Wave Nature of Matter

Q: Why do we not see the wave nature of everyday objects?

Their momentum is huge compared with h, so the wavelength is tiny: about 10⁻³⁴ m for a moving ball, far smaller than a nucleus. Wave effects such as diffraction appear only when the wavelength is comparable to the obstacle, which is never the case for everyday objects.

Q: What is the de Broglie wavelength of an electron accelerated through V volts?

λ = h/√(2meV) = 1.227/√V nm, or 12.27/√V Å. For 100 V it is 0.123 nm. The formula is only for electrons and only for voltages well below about 100 kV, where relativity can be ignored.

Q: With the same kinetic energy, which has the shortest wavelength: electron, proton or alpha particle?

The alpha particle. Since λ = h/√(2mK), at equal kinetic energy the wavelength is inversely proportional to the square root of the mass, and the alpha particle is the heaviest. The electron has the longest wavelength.

PhysicsDual Nature Of Radiation And MatterFor NEET aspirants

The wave nature of matter means that every moving particle has a wavelength, the de Broglie wavelength $λ = \frac{h}{p} = \frac{h}{m v}$ . It is noticeable only for tiny masses: an electron accelerated through $100 V$ has $λ \approx 0.123 nm$ , about the size of an atom, so crystals diffract electrons, as Davisson and Germer showed. This page covers the de Broglie relation, $λ$ from kinetic energy, voltage and temperature, particle comparisons, the uncertainty principle and the experiment. The wave nature of matter gives quick formula and ratio questions in JEE Main and NEET.

On this page1de Broglie relation2

λ

from

K

V

T

3Comparing particles4Electron vs photon5Bohr orbits6Uncertainty7Davisson-Germer8Revision map

Key Formulas - Quick Reference

★ Must learn de Broglie relation: $λ = \frac{h}{p} = \frac{h}{m v}$ (for photons too, with $p = \frac{E}{c}$ )
★ Must learn From kinetic energy or accelerating voltage: $λ = \frac{h}{2 m K} = \frac{h}{2 m q V}$
★ Must learn Electron through $V$ volts: $λ$ $= \frac{12.27}{V} \overset{˚}{A}$ $= \frac{1.227}{V} nm$
Proton $\frac{0.286}{V} \overset{˚}{A}$ ; alpha particle $\frac{0.101}{V} \overset{˚}{A}$ ( $V$ in volts); neutron $\frac{0.286}{K} \overset{˚}{A}$ ( $K$ in eV)
Thermal particle ( $K = \frac{3}{2} k T$ ): $λ = \frac{h}{3 mk T}$
★ Must learn Ratios: same $p$ : same $λ$ ; same $K$ : $λ \propto \frac{1}{m}$ ; same $V$ : $λ \propto \frac{1}{m q}$ , so $\frac{λ _{p}}{λ _{α}} = 22$
Same $λ$ : photon energy $E = \frac{h c}{λ}$ , electron kinetic energy $K = \frac{h ^{2}}{2 m λ ^{2}}$
★ Must learn Bohr orbit: $2 π r = nλ \Leftrightarrow m v r = \frac{nh}{2 π}$
Uncertainty principle: $Δ x Δ p \geq \frac{ℏ}{2}$ , $ℏ = \frac{h}{2 π}$ (order of magnitude $Δ x Δ p \approx ℏ$ )

1. The de Broglie Hypothesis

Light behaves as a wave in interference, diffraction and polarisation, and as particles (photons) in the photoelectric effect and Compton scattering. In 1924 de Broglie argued that nature should be symmetric: if waves can act as particles, particles should act as waves.

λ = \frac{h}{p} = \frac{h}{m v}

A particle of momentum $p$ has an associated matter wave (de Broglie wave) of wavelength $λ$ . The left side is a wave property, the right side a particle property; Planck's constant links them.

It works for photons too: $p = \frac{h ν}{c}$ gives $\frac{h}{p} = \frac{c}{ν} = λ$ , the ordinary wavelength of the light.
$λ$ depends only on momentum, not on charge or on what the particle is. Neutral particles (neutrons, atoms, molecules) have matter waves too.
The larger $m$ or $v$ , the smaller $λ$ . A $0.12 kg$ ball at $20 m s^{- 1}$ : $λ$ $= \frac{6.63 \times 1 0 ^{- 34}}{2.4}$ $= 2.8 \times 1 0^{- 34} m$ , far too small to observe.

