A Proton and an Alpha particle are accelerated by same potential difference.If they enter in a region of uniform electric field in a direction perpendicular to it, then deflection of which is more?

The deflections of both the proton and the alpha particle are same.

charge on the proton = e; the charge on the alpha particle = 2e

The mass of the proton = m; We can assume the mass of the alpha particle is very close to 4m.

In reality, mass of the proton = 1.00794 amu and mass of the alpha particle = 4.002602 amu

##Delta V =## the accelerating potential difference before they both enter (E).

##v_p ## = velocity of the proton when it enters the region of uniform electric field (E).

##v_alpha## = velocity of the alpha particle when it enters the uniform electric field (E).

Since they are both positively charged, they will both be deflected in the direction of the E field.

For the proton, done by the E field = the final kinetic ##eDeltaV = 1/2 m(v_p)^2##. Hence ##v_p = sqrt({2eDeltaV)/m}## Similarly for the alpha particle, ##2e DeltaV = 1/2 (4m) (v_alpha)^2 = 2m(v_alpha)^2##. Hence, ##v_alpha = sqrt((eDeltaV)/m)##

Now let us supose that both particles travel the same horizontal distance ##=Delta x## in the electric field (E).

Time for the proton to travel the distance, ##Delta x= t _p =(Delta x)/ v_p ##

Hence, ##t_p = (Delta x)/{sqrt((2eDeltaV)/m)} = Delta x sqrt{m/(2eDeltaV)}##

Similarly, ##t_alpha = (Delta x )/v_alpha = (Delta x )/{sqrt((eDeltaV)/m} ## =## Delta x sqrt{m/(eDeltaV)}##

From , the of the proton in the E field = ##a_p = (eE)/m## and the acceleration of the alpha particle = ##a_alpha = (2eE)/(4m) = (eE)/(2m)##

From kinematics, displacement of the proton

= ##x_p## = ##1/2a_p*(t_p)^2 = 1/2 ((eE)/m) *{(Delta x )^2 *m}/(2eDelta V ## = ##{E(Deltax)^2}/(4DeltaV)##

Similarly, the displacement of the alpha particle

= ## x_alpha = 1/2a_alpha* (t_alpha)^2 =1/2 ((eE)/(2m))* {(Delta x)^2*m}/(eDeltaV)= (E(Deltax)^2}/(4DeltaV)##

Hence, we see the deflections of both the proton and the alpha particle are identical.