Dual Character Of Matter And Radiation

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Modern quantum mechanics rests on three revolutionary ideas: de Broglie's hypothesis that matter (like light) has a wave nature; Heisenberg's uncertainty principle that you cannot simultaneously know an electron's exact position and momentum; and the Schrödinger equation whose solutions give us orbitals - probability clouds instead of fixed Bohr orbits. The photoelectric effect completes the picture by showing light itself has particle character (photons). Together these ideas explain why atoms behave the way they do at the smallest scale.

Key Formulas - Quick Reference

de Broglie: $λ = \frac{h}{p} = \frac{h}{m v}$
For KE $E$ : $λ = \frac{h}{2 m E}$
Electron accelerated through V volts: $λ = \frac{12.27}{V}$ Å
Heisenberg: $Δ x \cdot Δ p \geq \frac{h}{4 π}$ ; equivalently $Δ x \cdot Δ v \geq \frac{h}{4 π m}$
Photoelectric: $h ν = h ν_{0} + \frac{1}{2} m v^{2}$ (Einstein's equation)
Threshold frequency: $ν_{0} = ϕ / h$ ; threshold wavelength $λ_{0} = h c / ϕ$
Stopping potential: $e V_{s} = \frac{1}{2} m v_{ma x}^{2}$
Radial nodes = $n - ℓ - 1$ ; Angular nodes = $ℓ$ ; Total nodes = $n - 1$

1. de Broglie's Hypothesis (1924)

All matter has a wave nature. A particle of momentum

p

has an associated wavelength

λ = h / p = h / (m v)

de Broglie proposed that just as light shows both wave (interference, diffraction) and particle (photons in photoelectric effect) character, moving material particles should also show wave properties. The wavelength is inversely proportional to momentum - so heavy or fast objects have unmeasurably tiny wavelengths, but electrons show observable diffraction (Davisson-Germer, 1927).

Figure 1: de Broglie interpretation - an electron in the nth Bohr orbit forms a standing wave with

nλ = 2 π r

Connection to Bohr's model: If the electron behaves as a standing wave around the orbit, then

nλ = 2 π r

. Combining with

λ = h / (m v)

gives

m v r = nh /2 π

- Bohr's quantisation condition, now derived rather than assumed.

Solved Example 1

e^{-}

, a proton and an

α

-particle have KE of

16 E

4 E

, and

E

respectively. What is the qualitative order of their de Broglie wavelengths?

Solution:

$λ = h / 2 m E$ , so $λ \propto 1/ m E$ .

Compute $m E$ for each (mass in units of $m_{e}$ , energy in units of $E$ ):

Electron: $m_{e} \times 16 E = 16 m_{e} E$
Proton: $1836 m_{e} \times 4 E = 7344 m_{e} E$
$α$ : $4 \times 1836 m_{e} \times E = 7344 m_{e} E$

So $λ_{e} > λ_{p} = λ_{α}$ . Answer: $λ_{e^{-}} > λ_{p} = λ_{α}$ .

Solved Example 2

Calculate the de Broglie wavelength of a ball of mass 0.1 kg moving at 60 m/s.

Solution:

$λ = h / (m v) = (6.626 \times 1 0^{- 34}) / (0.1 \times 60) = 1.10 \times 1 0^{- 34} m$ .

This is unmeasurably small - which is why we don't see wave behaviour for everyday objects.

Solved Example 3

Calculate the number of waves made by an electron in the 4th Bohr orbit.

Solution:

By de Broglie standing-wave condition: $nλ = 2 π r_{n}$ . The number of waves equals $n$ . So for the 4th orbit, the electron makes 4 complete waves.

Solved Example 4

What is the de Broglie wavelength of an electron with KE = 5 eV?

Solution:

$K E = 5 \times 1.6 \times 1 0^{- 19} = 8 \times 1 0^{- 19}$ J
$p = 2 m E = 2 (9.11 \times 1 0^{- 31}) (8 \times 1 0^{- 19}) = 1.21 \times 1 0^{- 24}$ kg m/s
$λ = h / p = (6.626 \times 1 0^{- 34}) / (1.21 \times 1 0^{- 24}) \approx 5.48 \overset{˚}{A}$ .