Figure 1: Where matter waves matter. Electrons, neutrons and molecules have

λ

comparable to atomic spacings (

\sim 1 0^{- 10} m

), so crystals diffract them. A cricket ball's

λ

1 0^{19}

times smaller than a nucleus, so everyday objects never show wave behaviour.

Phenomenon	Explained by
Interference, diffraction, polarisation	wave picture only
Photoelectric effect, Compton effect, black-body radiation	particle (photon) picture only
Reflection, refraction, straight-line travel	both pictures
Electron and neutron diffraction	matter waves ( $λ = \frac{h}{p}$ )

Key idea

Every moving particle has

λ = \frac{h}{p}

; it matters only when

λ

is comparable to the size of the obstacle, which happens for atomic particles, never for everyday objects.

2. Wavelength from Kinetic Energy, Voltage and Temperature

Most questions give energy, voltage or temperature instead of speed. Write $p$ in terms of what is given:

Kinetic energy $K = \frac{p ^{2}}{2 m}$ , so $p = 2 m K$ and $λ = \frac{h}{2 m K}$ .
A charge $q$ accelerated from rest through $V$ gains $K = q V$ : $λ = \frac{h}{2 m q V}$ .
A gas particle at temperature $T$ has average $K = \frac{3}{2} k T$ : $λ = \frac{h}{3 mk T}$ .

Electron shortcut: putting $h$ , $m_{e}$ and $e$ into $\frac{h}{2 m _{e} e V}$ ,

λ_{e} = \frac{1.227}{V} nm = \frac{12.27}{V} \overset{˚}{A} (V in volts)

For example $V = 100 V$ gives $0.123 nm$ ; it is valid only for electrons (and only while $V$ is well below $1 0^{5} V$ ).

Particle	Mass, charge	$λ$ after $V$ volts	At $V = 100 V$
electron	$m_{e}$ , $e$	$\frac{12.27}{V} \overset{˚}{A}$	$1.23 \overset{˚}{A}$
proton	$1836 m_{e}$ , $e$	$\frac{0.286}{V} \overset{˚}{A}$	$0.0286 \overset{˚}{A}$
deuteron	$2 m_{p}$ , $e$	$\frac{0.202}{V} \overset{˚}{A}$	$0.0202 \overset{˚}{A}$
alpha ( $He^{2 +}$ )	$4 m_{p}$ , $2 e$	$\frac{0.101}{V} \overset{˚}{A}$	$0.0101 \overset{˚}{A}$

Figure 2:

λ = \frac{1.227}{V} nm

for an electron. A few tens to a few hundred volts give

λ

equal to atomic spacings, which is why Davisson and Germer used

54 V

electrons (

0.167 nm

Exam Trick

Scale from the electron. $λ \propto \frac{1}{m q V}$ , so for any particle $λ$ $= \frac{12.27}{V} \frac{m _{e} e}{m q} \overset{˚}{A}$ . For a neutron or any neutral particle there is no $q V$ : use $K$ directly, $λ_{n} = \frac{0.286}{K ( eV )} \overset{˚}{A}$ .

Quick Recall: tap to check

λ

of an electron accelerated through

400 V

\frac{12.27}{20} = 0.61 \overset{˚}{A}

The voltage is made 4 times. What happens to

λ

It halves (

λ \propto \frac{1}{V}

λ

of a thermal neutron at

300 K

\frac{h}{3 m _{n} k T} = 0.145 nm

, similar to atomic spacing, so neutrons diffract from crystals.

3. Comparing Particles

Condition	$λ$ depends on	Shortest wavelength
same speed $v$	$λ \propto \frac{1}{m}$	heaviest particle
same momentum $p$	$λ$ is the same for all	all equal
same kinetic energy $K$	$λ \propto \frac{1}{m}$	heaviest: $λ_{e} > λ_{p} > λ_{α}$
same accelerating voltage $V$	$λ \propto \frac{1}{m q}$	largest $m q$ : $\frac{λ _{p}}{λ _{α}}$ $= \frac{4 m _{p} \cdot 2 e}{m _{p} \cdot e}$ $= 22$
same temperature $T$	$λ \propto \frac{1}{m}$	heaviest particle

Figure 3: The four

λ

graphs.

