he radius of the innermost orbit of a hydrogen atom, r₁ = 5.3 x 10⁻¹¹ m.
Let r₂ be the radius of the orbit at n = 2. It is related to the radius of the innermost orbit as:
r₂ = (n)² r₁
= 4 x 5.3 x 10 ^-11 = 2.12 x 10 ^-10 m
For n = 3, we can write the corresponding electron radius as:
r₃ = (n)² r₁
= 9 x 5.3 x 10 ^-11 = 4.77 x 10 ^-10 m
Hence, the radii of an electron for n = 2 and n = 3 orbits are 2.12 x 10⁻¹⁰ m and 4.77 x 10⁻¹⁰ m respectively.

In the alpha-particle scattering experiment, if a thin sheet of solid hydrogen is used in place of a gold foil, then the scattering angle would not be large enough. This is because the mass of hydrogen (1.67 x 10⁻²⁷ kg) is less than the mass of incident α−particles (6.64 x 10⁻²⁷ kg). Thus, the mass of the scattering particle is more than the target nucleus (hydrogen). As a result, the α−particles would not bounce back if solid hydrogen is used in the α-particle scattering experiment.

The ratio of the radius of electron’s orbit to the radius of nucleus is (10^–10 m) /(10^–15 m) = 10⁵, that is, the radius of the electron’s orbit is 10⁵ times larger than the radius of nucleus. If the radius of the earth’s orbit around the sun were 10⁵ times larger than the radius of the sun, the radius of the earth’s orbit would be 10⁵ x 7 x 10⁸ m = 7 x 10¹³ m. This is more than 100 times greater than the actual orbital radius of earth. Thus, the earth would be much farther away from the sun. It implies that an atom contains a much greater fraction of empty space than our solar system does.

\(H_\alpha-\) line of the Balmer series in the emission spectrum of H-atom is obtained in visible region.

On removing one electron from He⁴ and He³, the energy levels, as worked out on the basis of Bohr's model will be very close as both the nuclei are very heavy as compared to electron mass. Also, after removing one electron, He⁴ and He³ atoms contain one electron and are hydrogen like atoms.

The radius R of atomic nucleus related to mass number A of the nucleus as \(R=R_0 A^{1 / 3} \text { where } R_0=1.2 \times 10^{-15} \mathrm{~m}\) an empty constant.

Ionisation energy is the minimum energy required to knock out an electron from an atom.Its value will be different for different atoms. Ionisation energy will also depend on the orbit from which electron is to be removed.
When an electron in hydrogen atom is replaced by a particle 200 times heavier than electron but having the same charge, ionisation energy will not change, as it depends only on charge and not on mass of particle.

(a)

(b) Number of atoms in 1 kg of pure Pu
\(=\frac{6.023 \times 10^{23}}{239} \times 1000=2.52 \times 10^{24}\)

As average energy released in fission is 18O MeV, the total energy released
= 2.52 \(\times\) 10²⁴ \(\times\)180 MeV = 4.53 \(\times\) 10²⁶ MeV

(a) Bohr's second postulate defines the stable orbits. This postulate states that the electron revolves around the nucleus only in those orbits for which the angular momentum is some integral multiple of h/2\(\pi\), where h is the Planck's constant (= 6.63 x 10^-34 J-s).

According to de-Broglie wavelength of moving electron \(\lambda=\frac{h}{m v_{n}}\)
where, v. is speed of electron revolving in nth orbit
As, 2 \(\pi\)r_n. = n\(\lambda\)
\(\therefore\) \(2 \pi r_{n}=\frac{n h}{m v_{n}}\)  
or \(m v_{n} r_{n}=\frac{n h}{2 \pi}=n(h / 2 \pi)\)
i.e, Angular momentum of electron revolving in nth orbit must be an integral multiple of  h/2\(\pi\),which is the quantum condition proposed by Bohr in his second postulate. 
(b) We know that, energy of electron in nth orbit is 
\(E_{n}=-\frac{13.6}{n^{2}} \mathrm{eV}\)
For n = 1, E₁ = -13.6 eV
Similarly, for n = 4,\(E_{4}=-\frac{13.6}{(4)^{2}} \mathrm{eV}\)
\(\therefore\) Energy difference, \(\triangle\)E = E₄ - E₁
\(=\left[-\frac{13.6}{16}-(-13.6)\right] \mathrm{eV}\)
Also, energy of photon is
\(\triangle\)E = hv \(\Rightarrow \quad v=\frac{\Delta E}{h}\)
From Eqs. (i) and (ii), we get
\(v=\left(-\frac{13.6}{16}+13.6\right) \times \frac{1.6 \times 10^{-19}}{6.63 \times 10^{-34}}\)
\(\therefore\) V = 3.1 x 10¹⁵ Hz

