Kinetic energy of particle of  mass m having momentum p is given by
\(k=\frac { p^{ 2 } }{ 2m } \quad p=\sqrt { 2mK } \)
The de-Broglie wavelength \(\lambda =\frac { h }{ p } =\frac { h }{ \sqrt { 2mK } }\)
\( \\ p=\frac { h }{ \lambda } \)
\(\\ K=\frac { h^{ 2 } }{ 2m\lambda ^{ 2 } } \)
If \(\lambda \) is constant then from Eq (i) we get
p = constant,i.e m_pv_p = m_ev_e
\(\text { or } \frac{v_{p}}{v_{e}}=\frac{m_{e}}{m_{p}}<1\)
or vp < ve
If \(\lambda \) is constant then from (ii)  \(K \propto \frac{1}{m}\)
\((\therefore \frac{K_{p}}{K_{e}}=\frac{m_{e}}{m_{p}}<1 \text { or } K_{p})\)
It means that the velocity of electron is greater than that of proton. Kinetic energy of electron is greater than that of proton.

Given, thershold frequency of A is
\({ v }_{ OA }=5\times { 1 }0^{ 14 }\quad Hz\)
and thershold frequency of B is \({ v }_{ OB }=10\times { 1 }0^{ 14 }\quad Hz\)
We know that work function,
\({ \phi }_{ 0 }=h{ v }_{ 0 } \ or \ { \phi }_{ 0 }\propto { v }_{ 0 }\)
\(So,\ \frac { { \phi }_{ OA } }{ { \phi }_{ OB } } =\frac { 5\times { 10 }^{ 14 } }{ 10\times { 10 }^{ 14 } } <1 \ \Rightarrow { \phi }_{ OA }<{ \phi }_{ OB }\)
Thus, work function of B is higher than A.
(ii) For metal A, slope
\(=\frac { h }{ e } =\frac { 2 }{ (10-5)\times { 10 }^{ 14 } } \)
\(or \ h=\ \frac { 2\times e }{ 5\times { 10 }^{ 14 } } =\frac { 2\times 1.6\times { 10 }^{ -19 } }{ 5\times { 10 }^{ 14 } } \)
\(=6.4\times { 10 }^{ -34 }\ J-s\) 
For metal B, slope \(=\frac { h }{ e } =\frac { 2.5 }{ (15-10)\times { 10 }^{ 14 } } \)
\(\\ or\quad h=\frac { 2.5\times e }{ 5\times { 10 }^{ 14 } } =\frac { 2.5\times 1.6\times { 10 }^{ -19 } }{ 5\times { 10 }^{ 14 } } =8\times { 10 }^{ -34 }J-s\)
Since, the value of b from experiment for metals A and B is different. Hence, experiment is not consistent with theory. 

(i) Important properties of photons which are used to establish Einstein's photoelectric equations are as follows:
(a) In interaction of radiation with matter, radition behaves as, if it is make up of particles called photons.
(b) All photons of light of a particular frequency v or wavelength \(\lambda \) have the same energy \(E(=hv=\frac { hc }{ \lambda } )\) and \(P(=\frac { hv }{ c } =\frac { h }{ \lambda } )\)
Whatever the indensity of radiation may be.
(ii) Since, Einstein's photoelectric equation is given by
\({ KE }_{ max }=\frac { 1 }{ 2 } { mv }^{ 2 }_{ max }=hv=h{ v }_{ 0 } =e{ V }_{ 0 }\)
(a) For a given material, there exist a certain minimum frequency of the incident radition, below which no emission of photoelectron takes place. This frequency is called threshold frequency (V₀). Above threshold frequency, the maximum kinetic energy of the emitted photoelectron or equivalent stopping potential is independent of the intensity of the incident light but depends only upon the frequency of the incident light.
(b) If the collecting plate in the photoelectric apparatus is made at high negative potenrial, then most of the high energetic electrons get repelled back along the same path and the photoelectric current in the circuit becomes zero. So, for a particular frequency of incident radiation, the minimum negative potential for which the electric current becomes zero is called cut-off or stopping potential(V₀).

de-Broglie matter wave equation,
\(\lambda =\frac { h }{ p } =\frac { h }{ \sqrt { 2mK } } \ \left[ \because K=\frac { { P }^{ 2 } }{ 2m } \right] \)
where K is kinetic energy and m is a mass of the particle.
\(K=\frac { { h }^{ 2 } }{ 2m{ \lambda }^{ 2 } } \)   [ for same wavelength \( \lambda] \)
\(K\propto \frac { 1 }{ m }\)
\(\Rightarrow \ { K }_{ e }:{ K }_{ \alpha }:{ K }_{ p }=\frac { 1 }{ { m }_{ e } } :\frac { 1 }{ { m }_{ \alpha } } :\frac { 1 }{ { m }_{ p } } \)
where \({ m }_{ e },{ m }_{ p } \ and \ { m }_{ \alpha }\) are masses of electron, proton and \(\alpha \) -particle, respectively.
Also, \({ K }_{ e },k_{ p } \ and \ K_{ \alpha }\) are their respective kinetic energies.
\(\because \ m_{ \alpha }>m_{ p }>m_{ e }\)
\(\\ \Rightarrow m_{ \alpha }m_{ p }>m_{ e }m_{ \alpha }>m_{ e }m_{ p }\)
\(\\ { K }_{ e }>k_{ p }>K_{ \alpha }\)
(i) \(\alpha \) particle possess minimum kinetic energy
(ii) The electron has maximum kinetic energy. The magnifying power of an electron microscope is inversely related to the wavelength of radiation used. The Smaller wavelength of the electron beam in comparison to visible light increases the magnifying power of the microscope.

Since, \(p=\frac { h }{ \lambda } \)
Given, ​​​​​​\(\lambda =0.4\mathring { A } =0.4\times 10^{ -10 }m\)
\(h=6.6\times { 10 }^{ -34 } \ J-s\)
\( \therefore \ p=\frac { 6.6\times { 10 }^{ -34 } }{ 0.4\times 10^{ -10 } } \)
\(\Rightarrow \ p=16.5\times { 10 }^{ -24 }\ kg-ms^{ -1 }\)