λ \propto \frac{1}{p}

gives a hyperbola (a) or a line through O (b);

λ \propto \frac{1}{K}

gives (c) and, on log scales, parallel lines of slope

- \frac{1}{2}

(d), the heaviest particle having the shortest

λ

Exam Trick

Change one thing, scale once. If $K$ becomes $n$ times, $λ$ becomes $\frac{1}{n}$ times. If $p$ doubles, $λ$ halves. If $λ$ must drop by $1%$ , $K$ must rise by about $2%$ ( $\frac{Δ λ}{λ}$ $= - \frac{1}{2} \frac{Δ K}{K}$ ).

Key idea

λ = \frac{h}{p}

: same momentum means the same wavelength whatever the particle; at the same energy or voltage the heavier (and more highly charged) particle has the shorter wavelength.

4. Electron and Photon with the Same Wavelength

Same $λ$ means the same momentum $p = \frac{h}{λ}$ , but the energies are very different, because a photon has $E = p c$ while a slow electron has $K = \frac{p ^{2}}{2 m}$ .

Photon

$E = p c = \frac{h c}{λ}$ , so $E \propto \frac{1}{λ}$ . Speed always $c$ ; no rest mass. At $1 nm$ : $E = 1240 eV$ .

Electron (non-relativistic)

$K = \frac{p ^{2}}{2 m} = \frac{h ^{2}}{2 m λ ^{2}}$ , so $K \propto \frac{1}{λ ^{2}}$ . Speed $v ≪ c$ . At $1 nm$ : $K = 1.5 eV$ .

Figure 4: Same

λ

, very different energies.

\frac{E _{p h o t o n}}{K _{e l ec t r o n}} = \frac{2 m c λ}{h}

, about

800

1 nm

. So for crystal diffraction (

λ \approx 0.1 nm

) an electron beam needs only

150 eV

, an X-ray photon

12.4 keV

This is why electron microscopes work: an electron of a few tens of keV has a wavelength of a few picometres, about $1 0^{5}$ times shorter than visible light, giving far higher resolution.

5. Matter Waves and Bohr Orbits

de Broglie's idea explains Bohr's quantisation rule. An electron wave going round an orbit must join itself smoothly, otherwise it cancels itself out. So the circumference must hold a whole number of wavelengths:

2 π r = nλ = \frac{nh}{m v} \Rightarrow m v r = \frac{nh}{2 π}

Figure 5: de Broglie's explanation of Bohr's quantisation.

2 π r = nλ = \frac{nh}{m v}

gives

m v r = \frac{nh}{2 π}

. In the ground state of hydrogen

λ = 2 π a_{0} = 0.332 nm

Useful result: in the $n$ th orbit $λ_{n}$ $= \frac{2 π r _{n}}{n} \propto \frac{n ^{2} / Z}{n}$ $= \frac{n}{Z}$ . For hydrogen, $λ_{1} = 2 π a_{0} = 0.332 nm$ .

6. The Uncertainty Principle and Probability Waves

A wave with one exact wavelength (exact momentum) fills all space, so the particle's position is completely unknown. A particle located in a small region is a wave packet, a mixture of many wavelengths, so its momentum is spread out. This trade-off is Heisenberg's uncertainty principle.

★ Must learn

Δ x Δ p \geq \frac{ℏ}{2}, ℏ = \frac{h}{2 π}

Position and momentum cannot both be known exactly at the same time. If $Δ p = 0$ then $Δ x \to \infty$ , and the reverse. (Many books write the order-of-magnitude form $Δ x Δ p \approx ℏ$ .)

Figure 6: Squeezing the packet (smaller

Δ x

) needs more wavelengths, hence a larger

Δ p

Δ x Δ p \geq \frac{ℏ}{2}

. The shaded

∣ ψ ∣^{2}

is the probability density (Born): the particle is most likely where the wave is strongest.

Born's probability interpretation: the square of the matter-wave amplitude, $∣ A ∣^{2}$ (or $∣ ψ ∣^{2}$ ), at a point is the probability per unit volume of finding the particle there. The wave is a probability wave, not a vibration of some material.
The wavelength of a matter wave is physical and measurable; the packet moves with the particle's velocity (group velocity).

