A full lesson: a cartoon movie to start, then de Broglie's bold idea λ = h/p, why the photon obeys it too, the wavelength of an electron accelerated through V, why you never see the wave of a cricket ball, electron vs proton vs alpha, NEET shortcuts, problems to solve yourself, and 12 MCQs.
Every moving thing has a de Broglie wavelength λ = h/p. For a cricket ball it is about 10⁻³⁴ m, hopelessly small. For an electron it is about the size of an atom, so its wave nature shows up, and we use it to see tiny things.
Sunday morning, a gully match. Kabir bowls, Arjun bats. Arjun has just learnt (Lesson 4) that light, a wave, also behaves like a stream of particles with momentum p = h/λ. He asks the obvious question: if a wave can act like a particle, can a particle act like a wave? Does Kabir's ball have a wavelength?
In 1924 Louis de Broglie answered yes. Nature likes symmetry: if radiation has a dual character, matter should have one too. Every moving particle has a wavelength λ = h/p. The catch is the size of h. For the ball, λ is so small that no experiment can ever notice it. For an electron, it is about the spacing of atoms in a crystal, and the wave nature becomes real and useful.
"Take Lesson 4's photon formula p = h/λ and simply turn it around: λ = h/p. That one line is the whole of de Broglie's idea. Everything in this lesson is just putting numbers into it."
"Work out your own de Broglie wavelength: you weigh 60 kg and walk at 1.5 m/s. Is it bigger or smaller than the nucleus of an atom (about 10⁻¹⁵ m)?"
λ = h/(mv) = 6.63 × 10⁻³⁴ / (60 × 1.5) ≈ 7.4 × 10⁻³⁶ m. That is about 10²⁰ times smaller than a nucleus. No slit or crystal could ever be small enough to show your wave nature.
A particle of mass m and speed v has momentum p = mv, and an associated wave of wavelength λ = h/p = h/(mv). The relation links a wave quantity (λ) with a particle quantity (p) through Planck's constant. Pick a particle, then change its speed with the slider or drag the dot on the graph.
Use SI units: mass in kg, speed in m/s.
h = 6.63 × 10⁻³⁴ J s. Double the speed (or the mass) and λ halves: λ ∝ 1/p.
For matter it began as a hypothesis whose validity could only be tested by experiment. Electron diffraction by crystals later confirmed it.
"Heavier or faster means shorter wavelength. An electron at a million metres per second has λ of about 0.7 nanometre. A proton at the same speed, being 1836 times heavier, has a wavelength 1836 times shorter."
For a photon, p = hν/c, so h/p = c/ν: its de Broglie wavelength is simply the wavelength of the light. The same relation works for light and matter. But a photon and an electron with the same λ (same p) carry very different energies: E = pc for the photon, K = p²/2m for the electron. Drag along the curve or use the slider.
Its de Broglie wavelength is its ordinary wavelength. E = pc = hc/λ.
For the same λ, much less energy than the photon. Ratio E/K = 2mc/p, which is large.
Use this when K is given instead of v. Valid for v much less than c.
"A photon and an electron both have λ = 1 nm. Which has the larger momentum?"
Neither: same λ means the same momentum, p = h/λ = 6.63 × 10⁻²⁵ kg m/s for both. Their energies differ: photon 1240 eV, electron only about 1.5 eV.
An electron starting from rest is accelerated through a potential difference V. It gains kinetic energy K = eV. Turn the voltage up and watch its matter wave squeeze. Try other particles too.
Work done by the field on a charge e through V. (For charge q: K = qV.)
From K = p²/2m (speeds much less than c).
Put in h, m and e of the electron.
V in volts. 100 V → 0.123 nm. 10 kV → 0.0123 nm. Four times the voltage halves λ.
"Remember 1.227 over root V nanometres, or 12.27 over root V angstrom. An electron through about 150 volts has λ of about one angstrom, the size of an atom. That is why crystals can diffract electrons."
A wave shows itself (by diffraction or interference) only when its wavelength is comparable to the size of an opening or obstacle. Pick an object, set its speed, and see where its λ falls on the ruler of sizes. Each mark on the ruler is a factor of 10.
| Object | m, v | λ = h/(mv) |
|---|---|---|
| You | 60 kg, 1.5 m/s | 7.4 × 10⁻³⁶ m |
| Ball | 0.15 kg, 30 m/s | 1.5 × 10⁻³⁴ m |
| Dust | 10⁻⁹ kg, 1 cm/s | 6.6 × 10⁻²³ m |
| Electron | 9.11 × 10⁻³¹ kg, 5.4 × 10⁶ m/s | 1.35 × 10⁻¹⁰ m |
"Planck's constant is tiny and everyday momenta are big, so λ comes out near 10⁻³⁴ metre. Nothing in the universe is that small to diffract it. And always change grams to kilograms first, or your answer is wrong by a factor of a thousand."
NEET loves ratio questions. First decide what is common: the kinetic energy K, the accelerating voltage V, or the momentum p. Then the formula tells you the ratio. Switch the condition and slide its value.
