The de Broglie wavelength associated with a particle of mass m and velocity v is
Aλ = m v / h
Bλ = h / (m v)
Cλ = h v / m
Dλ = h m v
Answer & Solution
Correct answer: B. λ = h / (m v)
1. de Broglie (1924) postulated wave-particle duality for matter.
2. Momentum p = m v; associated wavelength λ = h / p = h / (m v).
3. Confirmed for electrons by Davisson-Germer diffraction experiment.
4. h/(m v) is negligible for macroscopic objects but significant for electrons.
_Source: syllabus Chemistry, Ch 2 "Structure of Atom", §2.6_
Keep practising Structure of Atom
Browse all Structure of Atom questions for JEE Advanced, or explore more Chemistry topics.
Related questions
Test what each rule on light coversCatch the false line about lightAn electron and a proton have the same kinetic energy. Which has the larger de Broglie wavThe Aufbau principle states that electrons fill atomic orbitalsHeisenberg's uncertainty principle statesIn the photoelectric effect, the kinetic energy of ejected electrons is:The electronic configuration of Cr (Z=24) is an exception, written as:By the AUFBAU principle, the 3d subshell starts filling: