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\(\sqrt { \frac { 2eV }{ m } } \)
\(=\sqrt { 2\times 1.6\times 10^{ -19 }\times 56\times 9\times 10^{ -31 } }\)
\( =4.02\times 10^{ -24 }\ kg-m/s\)

de-Broglie wavelength is given by
\(\lambda =\frac { h }{ \sqrt { 2mK } } =\frac { h }{ \sqrt { 2mqV } }\)
\(\lambda \propto \frac { 1 }{ \sqrt { mqV } } \)
m = mass of charge particle, q = charge and V = potential difference
Raio of de-Broglie wavelengths of proton and a-particle is given by
\(\frac { \lambda _{ \alpha } }{ \lambda _{ p } } =\sqrt { \frac { m_{ \alpha }q_{ \alpha }V_{ \alpha } }{ m_{ p }q_{ p }V_{ p } } } =\sqrt { \left( \frac { m_{ \alpha } }{ m_{ p } } \right) \left( \frac { q_{ \alpha } }{ q_{ p } } \right) \left( \frac { V_{ \alpha } }{ V_{ p } } \right) } \)
\(\frac { m_{ \alpha } }{ m_{ p } } =4,\frac { q_{ \alpha } }{ q_{ p } } =2,\frac { V_{ \alpha } }{ V_{ p } } =\frac { 1 }{ 2 } \)
a-particle is 4 toimes heavier than proton and it has double the charge than that of proton
\(\frac { \lambda _{ p } }{ \lambda _{ \alpha } } =\sqrt { 4\times 2\times \frac { 1 }{ 2 } } =2\quad \Rightarrow \lambda _{ p }:\lambda _{ \alpha }=2:1\)

The variation of stopping potential with the frequency of radiation, incident on a metal plate is a straight line AB as shown in the figure.

Take two point C and D on the graph.
The corresponding frequency of radiation is v_1, v₂ and stopping potential is V₁,V₂.
Then,  \(e{ V }_{ 1 }=h{ v }_{ 1 }-{ \phi }_{ 0 }\) and \(e{ V }_{ 2 }=h{ v }_{ 2 }-{ \phi }_{ 0 }\) 
 \(\therefore \quad e\left( { V }_{ 2 }-{ V }_{ 1 } \right) =h\left( { V }_{ 2 }-{ V }_{ 1 } \right) \) or \(h=\frac { e\left( { V }_{ 2 }-{ V }_{ 1 } \right) }{ { V }_{ 2 }-{ V }_{ 1 } } \) 
Thus, Planck's constant can be determined.

During photelectric emission, the momentum of incident photon is transferred to the metal. At microscopic level, atoms of a metal absorb the photon and its momentum is transferred mainly to the nucleus and electrons.
The excited electron is emitted. Therefore, the conservation of momentum is to be considerd as the momentum of incident photon transferred to the nucleus and electrons.

(i) Here, it is given that, an electron absorbs 2 photons
each of frequency v, then v' = 2v
where, v' is the frequency of emitted electron
Given, \(E_{\max }=h v-\phi_{0}\)
Now, maximum energy for emitted electrons
\(E_{\max }^{\prime}=h(2 v)-\phi_{0}\)
\(=2 h v-\phi_{0}\)
(ii) The probability of absorbing two photons by the same electron is very low.
Hence, such emission will be negligible.