
Given here, kinetic energy 21e=21mux2⇒ux=me
Acceleration of electron due to electric field is a=meE, where, e is charge on electron.
Using equation of motion, s=uxt+21axt2 along x direction, we get
L=uxt+0(∵ax=0)⇒10×10−2=uxt⇒t=ux10−1....(1)
Using equation of motion, vy=uy+ayt along y direction, we get
vy=0+meEt=meEt...(2)
From equation (1)and(2), we have
vy=meE(ux10−1)
Now, tanθ=vxvy=meE(ux210−1)=m×mee×10×10−1=1
The angle of deviation of the path of electron as it comes out of the field is θ=45∘.