Physics · Dual Nature Of Radiation And Matter · NEET
Start with a photon. Its energy has two expressions: E = hc/lambda (wave picture, Planck) and E = mc^2 (particle picture, Einstein). Set them equal: hc/lambda = mc^2. Cancel one c: hc/lambda = mc x c, so h/lambda = mc = p, where p = mc is the photon momentum. Rearranging gives lambda = h/p. de Broglie's bold step: this same relation must hold for ALL matter, not just light. So for any particle, lambda = h/p = h/(mv).
They are the same thing. Momentum p = mv, so h/p = h/(mv). We write lambda = h/p as the general form because p is the true variable (it works even when you are given momentum directly or relativistic momentum). Use h/(mv) when you know mass and speed separately.
Light already showed dual nature: it behaves as a wave (interference, diffraction) and as particles called photons (photoelectric effect). de Broglie argued nature should be symmetric, so if waves can act like particles, particles should act like waves. He assigned the same relation p = h/lambda to electrons, protons, and all matter. This wave is called a matter wave.
For a non-relativistic particle, KE = p^2/(2m), so p = sqrt(2m x KE). Substitute into lambda = h/p to get lambda = h/sqrt(2m x KE). This is very useful in NEET problems where energy (in joules or eV) is given instead of speed.
The final formula lambda = h/p is identical for both. The difference is how you find p. For a photon, p = E/c = h/lambda directly. For a material particle like an electron, p = mv (or sqrt(2m x KE), or sqrt(2mqV) if accelerated through a potential). Same formula, different route to momentum.
Match List I with List II. List-I: A. E = h.nu B. Diffraction and interference C. lambda = h/p D. Compton effect. List-II: I. de Broglie wavelength II. Particle nature of light III. Wave nature of light IV. Energy of photon. Choose the correct answer.
The de Broglie wavelength of a neutron in thermal equilibrium with heavy water at temperature T (kelvin) and mass m is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
lambda = h/p = h/(mv), where h = 6.63 x 10^-34 J.s is Planck's constant, p is momentum, m is mass, and v is speed. In terms of kinetic energy it becomes lambda = h/sqrt(2m x KE).
It is a length, so the SI unit is the metre (m). For electrons and atoms the values are tiny, around 10^-10 m (angstrom scale), which is why matter waves are only noticeable for very small particles.
Louis de Broglie proposed it in 1924. He suggested that all moving matter has an associated wave of wavelength lambda = h/p. It was later confirmed by the Davisson-Germer experiment showing electron diffraction.
Because lambda = h/p and h is extremely small. A cricket ball has a large momentum p, making lambda around 10^-34 m, far too small to ever observe. Wave nature is only significant for tiny masses like electrons.
Yes, lambda = h/p holds for a photon too. For a photon momentum p = E/c, so its de Broglie wavelength equals its ordinary electromagnetic wavelength.
