Chapter 11 · Dual Nature of Radiation and Matter

11.3 The Wave Nature of Matter

If light waves can act as particles, can particles act as waves? Louis de Broglie said yes: every moving particle has a wavelength $\lambda=h/p$ — far too small to notice for a cricket ball, but comparable to the spacing of atoms for an electron accelerated through a few volts.

The images of the coronavirus spike protein that guided vaccine design in 2020 were made not with light but with electrons. A cryo-electron microscope fires electrons accelerated through $300\,000$ volts at a frozen sample and focuses them with magnetic lenses; it can resolve features a fraction of a nanometre across, a thousand times finer than the best light microscope. It works because electrons behave as waves, with a wavelength hundreds of thousands of times shorter than that of visible light.

The idea that matter has a wavelength came from a 1924 doctoral thesis by Louis de Broglie. Nature, he argued, loves symmetry: if light — a wave — also behaves as a stream of particles (Section 11.2), then particles such as electrons should also behave as waves. Three years later electron waves were seen diffracting from crystals exactly as he predicted. In this section we find the de Broglie wavelength of moving particles, work out the wavelength of electrons accelerated through a potential difference, and see why the waves of everyday objects can never be detected.

De Broglie's Hypothesis

For a photon, energy and momentum are linked to the wave's frequency and wavelength by $E=h\nu$ and $p=h/\lambda$. De Broglie proposed that the same relations hold for every particle: a particle of momentum $p$ is accompanied by a wave of wavelength $h/p$.

📐 Law 11.5 — The de Broglie relation

A particle of mass $m$ moving with speed $v$ (much less than $c$), and so with momentum $p=mv$, has an associated wave — a matter wave — of wavelength $$\lambda=\frac{h}{p}=\frac{h}{mv}.$$ This is the de Broglie wavelength of the particle. The same relation, $\lambda=h/p$, holds for a photon.

📘 Definition 11.6 — Matter waves and the de Broglie wavelength
Matter waves are the waves associated with moving material particles (electrons, protons, neutrons, atoms, molecules — and in principle cricket balls). Their wavelength, the de Broglie wavelength $\lambda=h/p$, decreases as the momentum increases. Matter waves show the wave behaviour — diffraction and interference — whenever they meet obstacles or openings comparable in size to $\lambda$.

Because $h$ is so small ($6.63\times10^{-34}$ J s), the wavelength is tiny unless the momentum is tiny too. That is why wave behaviour shows up for electrons, neutrons and atoms but never for anything you can see.

✏️ Example 1 — A cricket ball and an electron
Find the de Broglie wavelength of (a) a cricket ball of mass $0.16$ kg bowled at $40$ m/s and (b) an electron moving at $1.0\times10^6$ m/s. Compare each with the size of an atom (about $10^{-10}$ m). ($h=6.63\times10^{-34}$ J s, $m_e=9.1\times10^{-31}$ kg.)
(a) Ball $p=(0.16)(40)=6.4$ kg m/s; $\lambda=\dfrac{6.63\times10^{-34}}{6.4}\approx1.0\times10^{-34}$ m — about $10^{-24}$ of an atom's size, and $10^{-19}$ of a proton's. No slit, crack or wicket could ever diffract it.
(b) Electron $p=(9.1\times10^{-31})(1.0\times10^6)=9.1\times10^{-25}$ kg m/s; $$\lambda=\frac{6.63\times10^{-34}}{9.1\times10^{-25}}\approx7.3\times10^{-10}\ \text{m}=0.73\ \text{nm},$$ a few atomic diameters.
Lesson Matter waves matter only when $\lambda$ is comparable to something in the experiment — for electrons, the spacing of atoms in a crystal.

It is often more convenient to use the kinetic energy $K$. Since $K=\dfrac{p^2}{2m}$, the momentum is $p=\sqrt{2mK}$.

📐 Result 11.6 — De Broglie wavelength from kinetic energy

A particle of mass $m$ with kinetic energy $K$ (non-relativistic, $K\ll mc^2$) has $$\lambda=\frac{h}{\sqrt{2mK}}.$$ For particles with the same kinetic energy, $\lambda\propto\dfrac{1}{\sqrt m}$; for the same particle, $\lambda\propto\dfrac{1}{\sqrt K}$.