Or use shortcut: $λ = 12.27/ V$ Å $= 12.27/ 5 \approx 5.49$ Å.

Solved Example 5

Through what potential difference must an electron be accelerated to have a de Broglie wavelength of 1 Å?

Solution:

Use $λ = 12.27/ V$ Å with $λ = 1$ Å:
$V = 12.27$ , so $V \approx 150.5 volts$ .

2. Heisenberg's Uncertainty Principle (1927)

It is impossible to determine simultaneously the exact position and exact momentum of a small particle.

Δ x \cdot Δ p \geq \frac{h}{4 π}

Figure 2: Heisenberg's uncertainty principle - narrowing

Δ x

widens

Δ p

, and vice versa.

Δ x \cdot Δ p \geq h /4 π

The uncertainty is fundamental - not due to measurement error. It arises because at atomic scales, the act of measuring changes the system. For massive objects, $h /4 π \sim 1 0^{- 35}$ J s is negligible; for electrons it dominates.

Consequence for Bohr's model: Bohr assumes definite orbits (fixed

r

) and definite velocities (

v

). This violates the uncertainty principle - which is why quantum mechanics replaces "orbits" with "orbitals" (probability distributions).

Solved Example 6

Why can an electron not exist inside the nucleus according to Heisenberg's uncertainty principle?

Solution:

Nucleus radius $Δ x \sim 1 0^{- 15}$ m. Then minimum uncertainty in momentum:
$Δ p \geq \frac{h}{4 π Δ x} = \frac{6.626 \times 1 0 ^{- 34}}{4 π \times 1 0 ^{- 15}} \approx 5.27 \times 1 0^{- 20}$ kg m/s.

The corresponding electron KE $\approx p^{2} /2 m \approx 1.53 \times 1 0^{- 9}$ J $\approx 9.6$ GeV. This is enormously higher than typical nuclear binding energies ( $\sim$ MeV) - the electron would immediately escape. Hence electrons cannot exist inside the nucleus.

Solved Example 7

Uncertainty in an electron's velocity is 0.0058 m/s. Find the uncertainty in its position.

Solution:

$Δ x \geq \frac{h}{4 π m Δ v} = \frac{6.626 \times 1 0 ^{- 34}}{4 π ( 9.11 \times 1 0 ^{- 31} ) ( 0.0058 )} \approx 1.0 \times 1 0^{- 2} m$ = 1 cm.

Even a tiny velocity uncertainty gives large position uncertainty for an electron.

Solved Example 8

A 10 g ball moves at 100 m/s with 0.002% velocity uncertainty. Find

Δ x

Solution:

$Δ v = 100 \times 2 \times 1 0^{- 5} = 2 \times 1 0^{- 3}$ m/s
$Δ x = \frac{6.626 \times 1 0 ^{- 34}}{4 π ( 0.01 ) ( 2 \times 1 0 ^{- 3} )} \approx 2.6 \times 1 0^{- 30} m$ .

Completely undetectable - which is why classical mechanics works for everyday objects.

3. Quantum Mechanical Model of the Atom

Schrödinger (1926) wrote a wave equation whose solutions ( $ψ$ ) describe electrons. These solutions are the atomic orbitals.

The wave function $ψ$ : Solution to the Schrödinger equation. Has no direct physical meaning itself.
$∣ ψ ∣^{2}$ (probability density): The probability of finding the electron per unit volume at a given point.

∣ ψ ∣^{2} d V

= probability in volume element

d V

Orbitals vs orbits

Bohr orbit	Quantum-mechanical orbital
Fixed circular path	3D region of high probability
Exact position + velocity	Only probability - no definite path
Classical (violates uncertainty)	Consistent with uncertainty principle
2D concept	Truly 3D (shape depends on $ℓ, m$ )

Nodes in orbitals

A node is a surface (or point) where $ψ = 0$ - so the probability of finding the electron there is zero.

