MixedJEE Physics · Original learning card5 original chapter questions

Matter Wavelength in Electric and Magnetic Fields

Under a force, momentum evolves according to dp vector by dt = F vector, and the instantaneous wavelength is h divided by the momentum magnitude. A purely magnetic Lorentz force does no work, whereas an electric field can change kinetic energy.

Why this shows up in the exam

Charged-particle beam steering · Time-dependent matter-wave calculations · Separating electric and magnetic effects on wavelength

Learn the idea

Fields change de Broglie wavelength only by changing the particle's momentum magnitude. The wavelength tracks momentum, not merely the presence of a field. A magnetic field can bend a trajectory without changing speed, while an electric field can change both direction and speed.

🧠 Memory hook: Direction can bend while wavelength stays fixed; watch momentum magnitude.

Get this one clearly and it pays off every single time it shows up in the paper. 🎯

Formulas & facts to keep ready

  • d p_vector/dt = q(E_vector + v_vector x B_vector) — momentum evolution of a charged particle
  • lambda(t) = h / |p_vector(t)| — instantaneous de Broglie wavelength
  • dK/dt = q E_vector dot v_vector — magnetic force alone does not change kinetic energy

How to approach it

  1. 1Find the momentum vector as a function of time
  2. 2Take its magnitude before computing wavelength
  3. 3Use work-energy as a shortcut when only kinetic energy is needed

Common slip-ups that cost marks

  • •Assuming every magnetic field changes particle speed
  • •Adding velocity components as scalars instead of vectors
  • •Using lambda = h/(mv) after relativistic acceleration

🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.

Original chapter practice

Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.

Question 1 of 5

Photons of energy 5 eV illuminate a metal of work function 2 eV. Find the stopping potential.

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