For an isosceles prism of angle $$A$$ and refractive index $$\mu $$, it is found that the angle of minimum deviation $${\delta _m} = A.$$

Which of the following options is/are correct?

An electron in a hydrogen atom undergoes a transition from an orbit with quantum number $${n_i}$$ to another with quantum number $${n_f}$$. $${V_i}$$ and $${V_f}$$ are respectively the initial and final potential energies of the electron. If $${{{V_i}} \over {{V_f}}} = 6.25$$, then the smallest possible $${n_f}$$ is

A drop of liquid of radius $$R = {10^{ - 2}}\,m$$ having surface tension $$S = {{0.1} \over {4\pi }}N{m^{ - 1}}$$ divides itself into $$K$$ identical drops. In this process the total change in the surface energy $$\Delta U = {10^{ - 3}}\,J.$$ If $$K = {10^\alpha }$$ then the value of $$\alpha $$ is

A charged particle (electron or proton) is introduced at the origin (x=0,y=0,z=0) with a given initial velocity $$\overrightarrow v .$$ A uniform electric field $$\overrightarrow E $$ and a uniform magnetic field $$\overrightarrow B $$ exist everywhere. The velocity $$\overrightarrow v ,$$ electric field $$\overrightarrow E $$ and magnetic field $$\overrightarrow B $$ are given in column $$1,2$$ and $$3,$$ respectively. The quantities $${E_0},{B_0}$$ are positive in magnitude.

Column 1		Column 2		Column 3
(I)	Electron with $$\overrightarrow v = 2{{{E_0}} \over {{B_0}}}\widehat x$$	(i)	$$\overrightarrow E = {E_0}\widehat z$$	(P)	$$\overrightarrow B = - {B_0}\widehat x$$
(II)	Electron with $$\overrightarrow v = {{{E_0}} \over {{B_0}}}\widehat y$$	(ii)	$$\overrightarrow E = - {E_0}\widehat y$$	(Q)	$$\overrightarrow B = {B_0}\widehat x$$
(III)	Proton with $$\overrightarrow v = 0$$	(iii)	$$\overrightarrow E = - {E_0}\widehat x$$	(R)	$$\overrightarrow B = {B_0}\widehat y$$
(IV)	Proton with $$\overrightarrow v = 2{{{E_0}} \over {{B_0}}}\widehat x$$	(iv)	$$\overrightarrow E = {E_0}\widehat x$$	(S)	$$\overrightarrow B = {B_0}\widehat z$$

In which case will the particle move in a straight line with constant velocity?