Derive Bethe’s Law for electron refraction.

#### Solution

Region I has potential V1 and region II has potential V2. The plane surface AB constitutes one of the equipotential surfaces. Let an electron with velocity v1 enter region I making an angle I with the normal. As the electron passes through the equipotential surface AB, it experiences a force which alters its velocity. Because the electric field exists only in the y-direction, the vertical component (y-component) of electron changes while the tangential component (x-component) remains constant.

𝑣1𝑥=𝑣2𝑥

𝑣1sin𝑖=𝑣2sin𝑟

`sin i /sin r=v_2/v_1`

If V1>V2, 𝑣_{1𝑦} increases while if V2>V1, _{𝑣2𝑦} increases.

In our case we have taken V2>V1. As the electrons move through the electric field their kinetic energy is provided by the respective potential energy of the electric fields. Hence mv_{1}^{2}/2=qV_{1}

And mv_{2}^{2}/2=qV_{2}

Dividing the above equation we get,

v1^{2}/ v2^{2}= V_{1}/ V_{2}

v_{1}/ v_{2}=`sqrt(V1//V2)`

Hence we get,`sin i/sin r=(v2)/(v1)=sqrt((V2)/(V1))`

This is known as **Bethe’s law of electron refraction.**