de Broglie's Hypothesis (1924)

We've seen that light has a dual nature — it behaves as both a wave and a particle. Louis de Broglie asked a bold question: if radiation can behave as particles (photons), why can't particles of matter (like electrons) behave as waves?

The de Broglie Relation

de Broglie proposed that every moving particle has an associated wavelength, given by:

λ=hmv=hp\boxed{\lambda = \frac{h}{mv} = \frac{h}{p}}

where:

  • λ\lambda = de Broglie wavelength
  • hh = Planck's constant (6.626×10−346.626 \times 10^{-34} J s)
  • mm = mass of the particle (kg)
  • vv = velocity of the particle (m/s)
  • p=mvp = mv = momentum

Key Observations

  1. Wavelength is inversely proportional to momentum. Heavier or faster particles have shorter wavelengths.

  2. For macroscopic objects, the wavelength is incredibly tiny — undetectable:

  • A cricket ball (0.15 kg) moving at 30 m/s: λ=6.626×10−34/(0.15×30)=1.47×10−34\lambda = 6.626 \times 10^{-34}/(0.15 \times 30) = 1.47 \times 10^{-34} m
  • This is far too small to detect or have any observable effect.
  1. For microscopic particles like electrons, the wavelength is significant:
  • An electron (m=9.1×10−31m = 9.1 \times 10^{-31} kg) at v=106v = 10^6 m/s: λ=6.626×10−34/(9.1×10−31×106)=7.28×10−10\lambda = 6.626 \times 10^{-34}/(9.1 \times 10^{-31} \times 10^6) = 7.28 \times 10^{-10} m ≈7.3\approx 7.3 Å
  • This is comparable to atomic dimensions — so wave nature matters!

Key Point: Wave-particle duality is a universal property of matter, but its effects are only observable at the atomic and sub-atomic scale.

Relating de Broglie Wavelength to Kinetic Energy

Since kinetic energy KE=12mv2KE = \frac{1}{2}mv^2, we can write mv=2m⋅KEmv = \sqrt{2m \cdot KE}. Therefore:

λ=h2m⋅KE\lambda = \frac{h}{\sqrt{2m \cdot KE}}

For a charged particle accelerated through a potential difference VV: KE=eV(for an electron)KE = eV \quad \text{(for an electron)}

λ=h2meV\lambda = \frac{h}{\sqrt{2meV}}

For an electron specifically: λ=1.226V nm\lambda = \frac{1.226}{\sqrt{V}} \text{ nm}

where VV is the accelerating voltage in volts.

[JEE Tip] This formula (λ=1.226/V\lambda = 1.226/\sqrt{V} nm) is very handy for quick calculations involving electron diffraction problems.

Experimental Verification

de Broglie's hypothesis was experimentally verified by Davisson and Germer (1927), who demonstrated the diffraction of electrons by a nickel crystal. Electron diffraction patterns were exactly as predicted by de Broglie's formula, confirming that electrons indeed have wave properties.

G.P. Thomson independently verified this using thin metal foils. Both Davisson and Thomson shared the Nobel Prize in Physics (1937) for this work.

Heisenberg's Uncertainty Principle (1927)

Werner Heisenberg proposed one of the most profound principles in physics:

It is impossible to determine simultaneously, the exact position and exact momentum (or velocity) of a microscopic particle with absolute certainty.

Mathematically:

Δx⋅Δp≥h4π\boxed{\Delta x \cdot \Delta p \geq \frac{h}{4\pi}}

or equivalently:

Δx⋅mΔv≥h4π\boxed{\Delta x \cdot m \Delta v \geq \frac{h}{4\pi}}

where:

  • Δx\Delta x = uncertainty in position
  • Δp=mΔv\Delta p = m\Delta v = uncertainty in momentum
  • Δv\Delta v = uncertainty in velocity
  • h/4π=ℏ/2=5.27×10−35h/4\pi = \hbar/2 = 5.27 \times 10^{-35} J s

What Does This Mean?

  1. If you try to measure position very precisely (Δx\Delta x very small), then the uncertainty in momentum (Δp\Delta p) becomes very large — and vice versa.

  2. This is NOT about the limitations of our instruments. It is a fundamental property of nature — a consequence of the wave nature of matter.

  3. The uncertainty principle applies to any pair of conjugate variables: position-momentum, energy-time, etc.

Why is it Significant for Atoms?

  • It makes the concept of a definite orbit meaningless for electrons. You cannot say "the electron is at position xx moving with velocity vv."
  • This is why Bohr's model (with precise circular orbits) had to be replaced by the quantum mechanical model (with probability distributions called orbitals).

