Rydberg's formula \(\frac{1}{\lambda}=R\left(\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}}\right)\)
The transition corresponding to longest wavelength in the Lyman series,
\( \text { i.e. } n_{i}=2 \text { and } n_{f}=1 \)
\(\therefore \quad \frac{1}{\lambda} =R\left(1-\frac{1}{4}\right)=\frac{3}{4} R \)
\(\Rightarrow \quad \lambda =\frac{4}{3 R}=\frac{4}{3 \times 1.1 \times 10^{7}} \mathrm{~m} \)
\(=1.21 \times 10^{-7} \mathrm{~m} \)
\(\lambda =121 \mathrm{~nm} \)
The transition corresponding to longest wavelength in the Balmer series
\( \text { i.e. } n_{i}=3 \text { and } n_{f}=2 \)
\(\therefore \quad \frac{1}{\lambda} =R\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{5}{36} R \)
\(\Rightarrow \quad \lambda =\frac{36}{5 R}=\frac{36}{5 \times 1.1 \times 10^{7}} \)
\( =6.545 \times 10^{-7} \mathrm{~m} \)
\(\lambda =655 \mathrm{~nm} \)
The first transition lies in the ultraviolet region and the second one belongs to visible region.

According to the Rutherford's nuclear model of an atom, most of the space in an atom is empty with a positively charged nucleus occupying only a small space at its centre. Thus, for most of the α-particles, the impact parameter being very large, they pass through the atom without being affected by the positively charged nucleus and suffer scattering at an angle less than 90°. Only a few gets scattered at θ > 90°.

Conclusions:
(i) The mass of the atom is concentrated in a small volume.
(ii) The entire positive charge ofthe atom is with the nucleus and the negatively charged electrons are distributed around it.

\(_{ 3 }{ { Li }^{ 7 } }\)

0.511 MeV,  941.1 MeV
\({ E }_{ 1 }={ m }_{ e }{ c }^{ 2 }=9.1\times { 10 }^{ -31 }{ \left( 3\times { 10 }^{ 8 } \right) }^{ 2 }J\)
\(=\frac { 81.9\times { 10 }^{ -15 } }{ 1.6\times { 10 }^{ -13 } } MeV=0.511 \ MeV\)
\({ E }_{ 2 }={ m }_{ p }{ c }^{ 2 }=1.673\times { 10 }^{ -27 }{ \left( 3\times { 10 }^{ 8 } \right) }^{ 2 }J\)
\(=\frac { 1.673\times { 9\times 10 }^{ -11 } }{ 1.6\times { 10 }^{ -13 } } MeV=941.1 \ MeV\)

\(E_n=-\frac{13.6}{n^2} \mathrm{eV}\)
when \(n=\infty\)

Rydberg’s formula is given as
\(\frac{\mathrm{hc}}{\lambda}=21.76 \times 10^{-19}\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]\)
Where,
h = Planck’s constant = 6.6 x 10⁻³⁴ Js
c = Speed of light = 3 x 10⁸ m/s
(n₁ and n₂ are integers)
The shortest wavelength present in the Paschen series of the spectral lines is given for values n₁ = 3 and n₂ = ∞.
\(\frac{\mathrm{hc}}{\lambda}=21.76 \times 10^{-19}\left[\frac{1}{(3)^{2}}-\frac{1}{(\infty)^{2}}\right]\)
\(\lambda=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8} \times 9}{21.76 \times 10^{-19}}\)
= 8.189 x 10^-7 m
= 818.9 nm

(a) (i) Electron moving in a circular orbit around the nucleus would get accelerated, therefore it would spiral into the nucleus, as it looses its energy.
(ii) It must emit a continuous spectrum. According to Bohr's model of hydrogen atom
(i) Electron in an atom can revolve in certain stable orbits without the emission of radiant energy

Number of Uranium atoms fissioned
\(n=\frac { total \ energy \ released }{ energy \ released/fission } \)
\( =\frac { 7.6\times { 10 }^{ 13 } }{ 200\times 1.6\times { 10 }^{ -13 } } =2.375\times { 10 }^{ 24 }\)
\( =\frac { 7.6\times { 10 }^{ 13 } }{ 200\times 1.6\times { 10 }^{ -13 } } =2.375\times { 10 }^{ 24 }\)
\( Mass \ of \ Uranium=\frac { Mass \ number }{ Avogadro's \ number } \times n\)
\(=\frac { 235\times 2.375\times { 10 }^{ 24 } }{ 6.023\times { 10 }^{ 23 } } =926.66g\)

∵ Distance of closest approach, \(r_{0}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{1} q_{2}}{K . E}\)
As the kinetic energy is same and q₁ = e for a proton,
q' = 2e for an α-particle
q₂ = Ze, charge on nucleus.
Then \(\frac{\left(r_{0}\right)_{p}}{\left(r_{0}\right)_{\alpha}}=\frac{q_{1}}{q_{1}^{\prime}}=\frac{e}{2 e}=\frac{1}{2}\)

According to the Rutherford model, electrons are revolving round the nucleus in different orbits. These revolving electrons have some acceleration, so this will radiate energy in the form of EM wave, hence the orbits of electrons will go on decreasing and finally they will fall into the nucleus.
According to the Bohr model, electrons can revolve only in definite orbits without radiating any energy, where the angular momentum of an electron is an integral multiple of h/2π.

The distance of closest approach is given by
\(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{2 e \times Z_e}{r}=K\)         ...(i)
i.e.   \(r \propto \frac{1}{K}\)
Let r₀ be the new distance of closest approach for a twice energetic \(\alpha\)-particle.
\(\frac{r_0}{r}=\frac{K}{2 K}=\frac{1}{2} \Rightarrow r_0=\frac{r}{2}\)

Number of \(\alpha\)-particles scattered at an angle of \(\theta\)B is given by
\(N \propto \frac{1}{\sin ^{4} \theta / 2} \Rightarrow \frac{N_{1}}{N_{2}}=\left(\frac{\sin \frac{\theta_{2}}{2}}{\sin \frac{\theta_{1}}{2}}\right)^{4}\)
\(\Rightarrow \frac{100}{N_{2}}=\left(\frac{\sin 30^{\circ}}{\sin 45^{\circ}}\right)^{4} \Rightarrow \frac{100}{N_{2}}=\frac{4}{16} \Rightarrow\) N ₂ = 400

A larger number of alpha particle went through undeflected