✏️ Example 2 — A slow neutron
Neutrons from a nuclear reactor are slowed until their kinetic energy is about $0.025$ eV. Find their de Broglie wavelength. Take the neutron mass as $1.67\times10^{-27}$ kg.
Energy $K=0.025\times1.6\times10^{-19}=4.0\times10^{-21}$ J.
Momentum $p=\sqrt{2mK}=\sqrt{2(1.67\times10^{-27})(4.0\times10^{-21})}$ $\approx3.66\times10^{-24}$ kg m/s.
Wavelength $\lambda=\dfrac{6.63\times10^{-34}}{3.66\times10^{-24}}\approx1.8\times10^{-10}$ m $=0.18$ nm.
Comment Just right for diffraction by crystals: neutron beams are used worldwide to find where the atoms (including hydrogen, almost invisible to X-rays) sit in materials and proteins.

Electrons Accelerated through a Potential Difference

In an electron gun, electrons leave a hot cathode almost at rest and are accelerated through a potential difference $V$. Each gains kinetic energy $K=eV$ (Section 2.1), so its momentum is $p=\sqrt{2m_eeV}$.

📐 Result 11.7 — Wavelength of an accelerated electron

An electron accelerated from rest through a potential difference $V$ (non-relativistic, $eV\ll m_ec^2$: accurate to about $1\%$ for $V$ up to about $20$ kV) has de Broglie wavelength $$\lambda=\frac{h}{\sqrt{2m_eeV}}=\frac{1.227}{\sqrt V}\ \text{nm}\quad(V\ \text{in volts}).$$ The number $1.227$ comes from the precise values of $h$, $m_e$ and $e$; with our rounded constants it is $1.229$. For any particle of charge $q$ and mass $m$ accelerated through $V$: $\lambda=\dfrac{h}{\sqrt{2mqV}}$.

✏️ Example 3 — Electrons from an electron gun
Find the de Broglie wavelength of electrons accelerated from rest through (a) $100$ V, (b) $54$ V and (c) $10$ kV. (d) What accelerating voltage gives $\lambda=0.050$ nm?
(a) 100 V $\lambda=\dfrac{1.227}{\sqrt{100}}\approx0.123$ nm.
(b) 54 V $\lambda=\dfrac{1.227}{\sqrt{54}}\approx0.167$ nm.
(c) 10 kV $\lambda=\dfrac{1.227}{100}\approx0.0123$ nm $=12$ pm.
(d) Invert $V=\left(\dfrac{1.227}{0.050}\right)^2\approx6.0\times10^2$ V.
Check From first principles for (a): $\lambda=\dfrac{6.63\times10^{-34}}{\sqrt{2(9.1\times10^{-31})(1.6\times10^{-19})(100)}}$ $\approx1.23\times10^{-10}$ m ✓. Interactive 1 at $100$ V reads the same.
✏️ Example 4 — Electron, proton and alpha particle
An electron, a proton and an alpha particle (charge $2e$, mass $4m_p$) are each accelerated from rest through $100$ V. Find the ratio of their de Broglie wavelengths, and the wavelength of the proton. ($m_p=1.67\times10^{-27}$ kg.)
Scaling $\lambda=\dfrac{h}{\sqrt{2mqV}}\propto\dfrac{1}{\sqrt{mq}}$ at the same $V$.
Ratios $\dfrac{\lambda_e}{\lambda_p}=\sqrt{\dfrac{m_p}{m_e}}=\sqrt{\dfrac{1.67\times10^{-27}}{9.1\times10^{-31}}}\approx42.8$; $\dfrac{\lambda_p}{\lambda_\alpha}=\sqrt{\dfrac{(4m_p)(2e)}{m_pe}}=\sqrt8\approx2.83$.
Proton $\lambda_p=\dfrac{0.1227\ \text{nm}}{42.8}\approx2.9\times10^{-3}$ nm $=2.9\times10^{-12}$ m; then $\lambda_\alpha\approx1.0\times10^{-12}$ m.
Answer $\lambda_e:\lambda_p:\lambda_\alpha\approx121:2.83:1$ (or $42.8:1:0.354$).
✏️ Example 5 — An electron and a photon with the same wavelength
An electron and a photon both have wavelength $0.20$ nm. Compare (a) their momenta and (b) their energies (kinetic energy for the electron).
(a) Momenta Both $p=\dfrac h\lambda=\dfrac{6.63\times10^{-34}}{0.20\times10^{-9}}\approx3.3\times10^{-24}$ kg m/s — equal.
(b) Photon $E=\dfrac{1240}{0.20}=6200$ eV $=6.2$ keV (an X-ray).
(b) Electron $K=\dfrac{p^2}{2m_e}=\dfrac{(3.3\times10^{-24})^2}{2(9.1\times10^{-31})}$ $\approx6.0\times10^{-18}$ J $\approx38$ eV. (Check: $V=(1.227/0.20)^2\approx38$ V ✓.)
Lesson At the same wavelength the electron carries about $160$ times less energy — one reason electron diffraction is gentler on delicate surfaces than X-rays.
✏️ Example 6 — The electron microscope
An electron microscope accelerates electrons through $50$ kV. (a) Find their de Broglie wavelength from Result 11.7. (b) How many times shorter is it than green light of $550$ nm? (c) At this voltage the non-relativistic formula is about $2\%$ off. Does that change the conclusion?
(a) Wavelength $\lambda=\dfrac{1.227}{\sqrt{5.0\times10^4}}\approx5.5\times10^{-3}$ nm $=5.5$ pm.
(b) Ratio $\dfrac{550}{5.5\times10^{-3}}\approx1.0\times10^5$.
(c) Relativity The exact value is a little shorter, about $5.4$ pm; the ratio is still $\sim10^5$. The fine detail a microscope can show is limited roughly by the wavelength, so electron microscopes can in principle resolve far smaller objects — in practice the imperfections of magnetic lenses limit them to about $0.1$ nm, still a thousand times better than light.
Interactive 1 — Matter Waves Meet a Crystal

