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) 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.

(i) Most of the alpha particles are un-deflected through the gold foil and went straight.
(ii) Some of the alpha particles are deflected through a small angle.
(iii) A few alpha particles (one in thousand) are deflected through the angle more than 90^o.
(iv) Very few alpha particles returned back (ie) deflected back by 180^o.