A monochromatic light is incident on a metallic plate having work function $\phi$. An electron, emitted normally to the plate from a point A with maximum kinetic energy, enters a constant magnetic field, perpendicular to the initial velocity of electron. The electron passes through a curve and hits back the plate at a point $B$. The distance between $A$ and $B$ is : (Given : The magnitude of charge of an electron is e and mass is $\mathrm{m}, \mathrm{h}$ is Planck's constant and c is velocity of light. Take the magnetic field exists throughout the path of electron)

<p>To determine the distance between points $A$ and $B$ (where the electron re-enters the metallic plate), we first need to understand the electron's behavior in a magnetic field.</p> <p><p><strong>Maximum Kinetic Energy (KE)</strong>: The energy of the emitted electron can be given by the photoelectric equation:</p> <p>$ \text{KE}_{\max} = \frac{hc}{\lambda} - \phi $</p> <p>Here, $\frac{hc}{\lambda}$ is the energy of the incident photons, and $\phi$ is the work function of the metal.</p></p> <p><p><strong>Momentum (p)</strong>: The momentum of the electron is related to its kinetic energy by the equation:</p> <p>$ p = \sqrt{2m \cdot \text{KE}_{\max}} = \sqrt{2m\left(\frac{hc}{\lambda} - \phi\right)} $</p></p> <p><p><strong>Path of Electron in Magnetic Field</strong>: When an electron moves perpendicularly through a magnetic field, it follows a circular path. The radius $R$ of this path is given by:</p> <p>$ R = \frac{p}{eB} $</p> <p>where $e$ is the charge of the electron, and $B$ is the magnetic field strength.</p></p> <p><p><strong>Distance Between $A$ and $B$</strong>: Since the electron travels back to the plate, forming a complete semicircle, the distance between points $A$ and $B$ is twice the radius of this circular path:</p> <p>$ d_{A-B} = 2R = 2\left(\frac{p}{eB}\right) $</p> <p>Substituting the expression for momentum, we find:</p> <p>$ d_{A-B} = \frac{2 \sqrt{2m\left(\frac{hc}{\lambda} - \phi\right)}}{eB} = \frac{\sqrt{8m\left(\frac{hc}{\lambda} - \phi\right)}}{eB} $</p></p> <p>Therefore, the distance between $A$ and $B$ is $\frac{\sqrt{8m(\frac{hc}{\lambda} - \phi)}}{eB}$.</p>