1. **Problem Statement:** A monochromatic light with frequency $\nu$ shines on a metallic plate with work function $\phi$. An electron is emitted from point A with maximum kinetic energy and enters a magnetic field perpendicular to its velocity. The electron moves in a curved path and returns to the plate at point B. We need to find the distance between points A and B. 2. **Relevant Formulae and Concepts:** - Maximum kinetic energy of emitted electron: $$K_{max} = h\nu - \phi$$ where $h$ is Planck's constant. - Electron charge magnitude: $e$. - Electron mass: $m$. - Magnetic field $B$ is perpendicular to velocity, causing circular motion. - Radius of circular path: $$r = \frac{mv}{eB}$$ where $v$ is the velocity of the electron. 3. **Step-by-step Solution:** - The maximum kinetic energy is related to velocity by $$K_{max} = \frac{1}{2}mv^2$$. - Solve for velocity: $$v = \sqrt{\frac{2K_{max}}{m}} = \sqrt{\frac{2(h\nu - \phi)}{m}}$$. - Radius of the circular path in magnetic field: $$r = \frac{m v}{e B} = \frac{m}{e B} \sqrt{\frac{2(h\nu - \phi)}{m}} = \frac{\sqrt{2m(h\nu - \phi)}}{e B}$$. - Since the electron returns to the plate after traveling half a circle, the distance between points A and B is the diameter of the circular path: $$\text{Distance} = 2r = 2 \times \frac{\sqrt{2m(h\nu - \phi)}}{e B} = \frac{2 \sqrt{2m(h\nu - \phi)}}{e B}$$. 4. **Final Answer:** $$\boxed{\text{Distance between A and B} = \frac{2 \sqrt{2m(h\nu - \phi)}}{e B}}$$ This formula gives the distance between points A and B on the plate where the electron returns after moving in the magnetic field.