The following are the drawbacks of Rutherford's model
(i) When an electron revolves around the nucleus radiates electromagnetic energy and hence radius of the orbit of electron decreases gradually. Thus the electron will finally on the 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.
(ii) According to it, we should obtain radiation of all possible wavelength but in actual practice, the atomic spectrum is a line spectrum.

De-Broglie hypothesis states that atomic particles of matter moving with a given velocity, (or momentum) can display wave like properties.

Einstein's photoelectric equation,
K.E. of photoelectron = Incident energy of photons - Work function
or  K.E = hv - W₀
or  K.E = hv - hv₀
where v₀ is called threshold frequency
Threshold Frequency : For a given metal, there exists a certain minimum frequency of the incident radiation below which no emission of photoelectrons takes place. This frequency is called threshold frequency.

For a particular frequency of incident radiation, the minimum negative (retarding) potential V_o given to plate A for which the photoelectric current becomes zero, is called cut-off or stopping potential.

Given, Kinetic energy = KE = 120 eV
p=\(\sqrt { 2eVm } =\sqrt {2KE.m }\)  \([\because K E=e V]\)
\(P=\sqrt { 2\times 120\times 1.6\times 10^{ -19 }\times 9.1\times 10^{ -31 } } \)
\(=5.91\times 10^{ -24 }\ kg-m/s\)
(ii) We know that momentum, p = mv
or, \(v=\frac{p}{m}=\frac{5.91 \times 10^{-24}}{9.1 \times 10^{-31}}\)
\(=6.5 \times 10^{6} \mathrm{~m} / \mathrm{s}\)
(iii) de-Broglie wavelength associated with electron,
\(\lambda =\frac { 12.27 }{ \sqrt { V_{ } } } \mathring { A } =\frac { 12.27 }{ \sqrt { 120 } } \mathring { A=0.112\times 10^{ -9 } } \ m=0.112 \ nm\)

Protential difference V = 56 V
(i)Use the kinetic energy
eV = 1/2mv² 
= 2eV/m
= v²
v = \(\sqrt { \frac { 2eV }{ m } } \)
where m is mass v is velocity
p = mv = m\(\sqrt { \frac { 2eV }{ m } } \)
\(=\sqrt { 2\times 1.6\times 10^{ -19 }\times 56\times 9\times 10^{ -31 } }\)
\( =4.02\times 10^{ -24 }\ kg-m/s\)

Since, we Know that for Balmer series, 
 \(\frac { 1 }{ \lambda } =R\left( \cfrac { 1 }{ { 2 }^{ 2 } } -\cfrac { 1 }{ { n }_{ 2 }^{ 2 } } \right) ,{ n }_{ 2 }=\ 3,4,5,............\)
For shortest wavelength in Balmer series, the spectral series is given by
\({ n }_{ 1 }=2,{ n }_{ 2 }=\infty \Rightarrow \frac { 1 }{ \lambda } =R\left( \cfrac { 1 }{ { 2 }^{ 2 } } -\cfrac { 1 }{ { \infty }^{ 2 } } \right) \)    
\(\Rightarrow \frac { 1 }{ \lambda } =R\times \cfrac { 1 }{ 4 } \Rightarrow \cfrac { 1 }{ \lambda } =\cfrac { R }{ 4 } \Rightarrow \lambda =\cfrac { 4 }{ R } \)
\(\lambda =\cfrac { 4 }{ 1.097\times { 10 }^{ 7 } } [\because R=1.097\times { 10 }^{ 7 }{ m }^{ -1 }]\)
\(\Rightarrow \lambda =3.64\times { 10 }^{ -7 }m\)  
The lines of Balmer series are found in the visible part of the spectrum. 

de-Broglie wavelength, \(\lambda=\frac{b}{p}=\frac{b}{\sqrt{2 m q V}}\)

10.2 V; 12.09 V.

The work function of a metal is the minimum energy required by an electron to just escape from the metal surface so as to overcome the restraining forces at the surface.
The relation between work function \(\left(\phi_0\right)\) and threshold frequency \(\left(v_0\right) \text { is } \phi_0=h v_0\) where h is Plank's constant. 

This model could not explain scattering of \(\alpha\) particle through large angles.

