Whenever an electron jumps from higher energy orbit En₂ to lower energy orbit En_1, the frequency of the radiation emitted is given by hv = En₂ - En₂. It is called Bohr's frequency condition.

The transition representing the Balmer and Lyman series are shown in fig

\(2.25\times { 10 }^{ 6 }m{ s }^{ -1 }\)

(i) According to Bohr's principle, electrons revolve in a stationary orbit of which energy and momentum are fixed. The momentum of electrons in the fixed orbit is given by \(\frac { nh }{ 2\pi } \) (where n = the number of orbits). According to de-Broglie's hypothesis, the electron is associated with wave character. Hence, a circular orbit can be taken to be a stationary energy state only. if it contains an integral number of de-Broglie wavelengths, i.e., 2\(\pi\)r = n\(\lambda\)
(ii) According to question,

\({ E }_{ B }-{ E }_{ C }=\frac { hc }{ { \lambda }_{ 1 } } \quad \quad ...(i)\)
\({ E }_{ A }-{ E }_{ B }=\frac { hc }{ { \lambda }_{ 2 } } \quad \quad ...(ii)\)
\({ E }_{ C }-{ E }_{ A }=\frac { -hc }{ { \lambda }_{ 3 } } \quad ...(iii)\)
On adding Eqs. (i), (ii) and (iii), we get
E_B - E_C + E_A - E_B + E_C - E_A
\(=hc\left( \frac { 1 }{ { \lambda }_{ 1 } } +\frac { 1 }{ { \lambda }_{ 2 } } -\frac { 1 }{ { \lambda }_{ 3 } } \right) \)
\(\frac { 1 }{ { \lambda }_{ 3 } } =\frac { 1 }{ { \lambda }_{ 1 } } +\frac { 1 }{ { \lambda }_{ 2 } } \Rightarrow { \lambda }_{ 3 }=\frac { { { \lambda } }_{ 1 }{ \lambda }_{ 2 } }{ { \lambda }_{ 1 }+{ \lambda }_{ 2 } } \)

(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
\(\therefore \frac { 1 }{ \lambda } =\frac { 12.1\times 1.6\times { 10 }^{ -19 } }{ 6.62\times { 10 }^{ -34 }\times 3\times { 10 }^{ 8 } } \)
\(\Rightarrow \lambda =1.025\times { 10 }^{ -7 }m=1025\mathring { A } \)

(i) As radius of electron's n^th orbit in hydrogen atom
\({ r }_{ n }=\frac { { \varepsilon }_{ 0 }{ h }^{ 2 } }{ \pi m{ e }^{ 2 } } { n }^{ 2 }\quad \Rightarrow { r }_{ n }\propto { n }^{ 2 }\)
(ii) Also, the total energy of an electron belonging to n^th orbit,
\({ E }_{ n }=-\frac { m{ e }^{ 2 } }{ 8{ \varepsilon }_{ 0 }^{ 2 }{ n }^{ 2 }{ h }^{ 2 } } \Rightarrow \left| { E }_{ n } \right| \propto \frac { 1 }{ { n }^{ 2 } } \)
i.e. total energy of electron increases as \(\frac { 1 }{ { n }^{ 2 } } \)

Given figure shows a schematic diagram of Geiger -Marsden experiment.


The size of the nucleus can be obtained by finding impact parameter b using trajectories of n-particle. The impact parameter is the perpendicular distance of the initial velocity vector of o-particle from the central line of nucleus when it is far away from the atom.
Rutherford calculated impact parameter as
\(b=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } .\frac { Z{ e }^{ 2 }\cot { \left( { \theta }/{ 2 } \right) } }{ E } \)
where, 
E = KE of (\(\alpha\) -particle)
\(\theta\) = scattering angle
Z = atomic number of atom
The size of the nucleus is smaller than the impact parameter.
The idea of the size of the nucleus can also be obtained by finding the distance of closest approach.

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.

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, \(K{ E }_{ p }=\frac { 1 }{ 2 } K{ E }_{ \alpha } \)

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.

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.

\(Here,\ { r }_{ 0 }=2.8\times { 10 }^{ -14 }m,Z=79,\)
\( E=?\)
\( From\quad E=\frac { 1 }{ { 4\pi \varepsilon }_{ 0 } } \frac { (Ze)(2e) }{ { r }_{ 0 } } \)
\(E=\frac { 1 }{ { 4\pi \varepsilon }_{ 0 } } \frac { { 2Ze }^{ 2 } }{ { r }_{ 0 } } =\frac { 9\times { 10 }^{ 9 }\times 2\times 79{ \left( 1.6\times { 10 }^{ -19 } \right) }^{ 2 } }{ 2.8\times { 10 }^{ -14 } }\)
\(=1.300\times { 10 }^{ -12 }J\)
\( =\frac { 1.300\times { 10 }^{ -12 } }{ 1.6\times { 10 }^{ -13 } } MeV=8.125MeV\)
\(As \ E \ is \ doubled,{ r }_{ 0 } \ becomes \ half\)
\( \frac { { r }_{ 0 } }{ 2 } =1.4\times { 10 }^{ -14 }m\)

\(Here,\ { E }_{ n }=\frac { -13.6 }{ { n }^{ 2 } } eV\)
\( Putting\ n=1,2,3,......,\ we\ get\)
\( { E }_{ 1 }=\frac { -13.6 }{ { 1 }^{ 2 } } eV=-13.6eV\)
\({ E }_{ 2 }=\frac { -13.6 }{ { 2 }^{ 2 } } eV=-3.4eV\)
\( { E }_{ 3 }=\frac { -13.6 }{ { 3 }^{ 2 } } eV=-1.51eV\)
\( { E }_{ 4 }=\frac { -13.6 }{ { 4 }^{ 2 } } eV=-0.85eV;{ E }_{ \infty }=0\)
Clearly, an electron in a hydrogen atom cannot have the energy of -6.8 eV.
(ii) As the value of n increases, energy diff. between two consecutive energy levels decreases.

\(mvr = \frac {nh}{2\pi}\)
\(\frac { { mv }^{ 2 } }{ r } =\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { { e }^{ 2 } }{ { r }^{ 2 } } \)
\(r=\frac { { e }^{ 2 } }{ { 4\pi \varepsilon }_{ 0 }{ mv }^{ 2 } } \)
\(r=\frac { { Ze }^{ 2 } }{ { 4\pi \varepsilon }_{ 0 }m{ \left( \frac { nh }{ 2\pi mr } \right) }^{ 2 } } \)
\(\Rightarrow\)  \(r=\frac { { \epsilon }_{ 0 }{ n }^{ 2 }{ h }^{ 2 } }{ { \pi me }^{ 2 } } \)
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
\(\frac { 1 }{ \lambda } =R\left( \frac { 1 }{ { 2 }^{ 2 } } -\frac { 1 }{ { 3 }^{ 2 } } \right) \)
\(\lambda =\frac { 36 }{ 5R } \)