Physics Q44 – JEE-MAIN 2025

Home Physics Dual Nature of MatterQuestion #84

A monochromatic light is incident on a metallic plate having work function

ϕ

. 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)

Clue 1 of 3: Photoelectric Effect and Kinetic Energy

The energy of the incident photon is used to overcome the work function and provide kinetic energy to the emitted electron.

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● Video Walkthrough

Step 1: Calculate Maximum Kinetic Energy and Velocity✦ Active

According to the photoelectric effect, the maximum kinetic energy ( $K_{m a x}$ ) of the emitted electron is given by the energy of the incident photon minus the work function ( $ϕ$ ). The energy of a photon is $E = \frac{h c}{λ}$ .

K_{m a x} = \frac{h c}{λ} - ϕ

The kinetic energy is also related to the electron's mass ( $m$ ) and velocity ( $v$ ) by $K_{m a x} = \frac{1}{2} m v^{2}$ . Equating these, we find the velocity:

\frac{1}{2} m v^{2} = \frac{h c}{λ} - ϕ ⟹ v = \sqrt{\frac{2}{m} (\frac{h c}{λ} - ϕ)}

Step 2: Determine the Radius of the Circular Path○ Expand

The electron enters a constant magnetic field ( $B$ ) perpendicular to its initial velocity. This causes the electron to move in a circular path. The magnetic force ( $F_{B} = e v B$ ) provides the centripetal force ( $F_{c} = \frac{m v^{2}}{r}$ ). Here, $e$ is the charge of the electron.

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

💡 Teacher's Secret Hint

Remember that the magnetic force is perpendicular to both velocity and magnetic field, leading to circular motion.

Step 3: Calculate the Distance AB○ Expand

The electron is emitted from point A and hits the plate again at point B after passing through a curve. This implies the electron completes a semi-circular path, and the distance AB is the diameter of this semi-circle.

A B = 2 r = 2 (\frac{m v}{e B})

Substitute the expression for $v$ from Step 1 into this equation:

A B = \frac{2 m}{e B} \sqrt{\frac{2}{m} (\frac{h c}{λ} - ϕ)}

To simplify, bring $2 m$ inside the square root. When $2 m$ goes inside the square root, it becomes $(2 m)^{2} = 4 m^{2}$ :

A B = \frac{1}{e B} \sqrt{\frac{(2 m)^{2} \cdot 2}{m} (\frac{h c}{λ} - ϕ)} = \frac{1}{e B} \sqrt{\frac{4 m^{2} \cdot 2}{m} (\frac{h c}{λ} - ϕ)}

Simplifying the terms inside the square root:

A B = \frac{1}{e B} \sqrt{8 m (\frac{h c}{λ} - ϕ)}

This matches option 1.

💡 Teacher's Secret Hint

Ensure correct algebraic manipulation when bringing terms inside or outside the square root.

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Physics Question 44 – JEE-MAIN 2025

Clue 1 of 3: Photoelectric Effect and Kinetic Energy

🧠 Full Solution Path