JEE Advanced

Fast electrons need relativity. For $K$ comparable to $m_{0} c^{2} = 0.511 MeV$ use $p c = K (K + 2 m_{0} c^{2})$ , so

λ = \frac{h}{2 m _{0} K ( 1 + \frac{K}{2 m _{0} c ^{2}} )}

At $50 kV$ this gives $5.36 pm$ against the non-relativistic $5.49 pm$ (about $2.4%$ smaller). Below about $10 kV$ the correction is under $1%$ .

Phase velocity. With $E = h ν$ and $p = \frac{h}{λ}$ , the product $ν λ = \frac{E}{p}$ is not the particle's speed and has no physical meaning (relativistically $\frac{c ^{2}}{v} > c$ ). The packet, and so the particle, moves at the group velocity, which equals $v$ .

Quick Recall: tap to check

If the momentum of a particle is known exactly, what can you say about its position?

Completely uncertain:

Δ p = 0

means

Δ x \to \infty

What does

∣ ψ ∣^{2}

at a point tell you?

The probability per unit volume of finding the particle at that point.

de Broglie wavelength of the electron in the first Bohr orbit of hydrogen?

2 π a_{0} = 0.332 nm

7. Davisson-Germer Experiment

The first proof of matter waves (1927): electrons diffract from a crystal exactly as X-rays do.

Figure 7: The Davisson-Germer set-up (1927). Electrons from a heated filament are accelerated through

V

(44 to 68 V), strike a nickel crystal, and the collector measures the scattered intensity at each angle

θ

Figure 8: The diffraction peak (schematic, after Davisson and Germer's data). It grows, is sharpest at

V = 54 V

θ = 5 0^{\circ}

, then fades: a wave effect that particles alone cannot explain.

Electron gun: tungsten filament (coated with barium oxide), heated by a low-voltage supply; electrons accelerated by a high voltage $V$ (44 to 68 V).
Target: a nickel crystal; the scattered intensity is measured at each angle $θ$ (between the incident and scattered beams) by a movable collector and galvanometer.
A strong peak appears at $V = 54 V$ , $θ = 5 0^{\circ}$ .
Measured wavelength $0.165 nm$ ; de Broglie's prediction $\frac{1.227}{54} = 0.167 nm$ . Excellent agreement.
G. P. Thomson (1927) independently diffracted electrons through thin metal foils; Davisson and Thomson shared the 1937 Nobel Prize. Electron double-slit fringes (1989) and interference of iodine molecules (1994) confirmed it further.

JEE Advanced

Where 0.165 nm comes from. The surface atoms of nickel form rows $d = 0.215 nm$ apart and act like a reflection grating. Waves scattered by neighbouring atoms differ in path by $d sin θ$ (Figure 9). The first maximum needs $d sin θ = λ$ :

λ = 0.215 \times sin 5 0^{\circ} = 0.165 nm

Equivalently, Bragg reflection from planes $0.091 nm$ apart at glancing angle $6 5^{\circ}$ : $2 \times 0.091 \times sin 6 5^{\circ} = 0.165 nm$ .

Figure 9: The surface atoms act like a reflection grating. First-order maximum:

d sin θ = λ

, so

λ = 0.215 \times sin 5 0^{\circ} = 0.165 nm

, matching de Broglie's

\frac{1.227}{54} = 0.167 nm

Key idea

Davisson-Germer:

54 V

electrons, nickel crystal, peak at

5 0^{\circ}

λ = 0.165 nm

measured vs

0.167 nm

predicted.

8. Solving Matter-Wave Problems and Revision Map

Use the flowchart to pick the formula, then the mind map to revise the whole concept.

Figure 10: Choosing the right de Broglie formula. Every route is

λ = \frac{h}{p}

with

p

written in terms of what is given.

Figure 11: Mind map of this concept. Cover a branch, recall its three points, then check.

9. Solved Examples

Solved Example 1

Find the de Broglie wavelength of (a) an electron moving at

5.4 \times 1 0^{6} m s^{- 1}

and (b) a

150 g

ball moving at

30.0 m s^{- 1}

Solution:

(a) $p$ $= m v$ $= 9.11 \times 1 0^{- 31} \times 5.4 \times 1 0^{6}$ $= 4.92 \times 1 0^{- 24} kg m s^{- 1}$ ; $λ$ $= \frac{6.63 \times 1 0 ^{- 34}}{4.92 \times 1 0 ^{- 24}}$ $= 1.35 \times 1 0^{- 10} m$ .

