MixedJEE Physics · Original learning card10 original chapter questions

de Broglie Standing Waves in Bohr Orbits

Combining lambda = h/(mv) with 2 pi r_n = n lambda gives m v_n r_n = n hbar. In a Coulomb Bohr atom, the electron wavelength therefore scales as n/Z.

Why this shows up in the exam

Comparing orbital matter wavelengths · Deriving Bohr angular-momentum quantization · Relating orbit circumference to quantum number

Learn the idea

An allowed Bohr orbit fits an integer number of electron de Broglie wavelengths around its circumference. A wave closes smoothly around the orbit only when the circumference contains whole wavelengths. A fractional leftover would return with the wrong phase.

🧠 Memory hook: Whole waves close the orbit.

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

Formulas & facts to keep ready

  • 2 pi r_n = n lambda_n — standing-wave condition around a Bohr orbit
  • lambda_n = h/(m v_n) — non-relativistic electron de Broglie wavelength
  • lambda_n proportional to n/Z — hydrogen-like orbital wavelength scaling

How to approach it

  1. 1Identify which wavelength the question means
  2. 2Use circumference equals n wavelengths
  3. 3Cross-check with L = n hbar

Common slip-ups that cost marks

  • •Using lambda = circumference for every n
  • •Confusing emitted-photon wavelength with electron wavelength
  • •Applying h/(mv) relativistically

🌟 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 10

In hydrogen, an electron transitions from n = 2 to n = 1. Using E_n = -13.6/n^2 eV, find the emitted photon energy.

Take a timed JEE Physics sectional mock