Radial nodes =

n - ℓ - 1

(spherical shells of zero density)
Angular nodes =

ℓ

(nodal planes/cones through nucleus)
Total nodes =

n - 1

Figure 3: Nodes in orbitals - Radial nodes =

n - ℓ - 1

; Angular nodes =

ℓ

; Total nodes =

n - 1

Solved Example 9

Calculate the radial and angular nodes for the following: 3s, 3p, 3d, 4f.

Solution:

Orbital	$n$	$ℓ$	Radial ( $n - ℓ - 1$ )	Angular ( $ℓ$ )	Total ( $n - 1$ )
3s	3	0	2	0	2
3p	3	1	1	1	2
3d	3	2	0	2	2
4f	4	3	0	3	3

4. Photoelectric Effect

Discovered by Hertz (1887), and explained by Einstein (1905, Nobel Prize 1921).

Photoelectric effect: When light of sufficient frequency strikes a metal surface, electrons (photoelectrons) are ejected instantly.

Figure 4: Photoelectric effect - incident photons of frequency

ν > ν_{0}

eject photoelectrons.

h ν = ϕ + \frac{1}{2} m v^{2}

Experimental findings

Electrons are emitted instantaneously upon illumination (no lag)
Emission requires a minimum threshold frequency $ν_{0}$ . Below $ν_{0}$ , no ejection - no matter how bright the light
Number of photoelectrons increases with light intensity (photocurrent $\propto$ intensity)
KE of ejected electrons increases with frequency (not intensity!)

Einstein's photoelectric equation

Each photon carries energy $h ν$ . When it strikes the metal, part of the energy overcomes the binding (work function $ϕ = h ν_{0}$ ), and the rest becomes KE of the electron:

h ν = ϕ + \frac{1}{2} m v^{2}

\frac{1}{2} m v^{2} = h ν - h ν_{0} = h (ν - ν_{0})

Work function ( $ϕ$ ): Minimum energy needed to eject an electron. Different for each metal (Na: 2.5 eV, Cs: 2.14 eV, Cu: 4.7 eV, ...).

Stopping potential

Apply a reverse voltage to stop the fastest photoelectrons. When photocurrent just vanishes, that voltage is the stopping potential $V_{s}$ :

e V_{s} = \frac{1}{2} m v_{ma x}^{2} = h ν - h ν_{0}

A plot of $V_{s}$ vs $ν$ is a straight line with slope $h / e$ - a classic way to measure Planck's constant.

Solved Example 10

Work function of sodium is 2.5 eV. Predict whether wavelength 6500 Å is suitable for photoelectron ejection.

Solution:

Energy of incident photon: $E = h c / λ = (6.62 \times 1 0^{- 34}) (3 \times 1 0^{8}) / (6500 \times 1 0^{- 10})$
$= 3.055 \times 1 0^{- 19}$ J $= 3.055 \times 1 0^{- 19} / (1.6 \times 1 0^{- 19}) \approx 1.9$ eV.

This is less than 2.5 eV (the work function). Therefore no photoelectron ejection will occur.

Common Mistakes to Avoid

Watch out

Intensity doesn't overcome threshold. Increasing brightness of low-frequency light does NOT eject photoelectrons - each individual photon must have $h ν \geq ϕ$ .
de Broglie $λ = h / p$ , not $h / (m v^{2})$ . Always use momentum, not KE, in the denominator. If you have KE, use $p = 2 m E$ .
Uncertainty $Δ x \cdot Δ p \geq h /4 π$ - not $h /2 π$ . The $4 π$ is easy to misremember.
Nodes = surfaces, not points. Radial nodes are spherical shells; angular nodes are planes or cones passing through the nucleus.
$ψ$ can be positive or negative; $∣ ψ ∣^{2}$ is always non-negative. Signs of $ψ$ matter for bonding (constructive/destructive overlap), not for probability.
KE of photoelectron does not depend on intensity. More intense light $\Rightarrow$ more electrons (higher current), but each individual electron has the same maximum KE for a given frequency.