A particle from the gun on the left is accelerated through $V$ (logarithmic slider) and its matter wave travels towards a crystal whose rows of atoms are $0.25$ nm apart (an illustrative spacing) — wave and crystal are drawn to the same scale. Strong diffraction needs a wavelength comparable to the spacing. Compare electrons with protons and alpha particles at the same voltage. Computed with $\lambda=h/\sqrt{2mqV}$ (non-relativistic).

100 V
kinetic energy—
speed—
de Broglie $\lambda$—
$\lambda$ / atomic spacing—

Wave–Particle Duality

We now have a strange but consistent picture. Radiation shows interference and diffraction (wave behaviour) and the photoelectric effect (particle behaviour). Electrons are detected as point-like particles — one click of a counter, one dot on a screen — and they diffract from crystals. Neither "wave" nor "particle" alone describes either of them.

📘 Definition 11.7 — Wave–particle duality
Wave–particle duality is the fact that both radiation and matter show wave properties (interference, diffraction) in some experiments and particle properties (localised, countable impacts carrying definite energy and momentum) in others. The two descriptions are linked by $E=h\nu$ and $p=h/\lambda$. In any one experiment one aspect or the other shows up, never both at once.

What, then, is "waving" in a matter wave? The accepted answer is that the matter wave tells us where the particle is likely to be found: where the wave's intensity (the square of its amplitude) is large, the particle is likely to be detected. Send electrons one at a time at a crystal and each lands at a single point, but after thousands have arrived their dots build up the diffraction pattern predicted by the wave. Chapter 12 uses matter waves to explain why the electron in a hydrogen atom can only have certain energies.

Interactive 2 — The de Broglie Ruler

Choose an object and its speed (logarithmic slider). The ruler runs over more than thirty powers of ten of length; the red marker is the object's de Broglie wavelength $\lambda=h/mv$, and the orange marks are real sizes it can be compared with. Wave behaviour is observable only when $\lambda$ reaches the size of something in the experiment — atoms, slits, crystal spacings. The masses are typical values; the dust grain and bacterium are illustrative.

1.0 × 10⁶ m/s
momentum $p=mv$—
de Broglie $\lambda$—
kinetic energy—
⚠️ Common mistakes with matter waves

1. Using $\lambda=1.227/\sqrt V$ for any particle. It is for electrons only. For other particles use $\lambda=h/\sqrt{2mqV}$; an alpha particle has charge $2e$.

2. Using $E=hc/\lambda$ for an electron. That formula is for photons. For an electron, $K=\dfrac{p^2}{2m}=\dfrac{h^2}{2m\lambda^2}$.

3. Kinetic energy in eV inside $\sqrt{2mK}$. Convert $K$ to joules first.

4. "Doubling $K$ halves $\lambda$." $\lambda\propto1/\sqrt K$: doubling $K$ divides $\lambda$ by $\sqrt2$. To halve $\lambda$, quadruple $K$ (or $V$).

