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)

Solution & Explanation

Related Formula
K = (hc)/(λ) - φ p = 2m K R = (p)/(eB)

d = 2R

Core Logic

According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectron is:

K = (hc)/(λ) - φ

The momentum p corresponding to this kinetic energy is:

p = 2m K = √(2m ((hc)/(λ) - φ))

The electron is emitted normally to the plate and enters a perpendicular constant magnetic field B. It describes a circular arc (semicircle) and hits back the plate at point B. The distance between A and B is the diameter of this circular trajectory:

dAB = 2R = 2 ( (p)/(eB) ) = 2√(2m((hc)/(λ) - φ))eB

To align this with the options, move the factor of 2 inside the square root (2 = √(4)):

dAB = √(4 × 2m((hc)/(λ) - φ))eB = √(8m((hc)/(λ) - φ))eB
Step 1: Final Conclusion

The distance between points A and B is

Step 1: Final Conclusion

The distance between points A and B is $\sqrt{8m\left(\frac{hc}{\lambda} - \phi\right)} / eB.

Pattern Recognition

When a particle is launched perpendicularly from a flat boundary into a perpendicular magnetic field, it describes a semicircle and exits/re-hits the boundary at a distance equal to the diameter

Pattern Recognition

When a particle is launched perpendicularly from a flat boundary into a perpendicular magnetic field, it describes a semicircle and exits/re-hits the boundary at a distance equal to the diameter $2R = 2\frac{p}{qB}. Taking coefficients inside square roots converts2 \sqrt{2x}to\sqrt{8x}$.

Chapter Mix

Class 12 Physics: Dual Nature of Matter and Radiation Class 12 Physics: Moving Charges and Magnetism

Reference Study Guides

More Dual Nature of Matter and Radiation Previous-Year Questions — Page 7

Q50 jee_main_2024_31_jan_morning Photoelectric Effect
When a metal surface is illuminated by light of wavelength λ, the stopping potential is 8 V. When the same surface is illuminated by light of wavelength 3λ, stopping potential is 2 V. The threshold wavelength for this surface is:
  • A. 5λ
  • B. 3λ
  • C. 9λ
  • D. 4.5λ

Solution

Related Formula
(hc)/(λ) = φ + Kmax Kmax = eV₀ φ = (hc)/(λ₀)
Core Logic

Using Einstein's Photoelectric equation for the two cases:

Case 1 (Wavelength λ, Stopping Potential 8 V):

(hc)/(λ) = (hc)/(λ₀) + 8e (i)

Case 2 (Wavelength 3λ, Stopping Potential 2 V):

(hc)/(3λ) = (hc)/(λ₀) + 2e (ii)
Step 2: Solving the Equations

Multiply equation (ii) by 4 to eliminate e:

(4hc)/(3λ) = (4hc)/(λ₀) + 8e

Equating this to equation (i):

(hc)/(λ) - (hc)/(λ₀) = (4hc)/(3λ) - (4hc)/(λ₀)

Divide entirely by hc:

(1)/(λ) - (1)/(λ₀) = (4)/(3λ) - (4)/(λ₀) (4)/(λ₀) - (1)/(λ₀) = (4)/(3λ) - (1)/(λ) (3)/(λ₀) = (4 - 3)/(3λ) (3)/(λ₀) = (1)/(3λ) λ₀ = 9λ
Chapter Mix

Class 12 Physics: Dual Nature Of Radiation And Matter

More Dual Nature of Matter and Radiation Questions — jee_main_2025_02_april_morning

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