Physics · Dual Nature Of Radiation And Matter · NEET
The de Broglie wavelength is the wavelength of the "matter wave" linked to any moving particle. It is given by lambda = h/p = h/(mv), where h is Planck's constant and p = mv is the momentum. Memory hook: "heavier or faster = shorter wave" (bigger p means smaller lambda).
Every moving particle carries a matter wave. Its de Broglie wavelength lambda = h/p shrinks as momentum (mass times speed) grows, which is why big objects show no visible wave nature.
Your doubts, answered
Is the de Broglie wavelength only for electrons, or for every moving particle?
It is for every moving particle. Louis de Broglie said that ALL matter, an electron, a proton, a neutron, a cricket ball, even you, has a wave linked to it while moving. The formula lambda = h/p is the same for all of them. We only talk about electrons the most because their wavelength comes out big enough to actually measure and use (like in electron microscopes).
Why don't we see the wave nature of a cricket ball or any big object?
Because h is extremely small (6.63 x 10^-34 J s) and a big object has huge momentum p. So lambda = h/p becomes incredibly tiny. NCERT shows a 0.12 kg ball at 20 m/s has lambda = 2.76 x 10^-34 m, far too small to ever detect. Wave effects (diffraction) only show up when lambda is close to the size of the object it passes, so heavy things never show them. Wave nature is only visible for tiny particles like electrons.
Does the de Broglie wavelength depend on the charge of the particle?
No. The basic formula lambda = h/p uses only mass and speed (through momentum), not charge. Charge enters ONLY when the particle is accelerated by a voltage, because then the charge decides how much kinetic energy it gains. But once you know p, the wavelength does not care about charge.
What is the difference between de Broglie wavelength and photon wavelength?
A photon is light and always moves at speed c; its wavelength is lambda = c/nu = h/p, where its momentum is p = hnu/c. A matter particle (electron etc.) has rest mass and moves slower than c; its de Broglie wavelength is lambda = h/(mv). For the SAME energy E, they are different: photon lambda = hc/E, electron lambda = h/sqrt(2mE). So do not use the same shortcut for both.
Does a particle at rest have a de Broglie wavelength?
No. If v = 0 then p = 0, so lambda = h/p becomes infinite (undefined). A de Broglie wavelength exists only for a MOVING particle. This is a common NTA trick: the wave is a property of motion, not of the particle sitting still.
What are the units of the de Broglie wavelength?
It is a length, so its SI unit is the metre (m). In NEET it is usually given in nanometres (nm = 10^-9 m) or angstrom (A = 10^-10 m). For example, an electron's de Broglie wavelength is often around 0.1 nm = 1 A, which is about the size of an atom.
⚠️ The NEET trap ✗ Since an electron and a proton have the same kinetic energy, they must have the same de Broglie wavelength. ✓ Use lambda = h/sqrt(2mE). At equal energy E, lambda is proportional to 1/sqrt(m), so the heavier proton has the SHORTER wavelength. Wavelengths are equal only if the momenta p are equal, not the energies. 🧠 Same energy does NOT mean same wavelength.
Real NEET questions
2022
The graph which shows the variation of the de Broglie wavelength (lambda) of a particle and its associated momentum (p) is:
A · A straight line through the origin
B · A straight line with positive intercept
C · A parabola opening upward
D · A rectangular hyperbola (lambda decreases as p increases) ✓
Solution: Step 1: Write the de Broglie relation: lambda = h/p. Step 2: Here h is a constant, so lambda is proportional to 1/p. Step 3: A graph of a quantity versus its inverse is a rectangular hyperbola. As p increases, lambda falls but never reaches zero, and the curve never touches either axis. So the correct shape is the rectangular hyperbola.
2023
The de Broglie wavelength associated with an electron accelerated by a potential difference of 81 V is:
A · 1.36 nm
B · 0.136 nm ✓
C · 13.6 nm
D · 136 nm
Solution: Step 1: For an electron accelerated through V volts, the standard result is lambda = 12.27/sqrt(V) angstrom. Step 2: Put V = 81, so sqrt(V) = 9. Step 3: lambda = 12.27/9 = 1.36 angstrom. Step 4: Convert: 1 angstrom = 10^-10 m = 0.1 nm, so 1.36 angstrom = 0.136 nm. Answer: 0.136 nm.
2026
Match List-I with List-II. 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
A · A-IV, B-I, C-II, D-III
B · A-IV, B-III, C-II, D-I
C · A-I, B-IV, C-III, D-II
D · A-IV, B-III, C-I, D-II ✓
Solution: Step 1: E = h nu gives the energy of a photon, so A matches IV. Step 2: Diffraction and interference show the wave nature of light, so B matches III. Step 3: lambda = h/p is exactly the de Broglie wavelength relation, so C matches I. Step 4: The Compton effect (photon colliding like a particle) shows the particle nature of light, so D matches II. Final: A-IV, B-III, C-I, D-II.
Solved Dual Nature Of Radiation And Matter NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It is the wavelength of the matter wave linked to a moving particle, given by lambda = h/p = h/(mv).
Who proposed the de Broglie wavelength and when?
French physicist Louis Victor de Broglie proposed it in 1924. He got the 1929 Nobel Prize in Physics for the idea of the wave nature of matter.
What is the value of Planck's constant used here?
h = 6.63 x 10^-34 J s. It links the wave attribute (lambda) with the particle attribute (momentum p).
Which experiment proved the de Broglie wavelength is real?
The Davisson-Germer experiment (1927) showed electrons diffract like waves, confirming that moving electrons have the wavelength predicted by lambda = h/p.
How is de Broglie wavelength related to kinetic energy?
For a particle of mass m and kinetic energy E, lambda = h/sqrt(2mE), because p = sqrt(2mE). This is very common in NEET numericals.
