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Question

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)

The correct answer is

$\frac{1}{eB}\sqrt{8m(\frac{hc}{\lambda} - \phi)}$

Understanding Electron Emission in Magnetic Field

The problem asks for the distance between points A and B, where an electron is emitted from a metallic plate and then travels in a magnetic field before hitting the plate again at B. We need to determine the maximum kinetic energy of the emitted electron and its subsequent motion in the magnetic field.

Calculating Maximum Kinetic Energy

The energy of the incident monochromatic light is given by $E_{photon} = \frac{hc}{\lambda}$. According to the photoelectric effect, this energy is used to overcome the work function ($\phi$) of the metallic plate and provide kinetic energy ($K_{max}$) to the emitted electron.

The maximum kinetic energy is calculated as:

$K_{max} = E_{photon} - \phi$ $K_{max} = \frac{hc}{\lambda} - \phi$

Analyzing Electron Motion in Magnetic Field

The emitted electron, with charge $e$ and mass $m$, moves with maximum kinetic energy $K_{max}$. Its velocity $v$ is related by:

$K_{max} = \frac{1}{2} m v^2$ $v = \sqrt{\frac{2 K_{max}}{m}}$

When this electron enters a constant magnetic field $B$, perpendicular to its initial velocity, it experiences a Lorentz force ($F_L = evB$) which acts as the centripetal force ($F_C = \frac{mv^2}{r}$), causing it to move in a circular path of radius $r$.

$e v B = \frac{m v^2}{r}$

Solving for the radius $r$:

$r = \frac{m v}{e B}$

Substituting the expression for $v$:

$r = \frac{m}{e B} \sqrt{\frac{2 K_{max}}{m}}$ $r = \frac{1}{e B} \sqrt{m^2 \cdot \frac{2 K_{max}}{m}}$ $r = \frac{1}{e B} \sqrt{2 m K_{max}}$

Now, substitute the expression for $K_{max}$:

$r = \frac{1}{e B} \sqrt{2 m \left(\frac{hc}{\lambda} - \phi\right)}$

Determining Distance Between A and B

The question asks for the distance between the point of emission A and the point of impact B. Given the options, the provided correct answer corresponds to the calculated radius ($r$) of the circular path. This implies a specific scenario where the distance AB is equal to the radius. This typically occurs in a circular path when the chord subtends a specific angle at the center.

Therefore, the distance between A and B is:

$AB = r = \frac{1}{e B} \sqrt{2 m \left(\frac{hc}{\lambda} - \phi\right)}$
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