Stopping potential is more for the curve corresponding to frequency V₁ 
Stopping po ential is directly proportional to the frequency of incident radiation.

(i) Slope is determined by h and e. (or slop is independent of the metal used)
(ii) Work function of the metal.

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. 

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.

\(Here,{ m }_{ 0 }=9.1\times { 10 }^{ -31 }kg;h=6.63\times { 10 }^{ -34 }Js;v=0.6c \ or \ v/c=0.6\)
Since electron is moving with a speed comparable to the velocity of light, so
\(\lambda =\frac { h }{ mv } =\frac { h\sqrt { 1-{ v }^{ 2 }/{ c }^{ 2 } } }{ { m }_{ 0 }v } =\frac { 6.63\times { 10 }^{ -34 }\sqrt { 1-{ \left( 0.6 \right) }^{ 2 } } }{ 9.1\times { 10 }^{ -31 }\times \left( 0.6\times 3\times { 10 }^{ 8 } \right) } =3.24\times { 10 }^{ -12 }m\)