5. Thinking matter waves are electromagnetic. They are not radiated, do not travel at $c$, and exist for neutral particles such as neutrons too.

🚀 Beyond the board syllabus (enrichment only) — the Davisson–Germer experiment

Enrichment only: the Davisson–Germer experiment is not in the 2026 CBSE, JEE Main, NEET or JEE Advanced syllabi, all of which keep the de Broglie relation itself.

In 1927 Clinton Davisson and Lester Germer, in New York, fired electrons accelerated through $54$ V at a nickel crystal and measured how many scattered in each direction. Instead of a smooth spread they found a sharp peak at $50^\circ$ from the incident beam. The rows of nickel atoms on the surface, $d=0.215$ nm apart, act like a reflection grating (compare the grating idea in Chapter 10 — Wave Optics): a peak appears where waves from neighbouring rows differ in path by one wavelength, $d\sin\theta=\lambda$, giving $\lambda=0.215\sin50^\circ\approx0.165$ nm. De Broglie's formula predicts $1.227/\sqrt{54}\approx0.167$ nm — agreement within about $1\%$. In the same year G. P. Thomson in Aberdeen passed faster electrons through thin metal foils and photographed diffraction rings. The experiment's details are not in the board syllabus; JEE and NEET still ask for the $54$ V wavelength.