(b) $p = 0.150 \times 30.0 = 4.50 kg m s^{- 1}$ ; $λ = \frac{6.63 \times 1 0 ^{- 34}}{4.50}$ .

Answer: (a) $0.135 nm$ (X-ray range); (b) $1.47 \times 1 0^{- 34} m$ , about $1 0^{- 19}$ of a proton's size, impossible to detect.

Solved Example 2

An electron, a proton and an alpha particle have the same kinetic energy. Which has the shortest de Broglie wavelength?
(A) electron
(B) proton
(C) alpha particle
(D) all equal

Solution:

Answer: (C). $λ = \frac{h}{2 m K}$ , so at equal $K$ , $λ \propto \frac{1}{m}$ . The alpha particle is the heaviest ( $\approx 4 m_{p}$ ), so its wavelength is the shortest.

Solved Example 3

A particle moving three times as fast as an electron has a de Broglie wavelength

1.813 \times 1 0^{- 4}

times that of the electron. Find its mass and identify it.

Solution:

$m = \frac{h}{λ v}$ , so $\frac{m}{m _{e}}$ $= \frac{λ _{e}}{λ} \cdot \frac{v _{e}}{v}$ $= \frac{1}{1.813 \times 1 0 ^{- 4}} \times \frac{1}{3}$ .

$m$ $= \frac{9.11 \times 1 0 ^{- 31}}{3 \times 1.813 \times 1 0 ^{- 4}}$ $= 1.675 \times 1 0^{- 27} kg$ .

Answer: $1.675 \times 1 0^{- 27} kg$ , a proton or a neutron.

Solved Example 4

Electrons are accelerated from rest through

56 V

. Find their momentum and de Broglie wavelength.

Solution:

$p$ $= 2 m e V$ $= 2 \times 9.11 \times 1 0^{- 31} \times 1.6 \times 1 0^{- 19} \times 56$ $= 4.04 \times 1 0^{- 24} kg m s^{- 1}$ .

$λ$ $= \frac{h}{p}$ $= \frac{6.63 \times 1 0 ^{- 34}}{4.04 \times 1 0 ^{- 24}}$ , or quickly $\frac{1.227}{56} nm$ .

Answer: $p = 4.04 \times 1 0^{- 24} kg m s^{- 1}$ , $λ = 0.164 nm$ .

Solved Example 5

An electron has kinetic energy

120 eV

. Find its (a) momentum (b) speed (c) de Broglie wavelength.

Solution:

(a) $p$ $= 2 m K$ $= 2 \times 9.11 \times 1 0^{- 31} \times 120 \times 1.6 \times 1 0^{- 19}$ $= 5.91 \times 1 0^{- 24} kg m s^{- 1}$ .

(b) $v$ $= \frac{p}{m}$ $= \frac{5.91 \times 1 0 ^{- 24}}{9.11 \times 1 0 ^{- 31}}$ $= 6.49 \times 1 0^{6} m s^{- 1}$ .

Answer: (a) $5.91 \times 1 0^{- 24} kg m s^{- 1}$ ; (b) $6.49 \times 1 0^{6} m s^{- 1}$ ; (c) $0.112 nm$ .

Solved Example 6

An electron and a photon each have wavelength

1.00 nm

. Find (a) their momenta (b) the energy of the photon (c) the kinetic energy of the electron.

Solution:

(a) Both: $p$ $= \frac{h}{λ}$ $= \frac{6.63 \times 1 0 ^{- 34}}{1 0 ^{- 9}}$ $= 6.63 \times 1 0^{- 25} kg m s^{- 1}$ .

(b) $E$ $= p c$ $= 6.63 \times 1 0^{- 25} \times 3 \times 1 0^{8}$ $= 1.99 \times 1 0^{- 16} J$ $= 1.24 keV$ .

(c) $K$ $= \frac{p ^{2}}{2 m}$ $= \frac{( 6.63 \times 1 0 ^{- 25} ) ^{2}}{2 \times 9.11 \times 1 0 ^{- 31}}$ $= 2.41 \times 1 0^{- 19} J$ .

Answer: (a) $6.63 \times 1 0^{- 25} kg m s^{- 1}$ each; (b) $1.24 keV$ ; (c) $1.51 eV$ (about 800 times less).

