At the distance of closest approach \(\left( { r }_{ 0 } \right) \),
\(K{ E }_{ \alpha }=\frac { \left( Ze \right) \left( 2e \right) }{ 4\pi { \epsilon }_{ 0 }{ r }_{ 0 } } \)  and \(K{ E }_{ p }=\frac { \left( Ze \right) \left( e \right) }{ 4\pi { \epsilon }_{ 0 }{ r }_{ 0 } }\).
Clearly, \(K{ E }_{ p }=\frac { 1 }{ 2 } K{ E }_{ \alpha } \)

As radius, \(r\propto \frac { 1 }{ m } \)
\(\therefore \) When electron is replaced by a particle 200 times heavier, the radius would decrease to \(\frac { 1 }{ 200 } \) time the original radius.

When an electron of mass m is confined to move on a line of length l with velocity v, the de-Broglie wavelength \(\lambda \) associated with electron is  \(\lambda =\frac { h }{ mv } =\frac { h }{ p } \ \ or \ \ p=\frac { h }{ \lambda } =\frac { h }{ { 2l }/{ n } } =\frac { nh }{ 2l } \)
When electron revolves in a circular orbit of radius r; then \(2l=2\pi r\)
\( \ \therefore \ \ p=\frac { nh }{ 2\pi r } \\ \ \ or \ \ p\times r=\frac { nh }{ 2\pi } \)
i.e., angular momentum  \((p\times r)\) of electron is integral multiple of \({ h }/{ 2\pi }\) . This is Bohr's quantization condition of angular momentum.

(i) 10.2 eV;
(ii) The highest permitted energy level to the first permitted level,
\(\Delta E=E_{\infty}-E_1\) = 0 - (-13.6) = 13.6 eV
Ratio of energies of photon
\(=\frac{10.2}{13.6}=\frac{3}{4}=3: 4\)

(i) For hydrogen atom,
E₁ = -13.6 eV
E₂ = -3.4 eV
E₃ = -1.51 eV
E₄ = -0.85 eV
h = 6.63x10^-34 Js;
c = 3 x 10⁸ ms^-1
Photon Energy = \(\frac{hc}{\lambda}\)
=\(\frac{6.63\times 10^{-34}\times 3\times 10^{8}}{496\times 10{-9}\times 1.6\times10^{-19}}\)
= 2.5 eV
This equals (nearly) the difference (E₄- E₂).
Hence the required transition is (n = 4) to ( n = 2)
(ii) The transition n = 4 to n = 3 corresponds to emission of radiation of maximum wavelength.
It is so because this transmission gives out the photon of least energy.

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

\( 6562\mathring { A } \)
For \({ H }_{ \alpha }\) line,  \({ n }_{ 1 }=2\ and\ { n }_{ 2 }=3\) and for \({ H }_{ \beta }\) line \({ n }_{ 1 }=2\ and\ { n }_{ 2 }=4\)

\(\lambda=\frac{h}{p}=\frac{h}{mv}\)
\(2\pi r = n \lambda\)
\({2\pi r }= {n} \frac {h}{mv}\)
\(mvr = L = \frac {nh}{2n}\)
(b) kinetic energy = 13.6 eV
Potential energy = -27.2 eV

(i) The kinetic energy (E_K) of the electron in an orbit is equal to negative of its total energy (E).
 E_K = -E = -(-1.5) = 1.5 eV
(ii) The potential energy (E_p) of the electron in an orbit is equal to twice its total energy (E).
E_P = 2E = -1.5 x 2 = -3eV
(iii) As, a result of transition of electron from excited state to ground state.
Energy of radiation = -1.5 - (- 13.6)
(\(\because\) Ground state energy of H-atom = - 13.6 eV)
E = hv = h\(\frac { c }{ \lambda } \)
\(\Rightarrow\) \(\frac { hc }{ \lambda } =12.1eV\) = energy of radiation
\(\therefore \frac { 1 }{ \lambda } =\frac { 12.1\times 1.6\times { 10 }^{ -19 } }{ 6.62\times { 10 }^{ -34 }\times 3\times { 10 }^{ 8 } } \)
\(\Rightarrow \lambda =1.025\times { 10 }^{ -7 }m=1025\mathring { A } \)

Basic assumptions of Rutherford atomic model are given below:
(i) The atom consists of small central core called atomic nucleus in which whole mass and positive charge is assumed to be concentrated.
(ii) The size of the nucleus is much smaller than the size of the atom
(iii) The nucleus is surrounded by electrons. Atoms are electrically neutral as the total negative charge of electrons surrounding the nucleus is equal to the total positive charge on the nucleus.
(iv) Electrons revolve around the nucleus in various circular orbits and necessary centripetal force is provided by the electrostatic force of attraction between the positively charged nucleus and negatively charged electrons.
Stability of atom When an electron revolves around the nucleus, then it radiates electromagnetic energy and hence, the radius of the orbit of electron decreases gradually. Thus, electron revolve on spiral path of decreasing radius and finally, it should fall into the nucleus, but this does not happen. Thus, Rutherford atomic model cannot account for the stability of the atom.

\(mvr = \frac {nh}{2\pi}\)
\(\frac { { mv }^{ 2 } }{ r } =\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { { e }^{ 2 } }{ { r }^{ 2 } } \)
\(r=\frac { { e }^{ 2 } }{ { 4\pi \varepsilon }_{ 0 }{ mv }^{ 2 } } \)
\(r=\frac { { Ze }^{ 2 } }{ { 4\pi \varepsilon }_{ 0 }m{ \left( \frac { nh }{ 2\pi mr } \right) }^{ 2 } } \)
\(\Rightarrow\)  \(r=\frac { { \epsilon }_{ 0 }{ n }^{ 2 }{ h }^{ 2 } }{ { \pi me }^{ 2 } } \)
Potential energy U \(=-\frac { 1 }{ 4{ \pi \epsilon }_{ 0 } } .\frac { { e }^{ 2 } }{ { r } } \)
\(=\frac { { me }^{ 4 } }{ { 4\epsilon }_{ 0 }{ n }^{ 2 }{ h }^{ 2 } } \) 
\(KE=\frac { 1 }{ 2 } { mv }^{ 2 }=\frac { 1 }{ 2 } m{ \left( \frac { nh }{ 2\pi mr } \right) }^{ 2 }\)
\(=\frac { { n }^{ 2 }{ h }^{ 2 }{ \pi }^{ 2 }{ m }^{ 2 }{ e }^{ 4 } }{ { 8\pi }^{ 2 }{ me }_{ 0 }^{ 2 }{ n }^{ 2 }{ h }^{ 2 } } \)
\(KE=\frac { { me }^{ 4 } }{ { 8\varepsilon }_{ 0 }^{ 2 }{ n }^{ 2 }{ h }^{ 2 } } \)
TE = KE + PE
\(=-\frac { { me }^{ 4 } }{ { 8\epsilon }_{ 0 }^{ 2 }{ n }^{ 2 }{ h }^{ 2 } } \)
(b)Rydberg formula: For first member of Lyman series
  \(\frac { 1 }{ \lambda } =R\left( \frac { 1 }{ { 1 }^{ 2 } } -\frac { 1 }{ { 2 }^{ 2 } } \right) \)
\(=\frac { 4 }{ 3R } \)
For first member of Balmer Series
\(\frac { 1 }{ \lambda } =R\left( \frac { 1 }{ { 2 }^{ 2 } } -\frac { 1 }{ { 3 }^{ 2 } } \right) \)
\(\lambda =\frac { 36 }{ 5R } \)