The acceleration voltage generates an electric field \(E\) between cathode and anode:$$E=\frac{V_{\text a}}{\text d}$$In the electric field the electrons experience a force \(F_{\rm{el}}\):$$F_{\text{el}}=E\cdot e = \frac{V_{\text a}\cdot e}{\text d}$$So beetween cathode and anode work is done:$$W_{\text{el}}=F_{\text{el}}\cdot \text {d} = \frac{V_{\text a}\cdot e\cdot \text d}{\text d}= V_{\text a}\cdot e\quad (1)$$This work raises the kinetic energy of the electron. The kinetic energy Ekin of an electron is:$$E_{\text {kin}}= \frac {1}{2}mv^2\quad (2)$$Work done \( (1)\) and kinetic energy \( (2)\) are equal:$$\frac {1}{2}mv^2=e \cdot V_{\text a}$$So the final speed of the electrons is:$$\bbox[5px,border:2px solid red] {v_{\text{end}}=\sqrt{2\cdot \frac{e}{m}\cdot V_{\text a}}}$$
Classic
Relativistic
Exercises