Solved Example 7

(a) For what kinetic energy does a neutron have a de Broglie wavelength of

1.40 \times 1 0^{- 10} m

? (b) Find the wavelength of a thermal neutron at

300 K

(average

K = \frac{3}{2} k T

). (

m_{n} = 1.675 \times 1 0^{- 27} kg

k = 1.38 \times 1 0^{- 23} J K^{- 1}

)

Solution:

(a) $K$ $= \frac{h ^{2}}{2 m λ ^{2}}$ $= \frac{( 6.63 \times 1 0 ^{- 34} ) ^{2}}{2 \times 1.675 \times 1 0 ^{- 27} \times ( 1.40 \times 1 0 ^{- 10} ) ^{2}}$ $= 6.69 \times 1 0^{- 21} J$ $= 0.0418 eV$ .

(b) $λ = \frac{h}{3 mk T}$ , with $3 mk T$ $= 3 \times 1.675 \times 1 0^{- 27} \times 1.38 \times 1 0^{- 23} \times 300$ $= 2.08 \times 1 0^{- 47}$ , so $λ$ $= \frac{6.63 \times 1 0 ^{- 34}}{2.08 \times 1 0 ^{- 47}}$ $= \frac{6.63 \times 1 0 ^{- 34}}{4.56 \times 1 0 ^{- 24}}$ .

Answer: (a) $6.69 \times 1 0^{- 21} J$ ( $0.042 eV$ ); (b) $1.45 \times 1 0^{- 10} m$ , the atomic-spacing scale, which is why slow (thermal) neutrons are used for crystal diffraction.

Solved Example 8

Find the de Broglie wavelength of a nitrogen molecule in air at

300 K

, moving with the rms speed. (Molar mass of

N_{2}

= 28.0 g mol^{- 1}

)

Solution:

$m$ $= 28.0 \times 1.66 \times 1 0^{- 27}$ $= 4.65 \times 1 0^{- 26} kg$ ; $v_{r m s}$ $= \frac{3 k T}{m}$ $= \frac{3 \times 1.38 \times 1 0 ^{- 23} \times 300}{4.65 \times 1 0 ^{- 26}}$ $= 517 m s^{- 1}$ .

$λ$ $= \frac{h}{m v _{r m s}}$ $= \frac{6.63 \times 1 0 ^{- 34}}{4.65 \times 1 0 ^{- 26} \times 517}$ .

Answer: $λ \approx 2.76 \times 1 0^{- 11} m$ $= 0.0276 nm$ .

Solved Example 9

A proton and an alpha particle are accelerated from rest through the same potential difference. The ratio

\frac{λ _{p}}{λ _{α}}

of their de Broglie wavelengths is
(A)

1

(B)

2

(C)

22

(D)

\frac{1}{2 2}

Solution:

Answer: (C). $λ = \frac{h}{2 m q V}$ , so $\frac{λ _{p}}{λ _{α}}$ $= \frac{m _{α} q _{α}}{m _{p} q _{p}}$ $= \frac{4 m _{p} \times 2 e}{m _{p} \times e}$ $= 8$ $= 22$ .

Solved Example 10

The kinetic energy of an electron is doubled. Its de Broglie wavelength becomes
(A)

2

times
(B)

\frac{1}{2}

times
(C)

2

times
(D)

\frac{1}{2}

times

Solution:

Answer: (D). $λ \propto \frac{1}{K}$ , so doubling $K$ multiplies $λ$ by $\frac{1}{2}$ (a $29%$ decrease). Option (B) would need $K$ to become $4$ times.

Solved Example 11

Crystal diffraction can be done with X-rays or with electrons. For a wavelength of

1 \overset{˚}{A}

, which probe carries more energy, and how much more?

Solution:

X-ray photon: $E$ $= \frac{h c}{λ}$ $= \frac{12400 eV A ˚}{1 A ˚}$ $= 12.4 keV$ .

Electron: $K$ $= \frac{h ^{2}}{2 m λ ^{2}}$ $= \frac{( 6.63 \times 1 0 ^{- 34} ) ^{2}}{2 \times 9.11 \times 1 0 ^{- 31} \times 1 0 ^{- 20}}$ $= 2.41 \times 1 0^{- 17} J$ $= 151 eV$ .

Answer: the X-ray photon, about 80 times more energetic ( $12.4 keV$ against $151 eV$ ).