(a) The sizes of the atoms taken in Thomson’s model and Rutherford’s model have the same order of magnitude.
(b) In  the ground state of Thomson’s model, the electrons are in stable equilibrium. However, in Rutherford’s model, the electrons always experience a net force.
(c) A classical atom based on Rutherford’s model is doomed to collapse.
(d) An atom has a nearly continuous mass distribution in Thomson’s model, but has a highly non-uniform mass distribution in Rutherford’s model.
(e) The positively charged part of the atom possesses most of the mass in both the models.

we know that velocity of electron moving around a proton in hydrogen atom in an orbit of radius 5.3 × 10^–11 m is 2.2 × 10^–6 m/s. Thus, the frequency of the electron moving around the proton is
\(v=\frac{v}{2 \pi r}=\frac{2.2 \times 10^{6} \mathrm{~m} \mathrm{~s}^{-1}}{2 \pi\left(5.3 \times 10^{-11} \mathrm{~m}\right)}\)
\(\approx \) 6.6 × 10¹⁵ Hz
According to the classical electromagnetic theory we know that the frequency of the electromagnetic waves emitted by the revolving electrons is equal to the frequency of its revolution around the nucleus. Thus the initial frequency of the light emitted is 6.6 × 10¹⁵ Hz.

Total energy of the electron in hydrogen atom is –13.6 eV = –13.6 × 1.6 × 10^–19 J = –2.2 ×10^–18 J. Thus from Equation we have
\(-\frac{e^{2}}{8 \pi \varepsilon_{0} r}=-2.2 \times 10^{-18} \mathrm{~J}\)
This gives the orbital radius
\(r=-\frac{e^{2}}{8 \pi \varepsilon_{0} E}=-\frac{\left(9 \times 10^{9} \mathrm{~N} \mathrm{~m}^{2} / \mathrm{C}^{2}\right)\left(1.6 \times 10^{-19} \mathrm{C}\right)^{2}}{(2)\left(-2.2 \times 10^{-18} \mathrm{~J}\right)}\)
= 5.3 × 10^–11 m
The velocity of the revolving electron can be computed from Eq.uation with m = 9.1 × 10^–31 kg,
\(v=\frac{e}{\sqrt{4 \pi \varepsilon_{0} m r}}=2.2 \times 10^{6} \mathrm{~m} / \mathrm{s}\)

(i) According to the law of radioactive decay.
\(N={ N }_{ 0 }{ e }^{ -\lambda t }=\frac { { N }_{ 0 } }{ n } \) 
\(\therefore \frac { { N }_{ 0 } }{ n } ={ N }_{ 0 }{ e }^{ -\lambda t } \ \Rightarrow \ n={ e }^{ \lambda t } \ \Rightarrow \ \lambda =\frac { \log { \left( n \right) } }{ T } \) 
\(\therefore \)  Half-life, \({ T }_{ 1/2 }=\frac { 0.6931 }{ \lambda } =\frac { 0.693T }{ \log { \left( n \right) } } \)  
(ii) In  \(\beta ^{ - }\) - decay process, a nuclei emits a negative charge from the nucleus. A neutron is converted to a proton, causing the nuclide's atomic number to increase by one, but the atomic mass remains the same.
e.g. \(_{ 6 }^{ 14 }{ C\longrightarrow _{ 7 }^{ 14 }{ N+{ e }^{ - } } }+{ v }_{ e }^{ - }\).

\(3646.8\mathring { A } \)
\(Take \ { n }_{ 1 }=2 \ and \ { n }_{ 2 }=\infty \)

Radius of the orbit of the Earth around the Sun, r = 1.5 x 10¹¹ m
Orbital speed of the Earth, ν = 3 x 10⁴ m/s
Mass of the Earth, m = 6.0 x 10²⁴ kg
According to Bohr’s model, angular momentum is quantized and given as:
\(m v r=\frac{\mathrm{nh}}{2 \pi}\)
Where,
h = Planck’s constant = 6.62 x 10⁻³⁴ Js
n = Quantum number
\(\therefore n=\frac{\operatorname{mvr} 2 \pi}{h}\)
\(=\frac{2 \pi \times 6 \times 10^{24} \times 3 \times 10^{4} \times 1.5 \times 10^{11}}{6.62 \times 10^{-34}}\)
= 25.61 x 10⁷³ = 2.6 x 10⁷⁴
Hence, the quanta number that characterizes the Earth’ revolution is 2.6 x 10⁷⁴

