de Broglie Wavelength — λ = h/mv
Dual Nature of Matter
10
JEE Qs
8%
Hard
45
min
Master the various forms of the de Broglie wavelength formula for different scenarios (charged particles, thermal particles, relativistic speeds) and always ensure unit consistency in calculations.
🧮 Key Formulas
✅ Key Points for JEE
- 1The de Broglie wavelength (λ) is inversely proportional to the momentum (p) of the particle and applies to all matter, demonstrating wave-particle duality.
- 2Although universal, the wave nature described by de Broglie wavelength is only significant and experimentally observable for microscopic particles (e.g., electrons, protons, atoms) due to the extremely small value of Planck's constant (h).
- 3For a charged particle of charge 'q' and mass 'm' accelerated through a potential difference 'V', its kinetic energy is qV, leading to the wavelength formula λ = h / sqrt(2mqV).
- 4Always ensure unit consistency: use SI units (m in kg, v in m/s, h in J·s, KE in J) to get λ in meters. If using eV for energy, use appropriate conversion factors or specific values for h and electron charge.
- 5The de Broglie hypothesis successfully explained the quantization conditions (like Bohr's angular momentum postulate) by viewing electron orbits as standing matter waves.
⚠️ Common Mistakes
- ✕Inconsistent unit usage, particularly mixing SI units (J·s for h, kg for mass) with electron volts (eV) for energy without proper conversion.
- ✕Neglecting relativistic effects for particles moving at speeds comparable to the speed of light, where the momentum formula p = γmv should be used.
- ✕Incorrectly applying the kinetic energy formula KE = qV to uncharged particles (like neutrons) or failing to use the thermal energy formula KE = (3/2)kT for them.
- ✕Confusing de Broglie wavelength (for matter waves) with the wavelength of electromagnetic radiation (photons) or applying formulas specific to one to the other.
📝 Practice Questions
See allQ42.A light source of wavelength λ illuminates a metal surface and electrons are ejected with maximum kinetic energy of 2 eV . If the same surface is illuminated by a light source of wavelength λ , then the maximum 2 kinetic energy of ejected electrons will be (The work function of metal is 1 eV ) (1) 3 eV (2) 2 eV (3) 6 eV (4) 5 eV
Q45.In photoelectric effect an em-wave is incident on a metal surface and electrons are ejected from the surface. If the work function of the metal is 2.14 eV and stopping potential is 2 V , what is the wavelength of the em- wave ? (Given hc = 1242eVnm where h is the Planck's constant and c is the speed of light in vaccum.) 2025 (23 Jan Shift 2) JEE Main Previous Year Paper (1) 300 nm (2) 400 nm (3) 600 nm (4) 200 nm
Q30.An electron in the ground state of the hydrogen atom has the orbital radius of 5.3 × 10−11 m while that for the electron in third excited state is 8.48 × 10−10 m . The ratio of the de Broglie wavelengths of electron in the excited state to that in the ground state is (1) 3 (2) 16 (3) 9 (4) 4
Q40.The work functions of cesium (Cs) and lithium (Li) metals are 1.9 eV and 2.5 eV , respectively. If we incident a light of wavelength 550 nm on these two metal surfaces, then photo-electric effect is possible for the case of (1) Both Cs and Li (2) Neither Cs nor Li (3) Cs only (4) Li only
Q34.A sub-atomic particle of mass 10−30 kg is moving with a velocity 2.21 × 106 m/s. Under the matter wave consideration, the particle will behave closely like (h = 6.63 × 10−34 J. s) (1) Visible radiation (2) Gamma rays (3) Infra-red radiation (4) X-rays
Q32.An electron of mass ' m ' with an initial velocity →v = v0^i (v0 > 0) enters an electric field E = −Eo^k . If the initial de Broglie wavelength is λ0 , the value after time t would be (1) λ0 (2) e2E20t2 m2v2o √1+ e2E02t2m2v02 λo√1 + (3) λ0 (4) λ0 √1−e2Eo2t2m2v2o
NCERT Chapters
- Class 12 Physics Ch 11: Dual Nature of Radiation and Matter
- Class 11 Physics Ch 5: Laws of Motion
- Class 11 Physics Ch 6: Work, Energy and Power
- Class 12 Physics Ch 2: Electrostatic Potential and Capacitance