Exercises

I
The de Broglie relation. $h=6.63\times10^{-34}$ J s, $m_e=9.1\times10^{-31}$ kg, $m_p=1.67\times10^{-27}$ kg.
  • 1 Find the de Broglie wavelength of an electron moving at $2.0\times10^6$ m/s.
  • 2 Find the de Broglie wavelength of a proton moving at $2.0\times10^6$ m/s.
  • 3 For an electron and a proton with the same speed, find $\lambda_e/\lambda_p$.
  • 4 Find the de Broglie wavelength of a $0.15$ kg ball thrown at $30$ m/s.
  • 5 At what speed does an electron have a de Broglie wavelength of $0.50$ nm?
  • 6 The kinetic energy of a particle is made four times larger. What happens to its de Broglie wavelength?
  • 7 The momentum of a particle is doubled. What happens to its de Broglie wavelength?
  • 8 A person of mass $60$ kg walks through a doorway at $1.0$ m/s. Estimate the de Broglie wavelength and explain why the person is not diffracted.
1 $\lambda=\dfrac{6.63\times10^{-34}}{(9.1\times10^{-31})(2.0\times10^6)}\approx3.6\times10^{-10}$ m $=0.36$ nm.
2 $\lambda=\dfrac{6.63\times10^{-34}}{(1.67\times10^{-27})(2.0\times10^6)}\approx2.0\times10^{-13}$ m.
3 $\dfrac{\lambda_e}{\lambda_p}=\dfrac{m_p}{m_e}\approx1.8\times10^3$ (about $1835$).
4 $\lambda=\dfrac{6.63\times10^{-34}}{4.5}\approx1.5\times10^{-34}$ m.
5 $v=\dfrac{h}{m_e\lambda}$ $=\dfrac{6.63\times10^{-34}}{(9.1\times10^{-31})(5.0\times10^{-10})}$ $\approx1.5\times10^6$ m/s.
6 $\lambda\propto1/\sqrt K$: it halves.
7 $\lambda=h/p$: it halves.
8 $\lambda=\dfrac{6.63\times10^{-34}}{60}\approx1.1\times10^{-35}$ m — about $10^{-35}$ of the doorway's width, so the spreading is utterly negligible.
II
Accelerated particles. Use $\lambda=1.227/\sqrt V$ nm for electrons; $e=1.6\times10^{-19}$ C, $m_p=1.67\times10^{-27}$ kg.
  • 1 Find the de Broglie wavelength of electrons accelerated through $400$ V.
  • 2 Find the de Broglie wavelength of electrons accelerated through $25$ V.
  • 3 Through what potential difference must electrons be accelerated to have $\lambda=0.10$ nm?
  • 4 An electron has de Broglie wavelength $1.0$ nm. Find its kinetic energy in eV.
  • 5 Find the de Broglie wavelength of a proton accelerated from rest through $100$ V.
  • 6 An electron and a proton are accelerated through the same potential difference. Find $\lambda_e/\lambda_p$.
  • 7 A proton and an alpha particle are accelerated through the same potential difference. Find $\lambda_p/\lambda_\alpha$.
  • 8 By what factor must the accelerating voltage of an electron gun be increased to halve the wavelength of its electrons?
1 $\dfrac{1.227}{20}\approx0.061$ nm.
2 $\dfrac{1.227}{5}\approx0.245$ nm.
3 $V=\left(\dfrac{1.227}{0.10}\right)^2\approx1.5\times10^2$ V.
4 $V=\left(\dfrac{1.227}{1.0}\right)^2\approx1.5$ V, so $K\approx1.5$ eV.
5 $\lambda=\dfrac{h}{\sqrt{2m_peV}}$ $=\dfrac{6.63\times10^{-34}}{\sqrt{2(1.67\times10^{-27})(1.6\times10^{-19})(100)}}$ $\approx2.9\times10^{-12}$ m.
6 Same charge and $V$: $\dfrac{\lambda_e}{\lambda_p}=\sqrt{\dfrac{m_p}{m_e}}\approx43$.
7 $\dfrac{\lambda_p}{\lambda_\alpha}=\sqrt{\dfrac{m_\alpha q_\alpha}{m_pq_p}}=\sqrt{4\times2}=2\sqrt2\approx2.8$.
8 $\lambda\propto1/\sqrt V$: increase $V$ four times.
III
Comparisons and derivations. $hc=1240$ eV nm.
  • 1 An electron and a photon each have wavelength $0.50$ nm. Find the photon's energy and the electron's kinetic energy, both in eV.
  • 2 Do the electron and photon of part 1 have equal momenta? Find the momentum.
  • 3 An electron and a photon each have energy $100$ eV (kinetic energy for the electron). Find both wavelengths.
  • 4 Find the de Broglie wavelength of a neutron ($1.67\times10^{-27}$ kg) of kinetic energy $0.025$ eV.
  • 5 Starting from $K=\tfrac12mv^2$, show that $\lambda=\dfrac{h}{\sqrt{2mK}}$.
  • 6 Two particles of masses $m$ and $4m$ have the same kinetic energy. Find the ratio of their de Broglie wavelengths.
  • 7 A speck of dust of mass $1.0\times10^{-9}$ kg drifts at $2.0$ mm/s. Find its de Broglie wavelength and compare it with the size of a nucleus ($\sim10^{-15}$ m).
1 Photon: $\dfrac{1240}{0.50}=2480$ eV $\approx2.5$ keV. Electron: $V=\left(\dfrac{1.227}{0.50}\right)^2\approx6.0$, so $K\approx6.0$ eV.
2 Yes, $p=h/\lambda$ for both: $\dfrac{6.63\times10^{-34}}{5.0\times10^{-10}}\approx1.3\times10^{-24}$ kg m/s.
3 Electron: $\dfrac{1.227}{\sqrt{100}}\approx0.123$ nm. Photon: $\dfrac{1240}{100}=12.4$ nm — about $100$ times longer.
4 Same as Example 2: $\approx0.18$ nm.
5 $K=\dfrac{(mv)^2}{2m}=\dfrac{p^2}{2m}$, so $p=\sqrt{2mK}$ and $\lambda=\dfrac hp=\dfrac{h}{\sqrt{2mK}}$.
6 $\lambda\propto1/\sqrt m$ at equal $K$: $\lambda_m:\lambda_{4m}=2:1$.
7 $p=2.0\times10^{-12}$ kg m/s; $\lambda\approx3.3\times10^{-22}$ m — still about three million times smaller than a nucleus.
IV
Applications.
  • 1 Modelling. A scanning electron microscope uses $30$ kV electrons. Find their wavelength (non-relativistic) and how many times shorter it is than light of $500$ nm.
  • 2 A crystal has atomic planes $0.20$ nm apart. Roughly what accelerating voltage gives electrons a wavelength equal to this spacing?
  • 3 Neutron crystallographers want neutrons of wavelength $0.10$ nm. Find the neutron speed and kinetic energy in eV ($m_n=1.67\times10^{-27}$ kg).
  • 4 In a cathode-ray tube, electrons are accelerated through $2.0$ kV. Find $\lambda$, and explain why the beam can be treated as a stream of particles in designing the tube.
  • 5 An electron and a proton have the same de Broglie wavelength. Which has the larger kinetic energy, and by what factor?
  • 6 Explain, with one example each, what is meant by the wave nature of particles and the particle nature of waves.
1 $\lambda=\dfrac{1.227}{\sqrt{3.0\times10^4}}\approx7.1\times10^{-3}$ nm $=7.1$ pm; $\dfrac{500}{7.1\times10^{-3}}\approx7\times10^4$ times shorter.
2 $V=\left(\dfrac{1.227}{0.20}\right)^2\approx38$ V.
3 $v=\dfrac{h}{m\lambda}$ $=\dfrac{6.63\times10^{-34}}{(1.67\times10^{-27})(1.0\times10^{-10})}$ $\approx4.0\times10^3$ m/s; $K=\tfrac12mv^2\approx1.3\times10^{-20}$ J $\approx0.082$ eV.
4 $\lambda=\dfrac{1.227}{\sqrt{2000}}\approx0.027$ nm. The tube's openings and screen are millions of times larger, so diffraction is negligible and the electrons follow particle paths.
5 Equal $\lambda$ means equal $p$; $K=\dfrac{p^2}{2m}$, so the electron has $\dfrac{m_p}{m_e}\approx1.8\times10^3$ times more kinetic energy.
6 Wave nature of particles: electrons diffract from a crystal like X-rays (de Broglie $\lambda=h/p$). Particle nature of waves: light ejects electrons one photon at a time in the photoelectric effect ($E=h\nu$).
🎯 Exam corner — matter waves BoardJEE MainJEE AdvancedNEET