(a) Let ν₁ be the orbital speed of the electron in a hydrogen atom in the ground state level, n₁ = 1. For charge (e) of an electron, ν₁ is given by the relation,
\(v_{1}=\frac{e^{2}}{n_{14} \pi \in_{0}\left(\frac{h}{2 \pi}\right)}=\frac{e^{2}}{2 \epsilon_{0} h}\)
Where,e = 1.6 x 10⁻¹⁹ C`
∈₀ = Permittivity of free space = 8.85 x 10⁻¹² N⁻¹ C² m⁻²
h = Planck’s constant = 6.62 x 10⁻³⁴ Js
\(\therefore v_{1}=\frac{\left(1.6 \times 10^{-19}\right)^{2}}{2 \times 8.85 \times 10^{-12} \times 6.62 \times 10^{34}}\)
= 0.0218 x 10⁸ =2.18 x 10⁶ m/s
For level n₂ = 2, we can write the relation for the corresponding orbital speed as:
\(v_{2}=\frac{e^{2}}{n_{3} 2 \in_{0} h}\)
\(=\frac{1.6 \times 10^{-19}}{3 \times 2 \times 8.85 \times 10^{-12} \times 6.62 \times 10^{-34}}\)
= 7.27 x 10⁵ m/s
Hence, the speed of the electron in a hydrogen atom in n = 1, n = 2, and n = 3 is 2.18 x 10⁶ m/s, 1.09 x10⁶ m/s, 7.27x 10⁵ m/s respectively.
(b) Let T₁ be the orbital period of the electron when it is in level n₁ = 1.
Orbital period is related to orbital speed as:
\(T_{1}=\frac{2 \pi r_{1}}{v_{1}}\)
Where
r₁ = Radius of the orbit
\(\frac{n_{1}^{2} h^{2} \in_{0}}{\pi m e^{2}}\)
h = Planck’s constant = 6.62 x 10⁻³⁴ Js
e = Charge on an electron = 1.6 x 10⁻¹⁹ C
∈₀ = Permittivity of free space = 8.85 x 10⁻¹² N⁻¹ C² m⁻²
m = Mass of an electron = 9.1 x 10⁻³¹ kg
\(\therefore T_{1}=\frac{2 \pi r_{1}}{v_{1}}\)
\(=\frac{2 \pi \times(1)^{2} \times\left(6.62 \times 10^{-34}\right)^{2} \times 8.85 \times 10^{-12}}{2.18 \times 10^{6} \times \pi \times 9.1 \times 10^{-31} \times\left(1.6 \times 10^{-19}\right)^{2}}\)
= 15.27 x 10^{- 17} =1.527 x 10^-16s
For level n₂ = 2, we can write the period as:
\(T_{2}=\frac{2 \pi r_{2}}{v_{2}}\)
Where,
r₂ = Radius of the electron in n₂ = 2
\(=\frac{\left(n_{2}\right)^{2} h^{2} \in_{0}}{\pi m e^{2}}\)
\(\therefore T_{2}=\frac{2 \pi r^{2}}{v_{2}}\)
\(=\frac{2 \pi \times(2)^{2} \times\left(6.62 \times 10^{-34}\right)^{2} \times 8.85 \times 10^{-12}}{1.09 \times 10^{6} \times \pi \times 9.1 \times 10^{-31} \times\left(1.6 \times 10^{-19}\right)^{2}}\)
= 1.22 x 10^-15 s
And, for level n₃ = 3, we can write the period as:
\(T_{3}=\frac{2 \pi r_{3}}{v_{3}}\)
Where,
r₃ = Radius of the electron in n₃ = 3
\(=\frac{\left(n_{3}\right)^{2} h^{2} \in_{0}}{\pi m e^{2}}\)
\(\therefore T_{3}=\frac{2 \pi r_{3}}{v_{3}}\)
\(=\frac{2 \pi \times(3)^{2} \times\left(6.62 \times 10^{-34}\right)^{2} \times 8.85 \times 10^{-12}}{7.27 \times 10^{5} \times \pi \times 9.1 \times 10^{-31} \times\left(1.6 \times 10^{-19}\right)}\)
= 4.12 x 10^-15 s
Hence, the orbital period in each of these levels is 1.52 x 10⁻¹⁶ s, 1.22 x 10⁻¹⁵ s, and 4.12 x 10⁻¹⁵ s respectively.

