Solution

From the drawing

$$\tan\alpha = \frac{v_y}{v_x}$$

Law of conservation of energy

$$\frac{mv^2_x}{2}=eU$$

From where

$$v_x=\sqrt{\frac{2eU}{m}}$$

Force $\vec{F}$ acting on the particle:

$$F=\frac{eU}{d}=e\frac{U_0}{d}\sin\omega t$$

Second Newton's Laws

$$ma=\frac{eU_0}{d}\sin\omega t$$

By the definition of acceleration $a = \frac{dv}{dt}$

$$\frac{dv}{dt}=\frac{eU_0}{md}\sin\omega t$$

Let's regroup and integrate

$$\int _0^{v_y}dv=\frac{eU_0}{md}\int_0^t\sin\omega t\,dt$$

$$v_y=\frac{eU_0}{md\omega}(1-\cos\omega t)$$

Time for which the particle will move horizontally by the value $l$

$$t=\frac{l}{v_x}=l\sqrt{\frac{m}{2eU}}$$

Find the angle of velocity to the horizontal

$$\tan\theta = \frac{v_y}{v_x}=\frac{eU_0}{m\omega d}\sqrt{\frac{m}{2eU}} \left( 1-\cos\omega l \sqrt{\frac{m}{2eU}}\right)$$

Whence the angle $\angle \theta$

$$\boxed{\theta =\operatorname{\arctan } \left( \frac{U_0}{\omega d}\sqrt{\frac{e}{2mU}} \bigg[ 1-\cos\omega l \sqrt{\frac{m}{2eU}}\bigg] \right)}$$

A little extra. Given that $d \ll l$ we can use the approximation $\tan x \approx x$

$${\theta = \frac{U_0}{\omega d}\sqrt{\frac{e}{2mU}} \bigg[ 1-\cos\omega l \sqrt{\frac{m}{2eU}}\bigg] }$$

Find the required scattering angle

$${\Delta \alpha =2\theta = \frac{U_0}{\omega d}\sqrt{\frac{2e}{mU}} \bigg[ 1-\cos\omega l \sqrt{\frac{m}{2eU}}\bigg] }$$