E1. Board · 3 marks An electron starting from rest is accelerated through a potential difference $V$. Derive an expression for its de Broglie wavelength and show that $\lambda\approx\dfrac{1.227}{\sqrt V}$ nm. Find $\lambda$ for $V=100$ V.

Energy $K=eV=\dfrac{p^2}{2m_e}$, so $p=\sqrt{2m_eeV}$.
Wavelength $\lambda=\dfrac hp=\dfrac{h}{\sqrt{2m_eeV}}$. With precise constants $\dfrac{h}{\sqrt{2m_ee}}$ $=1.227\times10^{-9}$ m V$^{1/2}$, so $\lambda=\dfrac{1.227}{\sqrt V}$ nm.
Number $V=100$ V: $\lambda\approx0.123$ nm.
Exam tip State "non-relativistic" — examiners credit knowing where the formula applies.

E2. NEET An electron and a proton have the same kinetic energy. The ratio of their de Broglie wavelengths $\lambda_e/\lambda_p$ is (A) $1$ (B) $1836$ (C) $\sqrt{1836}$ (D) $1/\sqrt{1836}$.

Scaling $\lambda=\dfrac{h}{\sqrt{2mK}}\propto\dfrac{1}{\sqrt m}$, so $\dfrac{\lambda_e}{\lambda_p}=\sqrt{\dfrac{m_p}{m_e}}=\sqrt{1836}\approx43$.
Answer (C). Exam tip: (B) is the answer for equal speeds; read whether the question fixes $v$, $K$ or $p$.

E3. JEE Main A proton and an alpha particle are accelerated from rest through the same potential difference. The ratio $\lambda_p:\lambda_\alpha$ of their de Broglie wavelengths is (A) $1:1$ (B) $2:1$ (C) $2\sqrt2:1$ (D) $\sqrt2:1$.

Scaling $\lambda=\dfrac{h}{\sqrt{2mqV}}\propto\dfrac{1}{\sqrt{mq}}$; $m_\alpha q_\alpha=(4m_p)(2e)=8m_pe$.
Answer $\lambda_p/\lambda_\alpha=\sqrt8=2\sqrt2$: (C). Exam tip: (B) forgets the alpha particle's double charge.

E4. JEE Advanced (One or more correct.) An electron and a photon each have wavelength $1.0$ nm. Then (A) they have equal momenta (B) the photon has the larger energy (C) the electron's kinetic energy is about $1.5$ eV (D) the electron moves at the speed of light.

Options (A) True: $p=h/\lambda$ for both. (B) True: photon $1240$ eV against electron $1.5$ eV. (C) True: $K=(1.227/1.0)^2$ eV $\approx1.5$ eV. (D) False: $v=\dfrac{h}{m_e\lambda}\approx7.3\times10^5$ m/s.
Answer (A), (B), (C). Exam tip: equal $\lambda$ always means equal momentum, whatever the particles.