For ground level, n₁ = 1
Let E₁ be the energy of this level. It is known that E₁ is related with n₁ as:
\(E_{1}=\frac{-13.6}{n_{1}^{2}} e V\)
\(=-\frac{13.6}{1^{2}}=-13.6 \mathrm{eV}\)
The atom is excited to a higher level, n₂ = 4.
Let E₂ be the energy of this level.
\(\therefore E_{2}=\frac{-13.6}{n_{2}^{2}} e V\)
\(=\frac{-13.6}{4^{2}}=-\frac{13.6}{16} e V\)
The amount of energy absorbed by the photon is given as:
E = E₂ − E₁
\(=-\frac{13.6}{16}-\left(-\frac{13.6}{1}\right)\)
\(=\frac{13.6 \times 15}{16} e V\)
\(=\frac{13.6 \times 15}{16} \times 1.6 \times 10^{-19}=2.04 \times 10^{-18} J\)
For a photon of wavelength λ, the expression of energy is written as:
\(E=\frac{\mathrm{hc}}{\lambda}\)
\(=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{2.04 \times 10^{-18}}\)
= 9.7 x 10^-8 x m= 97nm
And, frequency of a photon is given by the relation,
\(v=\frac{v}{\lambda}\)
\(=\frac{3 \times 10^{8}}{9.7 \times 10^{-8}} \approx 3.1 \times 10^{15} \mathrm{~Hz}\)
Hence, the wavelength of the photon is 97 nm while the frequency is 3.1 x 10¹⁵ Hz.

Separation of two energy levels in an atom,
\(\begin{aligned} \Delta E & =2.3 \mathrm{eV}=2.3 \times 1.6 \times 10^{-19} \end{aligned}\)
\(\begin{aligned} =3.68 \times 10^{-19} \mathrm{~J} \end{aligned}\)
Let v be the frequency of radiation emitted, when the atom transits from the upper level to the lower level.
We have the relation for energy as, \(\Delta\) E = hv
where, \(h=6.63 \times 10^{-34} \mathrm{Js}\)
\(\therefore \quad v=\frac{\Delta E}{h}=\frac{3.68 \times 10^{-19}}{6.63 \times 10^{-34}}=5.6 \times 10^{14} \mathrm{~Hz}\)
Hence, the frequency of the radiation is 5.6 \(\times\) 10¹⁴ Hz.

The key idea here is that throughout the scattering process, the total mechanical energy of the system consisting of an\(\alpha\) -particle and a gold nucleus is conserved. The system’s initial mechanical energy is E_t, before the particle and nucleus interact, and it is equal to its mechanical energy E_f when the (-particle momentarily stops. The initial energy E_t is just the kinetic energy K of the incoming \(\alpha\)- particle. The final energy E_f is just the electric potential energy U of the system. The potential energy U can be calculated from equation. 
Let d be the centre-to-centre distance between the (-particle and the gold nucleus when the (-particle is at its stopping point. Then we can write the conservation of energy E_t = E_f as
\(K=\frac{1}{4 \pi \varepsilon_{0}} \frac{(2 e)(Z e)}{d}=\frac{2 Z e^{2}}{4 \pi \varepsilon_{0} d}\)
Thus the distance of closest approach d is given by
\(d=\frac{2 Z e^{2}}{4 \pi \varepsilon_{0} K}\)
The maximum kinetic energy found in\(\alpha\) -particles of natural origin is 7.7 MeV or 1.2 × 10^–12 J. Since 1/4 \(\pi\) E₀= 9.0 × 10⁹ N m²/C². Therefore with e = 1.6 × 10^–19 C, we have
\(d=\frac{(2)\left(9.0 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right)\left(1.6 \times 10^{-19} \mathrm{C}\right)^{2} \mathrm{Z}}{1.2 \times 10^{-12} \mathrm{~J}}\)
= 3.84 × 10^–16 Z m
The atomic number of foil material gold is Z = 79, so that d (Au) = 3.0 × 10^–14 m = 30 fm. (1 fm (i.e. fermi) = 10^–15 m.)
The radius of gold nucleus is, therefore, less than 3.0 × 10^–14 m. This is not in very good agreement with the observed result as the actual radius of gold nucleus is 6 fm. The cause of discrepancy is that the distance of closest approach is considerably larger than the sum of the radii of the gold nucleus and the \(\alpha\)-particle. Thus, the \(\alpha\)-particle reverses its motion without ever actually touching the gold nucleus.

\(T=1.5\times { 10 }^{ 17 }s;\)
\( R=?\)
\( N=2.5\times { 10 }^{ 21 }\)
\(R=\lambda N=\frac { 0.693N }{ T } =\frac { 0.693\times 2.5\times { 10 }^{ 21 } }{ 1.5\times { 10 }^{ 17 } } \)
\(=11550\ disintegrations/sec.\)