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 m, h is Planck's constant and c is velocity of light. Take the magnetic field exists throughout the path of electron)

When monochromatic light of wavelength $$\lambda$$ is incident on a metal surface of work function $$\phi$$, the photo-electric equation is
$$h\nu = \phi + K_{\max}$$
Using $$\nu = \dfrac{c}{\lambda}$$ we have

Maximum kinetic energy of an emitted electron:
$$K_{\max}= \dfrac{hc}{\lambda} - \phi$$ $$-(1)$$

Relating kinetic energy to speed:
$$K_{\max}= \dfrac{1}{2} m v^{2}$$
Hence the maximum speed of the electron is
$$v = \sqrt{\dfrac{2K_{\max}}{m}} = \sqrt{\dfrac{2}{m}\left(\dfrac{hc}{\lambda} - \phi\right)}$$ $$-(2)$$

The magnetic field $$\mathbf{B}$$ is perpendicular to the initial velocity, so the electron moves in a uniform circular path. For a charge $$e$$ moving with speed $$v$$ perpendicular to $$\mathbf{B}$$, the radius of the circular path is given by
$$r = \dfrac{mv}{eB}$$ $$-(3)$$

Substituting $$v$$ from $$(2)$$ into $$(3)$$:
$$r = \dfrac{m}{eB}\,\sqrt{\dfrac{2}{m}\left(\dfrac{hc}{\lambda} - \phi\right)} = \dfrac{\sqrt{2m\left(\dfrac{hc}{\lambda} - \phi\right)}}{eB}$$ $$-(4)$$

Geometry of the motion:
• The electron is emitted normally from point A on the plate (the plate is the initial plane, say $$z = 0$$).
• Because $$\mathbf{B}$$ is in the plane of the plate and $$\mathbf{v}$$ is perpendicular to the plate, the circular path lies in a plane perpendicular to $$\mathbf{B}$$ and intersects the plate at two points.
• Starting from A, the electron completes half a circle (angle $$\pi$$) before arriving back at the plate, now at point B.
• The chord joining the two end points of a semicircle is the diameter, whose length is $$2r$$.

Therefore the distance AB is
$$AB = 2r = 2 \times \dfrac{\sqrt{2m\left(\dfrac{hc}{\lambda} - \phi\right)}}{eB} = \dfrac{\sqrt{8m\left(\dfrac{hc}{\lambda} - \phi\right)}}{eB}$$ $$-(5)$$

The expression in $$(5)$$ matches Option C.

Final answer: Option C