Solved Example 12

An electron microscope uses electrons accelerated through

50 kV

. Find their de Broglie wavelength and compare the resolving power with an optical microscope using yellow light (

550 nm

Solution:

$λ$ $= \frac{12.27}{50000} \overset{˚}{A}$ $= 0.0549 \overset{˚}{A}$ $= 5.49 \times 1 0^{- 12} m$ (the relativistic value is $5.36 \times 1 0^{- 12} m$ ).

Resolving power $\propto \frac{1}{λ}$ : $\frac{550 \times 1 0 ^{- 9}}{5.49 \times 1 0 ^{- 12}} \approx 1 0^{5}$ .

Answer: $λ \approx 5.5 pm$ ; about $1 0^{5}$ times the resolving power of the optical microscope (with other factors equal).

Solved Example 13

Show that the de Broglie wavelength of the electron in the

n

th Bohr orbit of a hydrogen-like atom is proportional to

\frac{n}{Z}

, and find it for the ground state of hydrogen (

a_{0} = 0.529 \overset{˚}{A}

Solution:

From $2 π r_{n} = n λ_{n}$ : $λ_{n} = \frac{2 π r _{n}}{n}$ . With $r_{n} = \frac{n ^{2} a _{0}}{Z}$ : $λ_{n} = \frac{2 π a _{0} n}{Z} \propto \frac{n}{Z}$ .

Ground state of hydrogen: $λ_{1} = 2 π \times 0.529 \overset{˚}{A} = 3.32 \overset{˚}{A}$ . Check: $v_{1} = 2.18 \times 1 0^{6} m s^{- 1}$ gives $\frac{h}{m v _{1}} = 3.34 \overset{˚}{A}$ .

Answer: $λ_{n} \propto \frac{n}{Z}$ ; $λ_{1} \approx 0.332 nm$ .

Solved Example 14

An electron is confined to a region of size

1.0 \times 1 0^{- 10} m

(an atom). Estimate the minimum uncertainty in its speed.

Solution:

$Δ p \geq \frac{ℏ}{2Δ x}$ $= \frac{6.63 \times 1 0 ^{- 34}}{4 π \times 1.0 \times 1 0 ^{- 10}}$ $= 5.3 \times 1 0^{- 25} kg m s^{- 1}$ .

$Δ v$ $= \frac{Δ p}{m}$ $= \frac{5.3 \times 1 0 ^{- 25}}{9.11 \times 1 0 ^{- 31}}$ .

Answer: $Δ v \approx 5.8 \times 1 0^{5} m s^{- 1}$ , comparable to the electron's speed in an atom, so an exact orbit cannot be traced.

Practice Questions

Find the de Broglie wavelength of (a) a $0.040 kg$ bullet at $1.0 km s^{- 1}$ (b) a $0.060 kg$ ball at $1.0 m s^{- 1}$ (c) a $1.0 \times 1 0^{- 9} kg$ dust particle at $2.2 m s^{- 1}$ .Answer: (a) $1.7 \times 1 0^{- 35} m$ (b) $1.1 \times 1 0^{- 32} m$ (c) $3.0 \times 1 0^{- 25} m$
Show that the de Broglie wavelength of a photon equals the wavelength of the light.Answer: $p = \frac{h ν}{c}$ , so $\frac{h}{p} = \frac{c}{ν} = λ$
Find the de Broglie wavelength of an alpha particle accelerated through $100 V$ .Answer: $0.0101 \overset{˚}{A} \approx 1.0 pm$
A proton and an electron have the same de Broglie wavelength. Which has more kinetic energy, and by what factor?Answer: the electron, $\frac{m _{p}}{m _{e}} \approx 1836$ times
Find the wavelength of a $150 eV$ neutron. Would it suit crystal diffraction?Answer: $2.34 \times 1 0^{- 12} m$ ; no, far smaller than atomic spacing (slow it to thermal energies first)
Estimate $λ$ of a helium atom at $27^{\circ} C$ (use $K = \frac{3}{2} k T$ ) and compare with the mean spacing of gas atoms at $1 atm$ (about $3.4 nm$ ).Answer: $0.073 nm$ , much smaller than the spacing, so gas atoms behave as separate particles
Estimate $λ$ of a free electron in a metal at $27^{\circ} C$ and compare with the electron spacing, about $2 \times 1 0^{- 10} m$ .Answer: $6.2 nm$ , much larger, so electron waves in a metal overlap strongly
To probe the $1 0^{- 15} m$ structure inside protons, roughly what electron energy is needed? (Use $E \approx p c$ .)Answer: $E \approx \frac{h c}{λ} \approx 1.2 \times 1 0^{9} eV$ , of the order of GeV