λe : λp ≈ √1836 ≈ 43 : 1. The alpha (heaviest) has the shortest λ (NCERT example).
K = qV, so the charge matters: λp : λα = √(4 × 2) = 2√2 : 1.
λ = h/p depends only on momentum, whatever the mass.
Detail finer than the wavelength cannot be seen. Electron λ is thousands of times shorter than light's, so it shows viruses and even rows of atoms.
NCERT: the matter wave tells us the probability of finding the particle at a place (Born). It is not a wave of any material.
A particle with a sharp λ (sharp p) is spread out in space; a wave packet localises it but spreads its λ (Heisenberg).
Through voltage: λ = h/√(2mqV). Pick the form that matches the data given.
Same as 12.27/√V Å. 150 V → 1 Å; 100 V → 1.23 Å; 10 kV → 0.123 Å.
Alpha: λ = 0.0101/√V nm (m ≈ 4mₚ and q = 2e, so √8 ≈ 2.83 times shorter than a proton).
K × 4 → λ ÷ 2. K up by 21% → λ ÷ 1.1, a 9% decrease. K × 2 → λ ÷ √2.
At 1 nm: photon 1240 eV, electron 1.5 eV. Same λ always means same p.
150 g = 0.15 kg. 72 km/h = 20 m/s. eV → J: × 1.6 × 10⁻¹⁹.
"In a ratio question, never put in numbers first. Write which of K, V or p is common, write the proportionality, and the answer is often one line: root of the mass ratio, or root of mass times charge."
Cover the solution, write the formula, and solve on paper. Open the hint only if you are stuck for more than a minute. Take h = 6.63 × 10⁻³⁴ J s, mₑ = 9.11 × 10⁻³¹ kg, e = 1.6 × 10⁻¹⁹ C, mₚ = 1836 mₑ.
What is the de Broglie wavelength of (a) an electron moving with speed 5.4 × 10⁶ m/s, and (b) a ball of mass 150 g travelling at 30.0 m/s?
λ = h/(mv). Change 150 g to 0.150 kg.
Find the de Broglie wavelength of an electron accelerated from rest through (a) 100 V and (b) 10 kV.
λ = 1.227/√V nm.
A particle is moving three times as fast as an electron. The ratio of the de Broglie wavelength of the particle to that of the electron is 1.813 × 10⁻⁴. Calculate the particle's mass and identify it.
λ ∝ 1/(mv): λ/λₑ = (mₑ vₑ)/(m v), with v = 3vₑ.
An electron, a proton and an alpha particle have the same kinetic energy. Which has the shortest de Broglie wavelength? Find λₑ : λₚ.
λ = h/√(2mK); at the same K, λ ∝ 1/√m.
A photon and an electron each have a wavelength of 1.0 nm. Find the momentum of each, the energy of the photon and the kinetic energy of the electron (in eV).
p = h/λ for both. Photon: E = pc. Electron: K = p²/(2m).
Take h = 6.63 × 10⁻³⁴ J s, mₚ/mₑ = 1836, and λ(electron) = 1.227/√V nm. Write the formula first, then click.
Score: 0 / 12
| If the question says… | Write this first | The trap |
|---|---|---|
| "de Broglie wavelength of a particle of mass m, speed v" | λ = h/(mv) | Grams not changed to kg (answer off by 1000) |
| "kinetic energy K is given" | λ = h/√(2mK) | Using λ = h/K or forgetting the factor 2 |
| "electron accelerated through V volts" | λ = 1.227/√V nm | Using it for a proton or alpha particle |
| "same kinetic energy" ratio | λ ∝ 1/√m | Writing λ ∝ 1/m |
| "accelerated through the same potential" | λ ∝ 1/√(mq) | Forgetting the alpha particle's charge 2e |
| "same momentum" or "same speed and mass" | same λ | Comparing masses when only p matters |
| "de Broglie wavelength of a photon" | λ = h/p = c/ν | Inventing a mass for the photon |
| "why no wave nature of macroscopic bodies" | λ ≈ 10⁻³⁴ m, too small to detect | Saying "they have no wavelength" |
In the Atoms chapter, Bohr's quantised orbits get a beautiful meaning from this lesson: an electron's orbit holds a whole number of de Broglie wavelengths, 2πr = nλ, which gives mvr = nh/2π. Now test the whole chapter in the Dual Nature Challenge.
Values used: h = 6.63 × 10⁻³⁴ J s, mₑ = 9.11 × 10⁻³¹ kg, mₚ = 1.673 × 10⁻²⁷ kg (≈ 1836 mₑ), m(alpha) = 6.64 × 10⁻²⁷ kg, e = 1.6 × 10⁻¹⁹ C, c = 3 × 10⁸ m/s. The Davisson–Germer experiment is not part of the rationalised NCERT text (it is in this chapter's Beyond NCERT set).
Follow the four steps: understand it, remember it, try it, then prove it with past papers.