Frequently Asked Questions

What is de Broglie's hypothesis?

Louis de Broglie proposed in 1924 that all matter has a wave nature. Every moving particle has an associated wavelength $λ = h / p = h / (m v)$ , where $h$ is Planck's constant, $m$ is mass, and $v$ is velocity. For large objects the wavelength is unmeasurably small; for electrons it's comparable to atomic dimensions, producing observable wave effects.

What is Heisenberg's uncertainty principle?

It is impossible to simultaneously determine both the exact position and exact momentum of a small particle. Mathematically, $Δ x \cdot Δ p \geq h / (4 π)$ . This isn't due to poor measurement - it's a fundamental property of quantum systems. It's the reason we describe electrons with probability clouds (orbitals) rather than fixed orbits.

What is the difference between an orbit and an orbital?

An orbit (Bohr's model) is a fixed 2D circular path with definite radius and velocity. An orbital (quantum model) is a 3D region of space where the probability of finding the electron is high. Orbits violate uncertainty; orbitals are consistent with it. Orbitals have shapes (s = sphere, p = dumbbell, d = clover) depending on quantum numbers.

How do you calculate the number of nodes in an orbital?

For an orbital with quantum numbers $n$ and $ℓ$ : Radial nodes = $n - ℓ - 1$ ; Angular nodes (nodal planes) = $ℓ$ ; Total nodes = $n - 1$ . Example: 3p has $n = 3, ℓ = 1$ , so 1 radial + 1 angular = 2 total nodes.

What is the photoelectric effect?

The photoelectric effect is the ejection of electrons from a metal surface when light of sufficient frequency strikes it. Einstein's equation is $h ν = ϕ + \frac{1}{2} m v^{2}$ , where $ϕ$ is the work function of the metal. Key points: emission is instantaneous, needs $ν \geq ν_{0} = ϕ / h$ , and KE depends on frequency (not intensity).

What is the work function of a metal?

The work function ( $ϕ$ ) is the minimum energy required to eject an electron from a metal surface. It's a property of the metal. Related to the threshold frequency by $ϕ = h ν_{0}$ . Typical values: Cs 2.14 eV, Na 2.5 eV, Al 4.08 eV, Cu 4.7 eV. Lower $ϕ$ = easier to eject electrons.

Why can't an electron exist inside the nucleus?

By uncertainty: confining an electron to $Δ x \sim 1 0^{- 15}$ m (nucleus size) gives $Δ p \geq h / (4 π Δ x)$ , implying kinetic energy of billions of eV. This is vastly greater than any nuclear binding energy (~MeV), so an electron confined to the nucleus would immediately escape. So electrons must exist outside the nucleus.

What is the difference between $ψ$ and $∣ ψ ∣^{2}$ ?

$ψ$ (the wave function) is a solution to Schrödinger's equation. It can be positive, negative, or complex. It has no direct physical meaning by itself. $∣ ψ ∣^{2}$ is the probability density - the probability per unit volume of finding the electron at a given point. It's always non-negative and physically observable.

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1. de Broglie's Hypothesis (1924)

2. Heisenberg's Uncertainty Principle (1927)

3. Quantum Mechanical Model of the Atom

Orbitals vs orbits

Nodes in orbitals

4. Photoelectric Effect

Experimental findings

Einstein's photoelectric equation

Stopping potential

Common Mistakes to Avoid

Frequently Asked Questions

What is de Broglie's hypothesis?

What is Heisenberg's uncertainty principle?

What is the difference between an orbit and an orbital?

How do you calculate the number of nodes in an orbital?

What is the photoelectric effect?

What is the work function of a metal?

Why can't an electron exist inside the nucleus?

What is the difference between ψ and ∣ψ∣2?

Previous year questions on Dual Character Of Matter And Radiation

Ready to master Atomic Structure?

What is the difference between $ψ$ and $∣ ψ ∣^{2}$ ?