Common Mistakes to Avoid

Watch out

Using $λ = \frac{1.227}{V} nm$ for protons or alpha particles. It holds only for electrons; use $\frac{h}{2 m q V}$ otherwise.
Forgetting the charge of the alpha particle: $K = q V = 2 e V$ , not $e V$ .
Using $K = p c$ for an electron or $K = \frac{p ^{2}}{2 m}$ for a photon. Photon: $E = p c$ ; slow particle: $K = \frac{p ^{2}}{2 m}$ .
Thinking $λ$ depends on charge. It depends only on momentum; charge enters only through the energy $q V$ gained.
Putting $V$ in kilovolts or $K$ in keV into the shortcuts. $\frac{12.27}{V} \overset{˚}{A}$ needs volts; $\frac{0.286}{K} \overset{˚}{A}$ needs eV.
Using $k T$ instead of $\frac{3}{2} k T$ for a thermal particle (unless the question says so).
Saying heavy objects have no matter wave. They do, but $λ$ is far too small to detect.
Taking $θ = 5 0^{\circ}$ in Davisson-Germer as the glancing angle. It is the angle between the incident and scattered beams; the Bragg glancing angle is $6 5^{\circ}$ .

Frequently Asked Questions

What is the de Broglie wavelength?

It is the wavelength of the matter wave associated with a moving particle, $λ = \frac{h}{p} = \frac{h}{m v}$ . It links a wave property, wavelength, to a particle property, momentum, through Planck's constant. For light it gives the ordinary wavelength, since a photon has $p = \frac{h}{λ}$ .

Why do we not see the wave nature of everyday objects?

Their momentum is huge compared with $h$ , so the wavelength is tiny: about $1 0^{- 34} m$ for a moving ball, far smaller than a nucleus. Wave effects such as diffraction appear only when the wavelength is comparable to the obstacle, which is never the case for everyday objects.

What is the de Broglie wavelength of an electron accelerated through V volts?

$λ = \frac{h}{2 m e V} = \frac{1.227}{V} nm$ , or $\frac{12.27}{V} \overset{˚}{A}$ . For $100 V$ it is $0.123 nm$ . The formula is only for electrons and only for voltages well below about $100 kV$ , where relativity can be ignored.

With the same kinetic energy, which has the shortest wavelength: electron, proton or alpha particle?

The alpha particle. Since $λ = \frac{h}{2 m K}$ , at equal kinetic energy the wavelength is inversely proportional to the square root of the mass, and the alpha particle is the heaviest. The electron has the longest wavelength.

What did the Davisson-Germer experiment prove?

It showed that electrons are diffracted by a nickel crystal like waves. With $54 V$ electrons a strong scattered peak appeared at $5 0^{\circ}$ , giving a wavelength of $0.165 nm$ , matching de Broglie's value $0.167 nm$ . It was the first experimental proof of matter waves.

What is Heisenberg's uncertainty principle?

It states that the position and momentum of a particle cannot both be measured exactly at the same time: $Δ x Δ p \geq \frac{ℏ}{2}$ . A particle with an exact wavelength, and so exact momentum, is spread over all space; a localised wave packet has a spread of momenta.

Which wave nature of matter questions come in NEET?

NEET mostly asks direct formula and ratio questions: the wavelength of an electron through a given voltage, the ratio of wavelengths of a proton and an alpha particle, how the wavelength changes when kinetic energy or momentum changes, and graph questions such as wavelength against momentum. Davisson-Germer facts also appear.

How is the wave nature of matter tested in JEE Main and Advanced?

JEE Main asks ratios for different particles and voltages, thermal wavelengths, comparisons of an electron and a photon with the same wavelength, and links with Bohr orbits. JEE Advanced adds relativistic wavelengths, diffraction-grating calculations like Davisson-Germer, and combined problems with the photoelectric effect.

Previous year questions on Wave Nature of Matter

22 questions from past papers, each with a step-by-step solution.

Show all 22 questions

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