Dual Nature of Radiation & Matter Formula Sheet — NEET Physics

Every key Dual Nature of Radiation & Matter formula, definition and fact for NEET Physics in one place — with common examiner traps and worked examples. Free to read; blurt from memory, then check your gaps.

Syllabus — topics coveredNCERT · 2 sub-topics

Photoelectric effect, Einstein’s photoelectric equation
Particle nature of light, matter waves, de Broglie relation

Electron Emission & Work Function

Work function

ϕ_{0}

: The

minimum

energy a free electron needs to just escape a metal surface; in eV (

1 e V = 1.6 \times 1 0^{- 19}

J). Lower

ϕ_{0} \Rightarrow

electrons emitted more easily.

Energy to escape

E_{s u ppl i e d} \geq ϕ_{0} \Rightarrow emission; ϕ_{0} = h ν_{0} = \frac{h c}{λ _{0}}

ν_{0}, λ_{0} =

threshold frequency / wavelength of the metal

Four ways to supply the energy

▸ $Thermionic$ — heat the metal (CRT filament).
▸ $Field$ — strong external E-field ( $\sim 1 0^{8}$ V/m).
▸ $Photoelectric$ — light with $ν \geq ν_{0}$ .
▸ $Secondary$ — bombard with fast particles.

Potential-well picture: a free electron at the metal surface needs energy at least equal to the work function to reach the vacuum level and escape with kinetic energy — An electron escapes only if energy given $> ϕ_{0}$ .

Metal	$ϕ_{0}$ (eV)
Cs	2.14
Na	2.75
Ca	3.20
W	4.5
Cu	4.65
Ni	5.15
Pt	5.65

🚫 Examiner Trap · Work function

(1)

ϕ_{0}

is the

minimum

escape energy — given energy below it gives NO emission, however long you wait. (2) Alkali metals (Cs, Na, K) have the

lowest

ϕ_{0}

\Rightarrow

photosensitive to visible light (photocathodes). (3)

ϕ_{0} = h ν_{0}

links work function to threshold frequency. (4) Work in eV but convert to joules before using SI formulae.

Photoelectric Effect — Experiment & Laws

Photoelectric effect: Emission of electrons from a metal when light of

ν \geq ν_{0}

falls on it. Discovered by

Hertz

(1887); studied by Hallwachs & Lenard.

Max KE & stopping potential

K_{ma x} = \frac{1}{2} m v_{ma x}^{2} = e V_{0}

V_{0} =

stopping potential — reverse voltage that just halts the fastest electron

Experimental laws

▸ $Intensity$ : photocurrent $\propto$ intensity, but $K_{ma x}$ unchanged.
▸ $Frequency$ : $K_{ma x}$ (hence $V_{0}$ ) rises linearly with $ν$ ; independent of intensity.
▸ $Threshold$ : no emission below $ν_{0}$ , however intense.
▸ $Instantaneous$ : emission within $< 1 0^{- 9}$ s (no lag).

Left: photocurrent vs collector voltage for two intensities — same stopping potential, higher saturation current. Right: two frequencies at fixed intensity — same saturation current, larger stopping potential for higher frequency — Intensity $↑$ raises saturation current (same $V_{0}$ ); $ν ↑$ raises $V_{0}$ .

Comparative: vary intensity vs frequency

Increase	Saturation current	$V_{0}$ / $K_{ma x}$
Intensity (same $ν$ )	rises	unchanged
Frequency (same $I$ )	unchanged	rises

🚫 Examiner Trap · Photoelectric laws

(1)

K_{ma x}

depends on

frequency

, NOT intensity; intensity sets only the

number

of electrons (saturation current). (2) Below

ν_{0}

there is

zero

emission even at huge intensity. (3) Emission is

instantaneous

— no time lag. (4) Classical wave theory fails on all three counts — needs photons.

Einstein's Photoelectric Equation

Einstein's equation

h ν = ϕ_{0} + K_{ma x} \Rightarrow K_{ma x} = h ν - ϕ_{0} = h (ν - ν_{0})

one photon absorbed by one electron

Stopping potential

e V_{0} = h ν - ϕ_{0} \Rightarrow V_{0} = \frac{h}{e} ν - \frac{ϕ _{0}}{e}

a straight line

V_{0}

ν

, slope

h / e

Threshold

ν_{0} = \frac{ϕ _{0}}{h}, λ_{0} = \frac{h c}{ϕ _{0}} = \frac{1240 eV nm}{ϕ _{0} ( eV )}

✎ Example · Stopping potential

Light

λ = 400

nm on a metal of

ϕ_{0} = 2.0

eV. Find

V_{0}

$E = \frac{1240}{400} = 3.1$ eV
$K_{ma x} = E - ϕ_{0} = 3.1 - 2.0 = 1.1 e V = e V_{0}$

∴

V_{0} = 1.1

🚫 Examiner Trap · Einstein's equation

(1) Slope of

V_{0}

–

ν

h/e

, UNIVERSAL (Millikan) — only the intercept (

ν_{0}

) changes with metal. (2)

K_{ma x} = h ν - ϕ_{0}

— subtract the FULL work function, not

h ν_{0}

twice. (3) Use

h c = 1240

eV nm to convert

λ

(nm)

\to

energy(eV) fast. (4) One photon ejects at most one electron.