Why Uncertainty Doesn't Matter for Macroscopic Objects

The uncertainty principle applies to ALL objects, but for large objects, the uncertainty is negligibly small.

Example: For a ball of mass 1 kg with Δv=10−6\Delta v = 10^{-6} m/s: Δx≥h4πmΔv=6.626×10−344×3.14×1×10−6=5.3×10−29 m\Delta x \geq \frac{h}{4\pi m \Delta v} = \frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 1 \times 10^{-6}} = 5.3 \times 10^{-29} \text{ m}

This is absurdly small — far smaller than the nucleus! So for macroscopic objects, the uncertainty is meaningless in practice.

For an electron with the same Δv=10−6\Delta v = 10^{-6} m/s: Δx≥6.626×10−344×3.14×9.1×10−31×10−6=58 m\Delta x \geq \frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 9.1 \times 10^{-31} \times 10^{-6}} = 58 \text{ m}

The uncertainty in position is 58 metres! This means we really cannot pinpoint where the electron is.

Significance of Both Concepts

de Broglie's hypothesis + Heisenberg's uncertainty principle together require us to abandon the classical idea of electrons moving in fixed orbits. Instead:

  • Electrons are described by wave functions (ψ\psi)
  • We can only talk about the probability of finding an electron in a region of space
  • These probability regions are called orbitals (not orbits!)

Key Point: The dual nature of matter and the uncertainty principle are the foundations of the quantum mechanical model of the atom.

Solved Examples

Example 1: de Broglie Wavelength of an Electron

Calculate the de Broglie wavelength of an electron moving at 1×1071 \times 10^7 m/s. (me=9.1×10−31m_e = 9.1 \times 10^{-31} kg)

Solution: λ=hmv=6.626×10−349.1×10−31×107\lambda = \frac{h}{mv} = \frac{6.626 \times 10^{-34}}{9.1 \times 10^{-31} \times 10^7} =6.626×10−349.1×10−24=7.28×10−11 m=0.728 A˚= \frac{6.626 \times 10^{-34}}{9.1 \times 10^{-24}} = 7.28 \times 10^{-11} \text{ m} = 0.728 \text{ Å}

Answer: λ=7.28×10−11\lambda = 7.28 \times 10^{-11} m =72.8= 72.8 pm.

Example 2: de Broglie Wavelength of a Cricket Ball

A cricket ball of mass 0.15 kg is bowled at 140 km/h. Calculate its de Broglie wavelength.

Solution: v=140×10003600=38.89 m/sv = 140 \times \frac{1000}{3600} = 38.89 \text{ m/s}

λ=6.626×10−340.15×38.89=6.626×10−345.833=1.14×10−34 m\lambda = \frac{6.626 \times 10^{-34}}{0.15 \times 38.89} = \frac{6.626 \times 10^{-34}}{5.833} = 1.14 \times 10^{-34} \text{ m}

This is about 10−1910^{-19} times smaller than the diameter of an atomic nucleus. Completely undetectable!

Answer: λ=1.14×10−34\lambda = 1.14 \times 10^{-34} m.

Example 3: de Broglie Wavelength from Kinetic Energy

An electron has a kinetic energy of 4.55×10−254.55 \times 10^{-25} J. Find its de Broglie wavelength.

Solution: λ=h2mKE=6.626×10−342×9.1×10−31×4.55×10−25\lambda = \frac{h}{\sqrt{2mKE}} = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 4.55 \times 10^{-25}}}

Denominator: 8.281×10−55=9.10×10−28\sqrt{8.281 \times 10^{-55}} = 9.10 \times 10^{-28}

λ=6.626×10−349.10×10−28=7.28×10−7 m=728 nm\lambda = \frac{6.626 \times 10^{-34}}{9.10 \times 10^{-28}} = 7.28 \times 10^{-7} \text{ m} = 728 \text{ nm}

Answer: λ=728\lambda = 728 nm.

Example 4: Electron Accelerated Through a Potential

Calculate the de Broglie wavelength of an electron accelerated through a potential difference of 100 V.

Solution: Using the quick formula: λ=1.226V nm=1.226100=1.22610=0.1226 nm=1.226 A˚\lambda = \frac{1.226}{\sqrt{V}} \text{ nm} = \frac{1.226}{\sqrt{100}} = \frac{1.226}{10} = 0.1226 \text{ nm} = 1.226 \text{ Å}

Answer: λ=0.1226\lambda = 0.1226 nm = 1.226 Å.

Example 5: Uncertainty in Velocity

The uncertainty in position of an electron is 0.1 nm. Calculate the uncertainty in its velocity.

