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 the electron. The electron passes through a curve and hits back the plate at a point B. The distance between A and B is:

• A. \( \sqrt{\frac{2m \left( \frac{hc}{\lambda} - \phi \right)}{eB}} \)
• B. \( \frac{m \left( \frac{hc}{\lambda} - \phi \right)}{eB} \)
• C. \( \sqrt{8m \left( \frac{hc}{\lambda} - \phi \right)} \div eB \)
• D. \( 2 \frac{m \left( \frac{hc}{\lambda} - \phi \right)}{eB} \)

Hint:

In problems involving magnetic fields, the radius of the electron’s path is related to its momentum and the magnetic field. The formula for the distance is derived by equating the magnetic force to the centripetal force.

Answer 1

The Correct Option is C

The maximum kinetic energy \( K_E \) of the electron is given by: \[ K_E = \frac{hc}{\lambda} - \phi \] where \( p \) is the momentum of the electron, and the relation for momentum is: \[ p = \sqrt{2m K_E} = \sqrt{2m \left( \frac{hc}{\lambda} - \phi \right)} \] Since the motion is in a magnetic field, the radius of the circular path is: \[ d_{A-B} = 2R = \frac{p}{qB} \] Thus, the distance between A and B becomes: \[ d_{A-B} = \frac{2}{eB} \sqrt{2m \left( \frac{hc}{\lambda} - \phi \right)} = \frac{\sqrt{8m \left( \frac{hc}{\lambda} - \phi \right)}}{eB} \]

Answer 2

This problem involves two main concepts: the photoelectric effect, which determines the kinetic energy of the emitted electron, and the motion of a charged particle in a uniform magnetic field, which determines the trajectory of the electron.

Concept Used:

1. Einstein's Photoelectric Equation: When light of frequency \( \nu \) (or wavelength \( \lambda \)) is incident on a metal surface with work function \( \phi \), the maximum kinetic energy (\( K_{max} \)) of the emitted photoelectrons is given by:

\[ K_{max} = h\nu - \phi = \frac{hc}{\lambda} - \phi \]

where \( h \) is Planck's constant and \( c \) is the speed of light.

2. Motion in a Magnetic Field: When a charged particle (charge \( q \), mass \( m \)) moves with velocity \( v \) perpendicular to a uniform magnetic field \( B \), it experiences a magnetic force that acts as a centripetal force, causing it to move in a circular path. The radius \( r \) of this path is given by:

\[ qvB = \frac{mv^2}{r} \implies r = \frac{mv}{eB} \]

The kinetic energy is related to velocity by \( K_{max} = \frac{1}{2}mv^2 \).

Step-by-Step Solution:

Step 1: Determine the maximum kinetic energy of the emitted electron. Using Einstein's photoelectric equation, the maximum kinetic energy \( K_{max} \) of the electron is:

\[ K_{max} = \frac{hc}{\lambda} - \phi \]

Step 2: Find the velocity of the electron from its kinetic energy. The kinetic energy is also given by \( K_{max} = \frac{1}{2}mv^2 \), where \( m \) is the mass of the electron and \( v \) is its maximum velocity. Solving for \( v \):

\[ v = \sqrt{\frac{2K_{max}}{m}} \]

Substituting the expression for \( K_{max} \) from Step 1:

\[ v = \sqrt{\frac{2}{m} \left( \frac{hc}{\lambda} - \phi \right)} \]

Step 3: Analyze the electron's trajectory in the magnetic field. The electron is emitted normally (perpendicularly) from the plate and enters a magnetic field that is perpendicular to its velocity. The magnetic force \( \vec{F} = q(\vec{v} \times \vec{B}) \) will be perpendicular to \( \vec{v} \), causing the electron to follow a circular path. Since the electron hits the plate again, it must have completed a semi-circular trajectory. The distance between the point of emission (A) and the point of return (B) is the diameter of this semi-circle.

\[ \text{Distance } AB = 2r \]

where \( r \) is the radius of the circular path.

Step 4: Calculate the radius of the circular path. The radius of the circular path of an electron (charge \( e \)) in a magnetic field \( B \) is:

\[ r = \frac{mv}{eB} \]

Step 5: Substitute the expression for velocity (\(v\)) into the radius formula.

\[ r = \frac{m}{eB} \left[ \sqrt{\frac{2}{m} \left( \frac{hc}{\lambda} - \phi \right)} \right] \]

To simplify, we can bring the \( m \) inside the square root:

\[ r = \frac{1}{eB} \sqrt{m^2 \cdot \frac{2}{m} \left( \frac{hc}{\lambda} - \phi \right)} \] \[ r = \frac{1}{eB} \sqrt{2m \left( \frac{hc}{\lambda} - \phi \right)} \]

Step 6: Calculate the distance AB. The distance AB is the diameter of the semi-circle, which is \( 2r \).

\[ AB = 2r = 2 \times \frac{1}{eB} \sqrt{2m \left( \frac{hc}{\lambda} - \phi \right)} \]

Bringing the factor of 2 inside the square root (as \( \sqrt{4} \)) to match the options:

\[ AB = \frac{1}{eB} \sqrt{4 \times 2m \left( \frac{hc}{\lambda} - \phi \right)} \] \[ AB = \frac{\sqrt{8m \left( \frac{hc}{\lambda} - \phi \right)}}{eB} \]

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