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.

Maximum Kinetic Energy (KE): The energy of the emitted electron can be given by the photoelectric equation:

$ \text{KE}_{\max} = \frac{hc}{\lambda} - \phi $

Here, $\frac{hc}{\lambda}$ is the energy of the incident photons, and $\phi$ is the work function of the metal.

Momentum (p): The momentum of the electron is related to its kinetic energy by the equation:

$ p = \sqrt{2m \cdot \text{KE}_{\max}} = \sqrt{2m\left(\frac{hc}{\lambda} - \phi\right)} $

Path of Electron in Magnetic Field: When an electron moves perpendicularly through a magnetic field, it follows a circular path. The radius $R$ of this path is given by:

$ R = \frac{p}{eB} $

where $e$ is the charge of the electron, and $B$ is the magnetic field strength.

Distance Between $A$ and $B$: 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:

$ d_{A-B} = 2R = 2\left(\frac{p}{eB}\right) $

Substituting the expression for momentum, we find:

$ d_{A-B} = \frac{2 \sqrt{2m\left(\frac{hc}{\lambda} - \phi\right)}}{eB} = \frac{\sqrt{8m\left(\frac{hc}{\lambda} - \phi\right)}}{eB} $

Therefore, the distance between $A$ and $B$ is $\frac{\sqrt{8m(\frac{hc}{\lambda} - \phi)}}{eB}$.