Solution: Δx⋅mΔv≥h4π\Delta x \cdot m\Delta v \geq \frac{h}{4\pi} Δv≥h4πmΔx=6.626×10−344×3.14×9.1×10−31×10−10\Delta v \geq \frac{h}{4\pi m \Delta x} = \frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 9.1 \times 10^{-31} \times 10^{-10}} =6.626×10−341.143×10−39=5.79×105 m/s= \frac{6.626 \times 10^{-34}}{1.143 \times 10^{-39}} = 5.79 \times 10^5 \text{ m/s}

Answer: Δv≥5.79×105\Delta v \geq 5.79 \times 10^5 m/s.

Example 6: Uncertainty in Position of a Bullet

A bullet of mass 0.05 kg has a speed of 500 m/s with an uncertainty of 0.02%. Calculate the uncertainty in its position.

Solution: Δv=0.02% of 500=0.0002×500=0.1 m/s\Delta v = 0.02\% \text{ of } 500 = 0.0002 \times 500 = 0.1 \text{ m/s}

Δx≥h4πmΔv=6.626×10−344×3.14×0.05×0.1\Delta x \geq \frac{h}{4\pi m \Delta v} = \frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 0.05 \times 0.1} =6.626×10−346.28×10−2=1.055×10−32 m= \frac{6.626 \times 10^{-34}}{6.28 \times 10^{-2}} = 1.055 \times 10^{-32} \text{ m}

Answer: Δx≥1.06×10−32\Delta x \geq 1.06 \times 10^{-32} m.

Example 7: Comparing Wavelengths of Two Particles

A proton and an electron have the same kinetic energy. Which has the longer de Broglie wavelength?

Solution: λ=h2mKE\lambda = \frac{h}{\sqrt{2mKE}}

For the same kinetic energy, λ∝1m\lambda \propto \frac{1}{\sqrt{m}}.

Since mp≫mem_p \gg m_e, the electron has the longer wavelength.

λeλp=mpme=1836≈42.8\frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_p}{m_e}} = \sqrt{1836} \approx 42.8

Answer: The electron has a wavelength about 42.8 times longer than the proton.

Example 8: de Broglie Wavelength and Bohr Orbit

Show that the circumference of the nthn^{\text{th}} Bohr orbit equals nn times the de Broglie wavelength.

Solution: From Bohr's quantisation condition: mvrn=nh2πmvr_n = \frac{nh}{2\pi}

Rearranging: 2πrn=nhmv=nλ2\pi r_n = \frac{nh}{mv} = n\lambda

Therefore, 2πrn=nλ2\pi r_n = n\lambda

This means the circumference of the nthn^{\text{th}} orbit equals nn times the de Broglie wavelength.

Answer: 2πrn=nλ2\pi r_n = n\lambda.

Example 9: Two Particles with Same Wavelength

An electron and a photon each have a wavelength of 1 Å. Which has greater energy?

Solution: Photon energy: Ep=hcλ=6.626×10−34×3×10810−10=1.988×10−15 JE_p = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{10^{-10}} = 1.988 \times 10^{-15} \text{ J}

Electron kinetic energy: KE=h22mλ2=(6.626×10−34)22×9.1×10−31×10−20KE = \frac{h^2}{2m\lambda^2} = \frac{(6.626 \times 10^{-34})^2}{2 \times 9.1 \times 10^{-31} \times 10^{-20}} =4.39×10−671.82×10−50=2.41×10−17 J= \frac{4.39 \times 10^{-67}}{1.82 \times 10^{-50}} = 2.41 \times 10^{-17} \text{ J}

Thus photon energy is about 1.988×10−152.41×10−17≈82.5\frac{1.988 \times 10^{-15}}{2.41 \times 10^{-17}} \approx 82.5 times greater.

Answer: The photon has much greater energy than the electron at the same wavelength.

Example 10: Uncertainty in Momentum

If the uncertainty in the position of an electron is equal to its de Broglie wavelength, what is the minimum uncertainty in its velocity?

Solution: Given: Δx=λ=hmv\Delta x = \lambda = \frac{h}{mv}

Using Heisenberg's principle: Δx⋅mΔv≥h4π\Delta x \cdot m\Delta v \geq \frac{h}{4\pi}

Substituting Δx=hmv\Delta x = \frac{h}{mv}: hmv⋅mΔv≥h4π\frac{h}{mv} \cdot m\Delta v \geq \frac{h}{4\pi}

h⋅Δvv≥h4πh \cdot \frac{\Delta v}{v} \geq \frac{h}{4\pi}

Δv≥v4π\Delta v \geq \frac{v}{4\pi}

Answer: Δv≥v/4π\Delta v \geq v/4\pi.