{o} \mathrm{e}^{-50.5}=\mathrm{R}_{o} \times 1.17 \times 10^{-22} \)
\(\mathrm{R}=1 \mu \mathrm{Ci} \times 1.17 \times 10^{-22}=1.17 \times 10^{-22} \mu \mathrm{Ci} \)
\(\mathrm{R}=1.17 \times 10^{-22} \mu \mathrm{Ci} \)
b) \(\lambda=\frac{0.6931}{T_{\frac{1}{2}}}=\frac{0.6931}{5.01}=0.1383 \text { day }^{-1} \)
\(\lambda=\frac{0.6931}{T_{\frac{1}{2}}}=\frac{0.6931}{5.01 \times 24 \times 60 \times 60}=1.6 \times 10^{-6} \mathrm{~s}^{-1} \)
\(\lambda=1.6 \times 10^{-6} \mathrm{~s}^{-1} \)
c) \(\tau=\frac{1}{\lambda}=\frac{1}{0.1383}=7.23 \text { days } \)
\(\tau=7.23 \text { days } \)
\(R_{o}=\lambda N_{o} \)
\(N_{o}=\frac{R_{o}}{\lambda}=\frac{3.7 \times 10^{10} \times 10^{-6}}{1.6 \times 10^{-6}}=2.3 \times 10^{10} \)
\(N_{o}=2.3 \times 10^{10} \)

For A, Half life of  \(, \mathrm{T}_{\mathrm{A}} \) = 20 minutes, n = 4
For B, Half life of  \(, T_{B}\) = 40 minutes, n = 2
\(\mathrm{N}_{01}=\mathrm{N}_{02}=\mathrm{N}_{0}\)
sample left A, \(\frac{N_{1}}{N_{0}}=\left(\frac{1}{2}\right)^{n}=\left(\frac{1}{2}\right)^{4}=\frac{1}{16}\)
A-sample decayed \(=1-\frac{N_{1}}{N_{0}}=1-\frac{1}{16}=\frac{15}{16}\)
sample Ieft B, \(\frac{N_{1}}{N_{0}}=\left(\frac{1}{2}\right)^{n}=\left(\frac{1}{2}\right)^{2}=\frac{1}{4}\)
B-sample decayed \(=1-\frac{N_{2}}{N_{0}}=1-\frac{1}{4}=\frac{3}{4}\)
Ratio of decayed number of A and B \(=\frac{\frac{15}{16}}{\frac{1}{4}}=\frac{15}{16} \times \frac{4}{3}=\frac{5}{4}\)
Ratio of decayed number of A and |B = 5:4

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

Wavelength of incident radiation = 97.5 nm = 97.5 x 10^-9 m
Energy of hydrogen atom in its ground state = -13.6 eV
(a) Principal quantum number n = ?
(b) (i) Number of possible transitions = ?
(ii) Total number possible lines = ?
(a) Energy absorbed by Hydrogen atom
\(E=\frac{h c}{\lambda}=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{97.5 \times 10^{-9}} \mathrm{~J} \)
\(E=\frac{h c}{\lambda}=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{97.5 \times 10^{-9} \times 1.6 \times 10^{-19}} \mathrm{eV} \)
E = 12.74 eV
Energy of the electron in first orbit of Hydrogen is -13.6 ev
E_n = -13.6 + 12.74 = -0.86 eV
We know that
\(\mathrm{E}_{\mathrm{n}} =-\frac{13.6}{\mathrm{n}^{2}} \)
\(-0.86 =-\frac{13.6}{\mathrm{n}^{2}} \)
\(\mathrm{n}^{2} =15.88 \)
\(\mathrm{n} \cong 4 \)
(b) (i) By using arithmetic progression, For the principle quantum number "n",
Total number of possible transition form level n is \(\frac{\mathrm{n}(\mathrm{n}-1)}{2}\)
(ii) Total number of possible transitions form level 4 is 3
Total number of possible transitions form level 3 is 2
Total number of possible transition form level 2 is 1
Hence total number of possible transitions is 3 + 2 + 1 = 6

(i) An atom has a nucleus surrounded by electrons
(ii) Nucleus is made up of protons and neutrons.
(iii) Till 1960s, it was thought that protons, neutrons and electrons are fundamental building blocks of matter.
(iv) In 1964, physicists Murray Gellman and George Zweig theoretically proposed that protons and neutrons are not fundamental particles in fact they are made up of quarks.
(v) These quarks are now considered elementary particles of nature.
(vi) Electrons are fundamental or elementary particles because they are not made up of anything.
(vii) In the year 1968, the quarks were discovered experimentally by Stanford Linear Accelerator Center (SLAC), USA.
(viii) There are six quarks namely, up, down, charm, strange, top and bottom and their antiparticles.
(ix) All these quarks have fractional charges. For example, charge of up quark is + \(\frac23\) e and that of down quark is \(\frac13\) e.
(x) According to quark model, proton is made up of two up quarks and one down quark and neutron is made up of one up quark and two down quarks.