E5. JEE Main NEET Electrons accelerated from rest through $54$ V are scattered from a nickel crystal. Their de Broglie wavelength is about (A) $0.017$ nm (B) $0.167$ nm (C) $0.227$ nm (D) $1.67$ nm.

Formula $\lambda=\dfrac{1.227}{\sqrt{54}}\approx0.167$ nm.
Answer (B). These are the conditions of the Davisson–Germer experiment (🚀 box above; its details are beyond NCERT, the wavelength calculation is not). Exam tip: atomic spacings are $0.1$–$0.3$ nm; an answer far outside that range would not diffract.

E6. Board · Case study A university electron microscope accelerates electrons through $40$ kV. (i) (2 marks) Find the de Broglie wavelength of the electrons (ignore relativity) and how many times shorter it is than green light of $550$ nm. (ii) (1 mark) What happens to the wavelength if the voltage is raised to $160$ kV? (iii) (1 mark) Why can an electron microscope show much finer detail than a light microscope?

(i) $\lambda=\dfrac{1.227}{\sqrt{40\,000}}=\dfrac{1.227}{200}\approx6.1\times10^{-3}$ nm; $\dfrac{550}{6.1\times10^{-3}}\approx9\times10^4$.
(ii) $V$ is four times larger, so $\lambda$ halves: $\approx3.1\times10^{-3}$ nm (before relativistic correction).
(iii) The finest detail an instrument can resolve is limited roughly by the wavelength it uses; the electrons' wavelength is about $10^5$ times shorter than that of light. Exam tip: mention "wavelength" explicitly — "electrons are smaller" earns no marks.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. A cryo-electron microscope uses $300$ kV electrons. How short is their wavelength — and if it is so short, why can such microscopes "only" resolve about $0.2$ nm?

Why it matters. Cryo-electron microscopy earned the 2017 Nobel Prize in Chemistry (Dubochet, Frank and Henderson — working in Switzerland, the USA and the UK) and is now used in laboratories on every continent to map proteins, viruses and drug targets atom by atom. Knowing what limits its resolution tells engineers where improvements must come from.

What to do.

  1. Use Result 11.7 to find the non-relativistic wavelength at $300$ kV.
  2. At such voltages the electrons move at over half the speed of light, and the correct formula is $\lambda=\dfrac{h}{\sqrt{2m_eeV\left(1+\dfrac{eV}{2m_ec^2}\right)}}$, with $m_ec^2=0.511$ MeV. Find the corrected wavelength and the percentage change.
  3. A light microscope resolves at best about $200$ nm. By what factor is the electron wavelength smaller than this? By what factor is the achieved resolution ($0.2$ nm) better?
  4. The two factors differ enormously. Suggest two reasons why cryo-EM cannot reach a resolution close to its wavelength (think about the lenses, and about what the electron beam does to a delicate frozen protein).

Tools: Calculator; optional reading on magnetic-lens aberrations and radiation damage.

Hand in: The two wavelengths, the comparison factors and a paragraph on the limits of resolution.

Going further. Repeat step 2 for a $1$ MV microscope. How does the size of the relativistic correction change with voltage?

Instructor note: non-relativistic $\lambda=1.227/\sqrt{3\times10^5}\approx2.24\times10^{-3}$ nm; $eV/2m_ec^2=300/1022\approx0.294$, so $\lambda\approx2.24/\sqrt{1.294}\approx1.97$ pm, about $12\%$ shorter. Wavelength $\sim10^5$ times smaller than $200$ nm; achieved resolution only $10^3$ times better. Limits: spherical and chromatic aberration of magnetic lenses forces tiny apertures; radiation damage limits the dose, so images are noisy and must be averaged over thousands of particles; specimen motion and detector noise. At $1$ MV: $\lambda\approx0.87$ pm, correction factor $\sqrt{1.98}\approx1.41$ ($\approx29\%$ shorter).

🧪 Self-check quiz

0/5 correct

Q1. The de Broglie wavelength of a particle of momentum $p$ is

Q2. Electrons accelerated from rest through $100$ V have a de Broglie wavelength of about

Q3. The kinetic energy of an electron is increased four times. Its de Broglie wavelength

Q4. An electron, a proton, a neutron and an alpha particle all move at the same speed. Which has the shortest de Broglie wavelength?

Q5. We never notice the wave nature of a moving cricket ball because

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