(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 \(\frac { 1 }{ { n }^{ 2 } } \)

This experiment confirms the wave nature of electron.
(b)   \(\lambda =\frac { h }{ p } =\frac { h }{ \sqrt { 2mK } }\)
\( K=K.E=eV\)
\(\lambda =\frac { h }{ \sqrt { 2meV } } \)

Potential difference (V)=100 V
de-Broglie Wavelength \(\lambda =\frac { 12.27 }{ \sqrt { V } } =\frac { 12.27 }{ \sqrt { 100 } } =1.227 \ \mathring { A } \)
In this case wavelength associated with an electron is of the order of wavelength of X-ray.

According to photon picture:
(i) Each quantum of radiation has energy hv
(ii) In photo - electric effect the electrons in the metal absorbs this quantum of energy (hv).
(iii) When this energy exceeds the minimum energy needed for the ejection of photoelectron, flow of photo current starts.

Potential difference, V = 56 V
Planck’s constant, h = 6.6 x 10⁻³⁴ Js
Mass of an electron, m = 9.1 x 10⁻³¹ kg
Charge on an electron, e = 1.6 x 10⁻¹⁹ C
At equilibrium, the kinetic energy of each electron is equal to the accelerating potential, i.e., we can write the relation for velocity (v) of each electron as:
\(\frac{1}{2} m v^{2}=e V\)
\(v^{2}=2 e \frac{V}{m}\)
\(\therefore v=\sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 56}{9.1 \times 10^{-31}}}\)
\(=\sqrt{19.69 \times 10^{12}}=4.44 \times 10^{6} \mathrm{~m} / \mathrm{s}\)
The momentum of each accelerated electron is given as:
p = mv
= 9.1 x 10⁻³¹ x 4.44 x 10⁶
= 4.04 x 10⁻²⁴ kg m s⁻¹
Therefore, the momentum of each electron is 4.04 x 10⁻²⁴ kg m s⁻¹_.
Potential difference, V = 56 V
Planck’s constant, h = 6.6 x 10⁻³⁴ Js
Mass of an electron, m = 9.1 x 10⁻³¹ kg
Charge on an electron, e = 1.6 x 10⁻¹⁹ C
De Broglie wavelength of an electron accelerating through a potential V, is given by the relation:
\(\lambda=\frac{12.27}{\sqrt{V}} \dot A\)
\(=\frac{12.27}{\sqrt{56}} \times 10^{-10} \mathrm{~m}\)
= 0.1639 nm
Therefore, the de Broglie wavelength of each electron is 0.1639 nm.

(a) Work function of caesium metal, ϕ₀ = 2.14eV
Frequency of light, v = 6.0 x 10¹⁴Hz
(a) The maximum kinetic energy is given by the photoelectric effect as:
= \(K=hv-{ \phi }_{ 0 }\)
Where,
h = Planck’s constant = 6.626 x 10⁻³⁴ Js
\(\therefore K=\frac{6.626 \times 10^{34} \times 6 \times 10^{14}}{1.6 \times 10^{-19}}-2.14\)
\(=2.485-2.140=0.345 \mathrm{eV}\)
Hence, the maximum kinetic energy of the emitted electrons is  0.345 eV.
(b) For stopping potential V₀, we can write the equation for kinetic energy as:
\(K=e V_{o}\)
\(\therefore V_{o}=\frac{K}{e}\)
\(=\frac{0.345 \times 1.6 \times 10^{-19}}{1.6 \times 10^{-19}}=0.345 \mathrm{~V}\)
Hence, the stopping potential of the material is 0.345 V.
c) Maximum speed of the emitted photoelectrons = v
Hence, the relation for kinetic energy can be written as:
\(K=\frac{1}{2} m v^{2}\)
Where,
m = Mass of an electron = 9.1 x 10⁻³¹ kg
\(v^{2}=\frac{2 K}{m}\)
\(=\frac{2 \times 0.345 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}=0.1104 \times 10^{12}\)
\(\therefore v=3.323 \times 10^{5} \mathrm{~m} / \mathrm{s}=332.3 \mathrm{~km} / \mathrm{s}\)
Hence, the maximum speed of the emitted photoelectrons is  332.3 km/s.