Heisenberg Uncertainty Principle
Dual Nature of Matter
10
JEE Qs
8%
Hard
45
min
Master the formulas for position-momentum and energy-time uncertainty and understand the conceptual implications for microscopic systems, particularly distinguishing it from classical measurement errors.
🧮 Key Formulas
✅ Key Points for JEE
- 1The Heisenberg Uncertainty Principle states that it is fundamentally impossible to precisely know both the position (Δx) and momentum (Δp_x) of a particle simultaneously.
- 2It is an inherent property of quantum systems, arising from wave-particle duality, not a limitation of measurement instruments or observational disturbances.
- 3The minimum product of uncertainties for conjugate variables (like position-momentum or energy-time) is approximately `h/(4π)` or `ħ/2`.
- 4This principle is significant only for microscopic particles; its effects are negligible and unobservable for macroscopic objects due to the small value of Planck's constant.
- 5The 'uncertainty' (Δ) refers to the inherent spread or standard deviation in the possible values of a physical quantity, reflecting the probabilistic nature of quantum mechanics.
⚠️ Common Mistakes
- ✕Confusing the uncertainty principle with errors in measurement or disturbance caused by the act of measurement itself.
- ✕Incorrectly using Planck's constant 'h' directly instead of `h/(4π)` or `ħ/2` for the minimum uncertainty product.
- ✕Applying the principle to macroscopic objects, failing to recognize that its effects are imperceptible at larger scales.
- ✕Misinterpreting `Δx`, `Δp`, `ΔE`, `Δt` as mere errors rather than inherent indefiniteness or spread in quantum states.
📝 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