Fundamental forces of nature:
(i) It is known that there exists gravitational force between two masses and it is universal in nature. Our planets are bound to the Sun through gravitational force of the Sun.
(ii) ''Force is the external agency applied on a body to change its state of rest and motion"
There are four basic forces in nature.
(a) Gravitational force
(b) Electromagnetic force
(c) Strong nuclear force
(d) Weak nuclear force.
(a) Gravitational force :
(i) It is the force between any two objects in the universe.
(ii) It is an attractive force by virtue of their masses
(iii) By Newton's law of gravitation, the gravitational force is directly proportional to the product of the masses and inversely proportional to the square of the distance between them.
(iv) Gravitational force is the weakest force among the fundamental forces of nature but has the greatest large-scale impact on the universe.
(v) Unlike the other forces, gravity works universally on all matter and energy, and is universally attractive.
(b) Electromagnetic force :
(i) It is the force between charged particles or the force between two current carrying wires.
(ii) It is attractive for unlike charges and repulsive for like charges.
(iii) The electromagnetic force obeys inverse square law.
(iv) It is very strong compared to the gravitational force.
(v) It is the combination of electrostatic and magnetic forces.
(c) Strong nuclear force :
(i) It is the strongest of all the basic forces of nature.
(ii) It, however, has the shortest range, of the order of 10^-15 m.
(iii) This force holds the protons and neutrons together in the nucleus of an atom.
(d) Weak nuclear force :
(i) Weak nuclear force is even shorter in range than nuclear force.
(ii) This force plays an important role in beta decay and energy production of stars
(iii) During the fusion of hydrogen into helium in sun, neutrinos and enormous radiations are produced through weak force.
(iv) In our day to - day life, we require these four fundamental forces.
To put it in simple words :
(a) we are in the Earth because of Earth's gravitational attraction on our body.
(b) We are standing on the surface of the earth because of the electromagnetic force between atoms of the surface of the earth with atoms in our foot.
(c) The atoms in our body are stable because of strong nuclear force.
(d) Finally, the lives of species in the earth depend on the solar energy from the sun and it is due to weak force which plays vital role during nuclear fusion reactions going on in the core of the sun.

(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.

(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,
\(_{ 5 }^{ 12 }{ B }\rightarrow _{ 6 }^{ 12 }{ C+ }{ e }^{ - }+\overline { v } \)
\(_{ 6 }^{ 12 }{ C^* }\rightarrow _{ 6 }^{ 12 }{ C }+\gamma -rays\) 

(i) In beta decay, a radioactive nucleus emits either electron or positron. If electron (e^-) is emitted, it is called β^- decay and if positron (e⁺) is emitted, it is called β^- decay
(ii) The positron is an anti-particle of an electron whose mass is same as that of electron and charge is opposite to that of electron - that is, +e. Both positron and electron are referred to as beta particles.
β^- decay:
(iii) β^- decay: In β^- decay, the atomic number of the nucleus increases by one but mass number remains the same. This decay is represented by
\(_{ Z }^{ A }{ X }\rightarrow _{ Z+1 }^{ A }{ Y+ }{ e }^{ - }+\bar { v } \) ...(1)
(iv) It implies that the element X becomes Y by giving out an electron and antineutrino (⊽). 
(v) In other words, In each β^- decay, one neutron (n) in the nucleus of X is converted into a proton(p) by emitting an electron (e^-) and antineutrino(⊽). It is given by
\(n\rightarrow p+{ e }^{ - }+\bar { v } \) 
Example :
\(_{ 6 }^{ 14 }{ C }\rightarrow _{ 7 }^{ 14 }{ N+ }{ e }^{ - }+\overline { v } \)
β⁺ decay:
(vi) In β⁺ decay, the atomic number is decreased by one and the mass number remains the same. This decay is represented by
\(_{ Z }^{ A }{ X }\rightarrow _{ Z-1 }^{ A }{ Y+ }{ e }^{ + }+v\)
(vii) It implies that the element X becomes Y by giving out an positron (e⁺) and neutrino (v),
In otherwords, in each β⁺ decay, one proton(p) in the nucleus of X is converted into a neutron by emitting a positron (e⁺) and a neutrino. It is given by
\(p\rightarrow n+{ e }^{ +}+{ v } \)
Example: 
\(_{ 11 }^{ 22 }{ Na }\rightarrow _{ 10 }^{ 22 }{ Ne }+{ e }^{ + }+v\)
(viii) However a single proton (not inside any nucleus) cannot have β⁺ decay due to energy conservation, because neutron mass is larger than proton mass.
(ix) But a single neutron (not inside any nucleus) can have  β^- decay.
(x) It is important to note that the electron or positron which comes out from nuclei during beta decay never present inside the nuclei rather they are produced during the conversion of neutron into proton or proton into neutron inside the nucleus.

(i) When unstable nuclei decay by emitting an \(\alpha\) - particle (\(_{ 4 }^{ 2 }{ He }\) nucleus), it loses two protons and two neutrons. As a result, its atomic number Z decreases by 2, the mass number decreases by 4. 
\(_{ Z }^{ A }{ X }\rightarrow _{ Z-2 }^{ A-4 }{ Y+ }_{ 2 }^{ 4 }{ He }\)
(ii) X is called the parent nucleus and Y is called the daughter nucleus.
Example : when uranium \(_{ 92 }^{ 238 }U\) emit (α - particle) it is converted into thorium  \(_{ 90 }^{ 234 }{ Th }\) .
\(_{ 92 }^{ 238 }{ U\rightarrow }_{ 90 }^{ 234 }{ Th }+_{ 2 }^{ 4 }{ He }\)
(iii) Total mass of the daughter nucleus and \({ }_{2}^{4} \mathrm{He}\) nucleus is always less than that of parent nucleus.
(iv) The difference in mass \(\left(\Delta m=m_{X}-m_{Y}-m_{\alpha}\right)\) is released as energy called disintegration energy Q 
\(\mathrm{Q}=\Delta \mathrm{m} \times \mathrm{c}^{2}=\left(\mathrm{m}_{\mathrm{X}}-\mathrm{m}_{\mathrm{Y}}-\mathrm{m}_{\alpha}\right) \mathrm{c}^{2}\) 
(v) For spontaneous decay (natural radioactivity) Q > 0. In alpha decay process.
(a) The disintegration energy is positive (Q > 0).
(i) if the parent nucleus is not at rest, Q is equal to the net kinetic energy gained in the decay process.
(ii) if the parent nucleus is at rest, Q is equal to the total kinetic energy of daughter nucleus and the \({ }_{2}^{4} \mathrm{He}\) nucleus.
(b) The disintegration energy is negative (Q < 0).
The decay process cannot occur spontaneously and energy must be supplied to induce the decay.

(i) The strong nuclear force is of very short range, acting only up to a distance of a few Fermi. But inside the nucleus, the repulsive Coulomb force or attractive gravitational forces between two protons are much weaker than the strong nuclear force .between two protons. Similarly, the gravitational force between two neurons is. also much weaker than strong nuclear force between the neutrons. So nuclear force is the strongest force in nature.
(ii) The strong nuclear force is attractive and acts with an equal strength between proton-proton, proton-neutron, and neutron-neutron.
(iii) Nuclear force does not act on the electrons. So it does not alter the chemical properties of the atom.

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.

According to Einstein mass - energy equivalence, E = mc²
Here \(m=l u=1.66 \times 10^{-27} \mathrm{~kg} \)
\(E =1 \mathrm{u} \times \mathrm{c}^{2}=1.66 \times 10^{-27} \times\left(3 \times 10^{8}\right)^{2} \mathrm{~J} \)
\(E =\frac{1.66 \times 10^{-27} \times\left(3 \times 10^{8}\right)^{2}}{1.6 \times 10^{-19}} \mathrm{eV}=931 \times 10^{6} \mathrm{eV}=931 \mathrm{MeV} \)

(i) Bohr atom model is valid only for hydrogen atom or hydrogen-like atoms but not for complex atoms.
(ii) Bohr atom model does not explain fine structure of spectral lines.
(iii) Bohr atom model does not explain the intensity variations in the spectral lines.
(iv) The distribution of electrons in atoms is not completely explained by Bohr atom model.

(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.