Photon — Particle Nature of Light

Photon: A quantum of EM radiation: a packet of energy

E = h ν

and momentum

p = h / λ

, moving at c, electrically

neutral

, with

zero rest mass

Photon energy

E = h ν = \frac{h c}{λ}; E (eV) = \frac{1240}{λ ( nm )}

Photon momentum

p = \frac{h ν}{c} = \frac{h}{λ} = \frac{E}{c}

momentum despite zero rest mass

Photon flux

n = \frac{P}{h ν} (photons/s); I = \frac{P}{A}

P =

source power,

A =

area

A photon depicted as a localized wave packet moving at speed c, labelled with energy E=hν=hc/λ and momentum p=hν/c=h/λ — Photon: energy & momentum in one neutral, massless quantum.

✎ Example · Photons per second

2

mW laser emits

λ = 500

nm. Photons/s?

$E = \frac{1240}{500} = 2.48 e V = 3.97 \times 1 0^{- 19}$ J
$n = \frac{P}{E} = \frac{2 \times 1 0 ^{- 3}}{3.97 \times 1 0 ^{- 19}}$

∴

n \approx 5.0 \times 1 0^{15}

photons/s

🚫 Examiner Trap · Photon

(1) Photon has momentum

p = E / c = h / λ

despite

zero rest mass

— don't use

p = m v

. (2) Raising intensity at fixed

ν

adds more photons, NOT more energy per photon. (3) Photon is electrically neutral; its energy/momentum are unaffected by E or B fields. (4) In photon–electron collisions both energy AND momentum are conserved.

de Broglie Matter Waves

Matter wave (de Broglie, 1924): Every moving particle has a wave of wavelength

λ = h / p

. Significant only for light particles (electrons); negligible for macroscopic bodies.

de Broglie wavelength

λ = \frac{h}{p} = \frac{h}{m v} = \frac{h}{2 m K} = \frac{h}{2 m q V}

K = q V

for charge

q

accelerated through

V

Accelerated electron

λ_{e} = \frac{1.227}{V} nm = \frac{12.27}{V} \overset{˚}{A}

V in volts;

V = 100 V \Rightarrow λ = 0.123

Left: λ versus momentum showing inverse relation λ=h/p. Right: a sinusoidal matter wave with wavelength marked, with the accelerated-electron formula λ=1.227/√V nm — $λ \propto 1/ p$ — heavier/faster particles have shorter waves.

✎ Example · Shortest wavelength

Electron, proton,

α

with the SAME KE — shortest

λ

$λ = \frac{h}{2 m K} \Rightarrow λ \propto 1/ m$
heaviest $\Rightarrow$ shortest; $m_{α} > m_{p} > m_{e}$

∴

α

-particle has the shortest

λ

🚫 Examiner Trap · de Broglie waves

(1) Same

KE

λ \propto 1/ m

(heavier

\to

shorter). Same

momentum

: equal

λ

. Same

V

λ \propto 1/ m q

. (2) Use

λ = 1.227/ V

nm ONLY for electrons (in volts). (3) Macroscopic bodies have unmeasurably tiny

λ

. (4) Thermal:

λ = h / 3 mk T

Davisson–Germer Experiment

First direct proof of electron waves (Davisson & Germer, 1927): an accelerated electron beam strikes a

nickel crystal

; scattered intensity is read by a movable detector at angle

θ

Observation

▸Sharp $diffraction peak$ at $V = 54$ V, $θ = 5 0^{\circ}$ .
▸Measured wavelength $λ = 0.165$ nm.

Agreement with de Broglie

λ = \frac{1.227}{54} = 0.167 nm \approx 0.165 nm (measured)

theory matches experiment

\Rightarrow

confirms

λ = h / p

Davisson–Germer apparatus: electron gun directs an incident beam onto a nickel crystal; the diffracted beam at angle θ is detected by a movable collector on a circular scale; peak at 54 V, 50° — Diffraction peak at $54$ V, $5 0^{\circ}$ confirms electron waves.

Further confirmations

▸ $G. P. Thomson$ — electron diffraction through foils (1937 Nobel).
▸Electron microscope exploits matter-wave optics (high resolution).
▸Later: electron & molecule double-slit interference.

🚫 Examiner Trap · Davisson–Germer

(1) The peak appears at a

specific

V and

θ

(54 V, 50

^{\circ}

) — constructive interference of electron waves off the Ni lattice. (2) The measured

λ

matches

1.227/ V

— this is the

verification

of de Broglie. (3) It proves

electrons

(matter) diffract, the matter-wave analogue of light diffraction.

Wave–Particle Duality & Uncertainty

Wave–particle duality: Radiation and matter both show wave behaviour (interference, diffraction)

and

particle behaviour (photoelectric, collisions). Which shows up depends on the experiment — never both at once.

Heisenberg uncertainty

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

position and momentum cannot both be exact

Energy–time form

Δ E Δ t \geq \frac{ℏ}{2}

Top: a localized wave packet — small position spread but large momentum spread. Bottom: an extended wave of definite wavelength — momentum spread zero but position spread infinite — Localized packet vs definite- $λ$ wave — the $Δ x Δ p$ trade-off.

Quick constants

Constant	Value
$h$	$6.63 \times 1 0^{- 34}$ J s
$ℏ$	$1.05 \times 1 0^{- 34}$ J s
$h c$	$1240$ eV nm
$m_{e}$	$9.11 \times 1 0^{- 31}$ kg
$1$ eV	$1.6 \times 1 0^{- 19}$ J

🚫 Examiner Trap · Duality & uncertainty

(1) Use the

wave

picture for propagation/interference; the

photon

picture for emission/absorption/collisions — never both in one measurement. (2) A definite momentum (

Δ p = 0

) means a wave spread over

all space

(

Δ x \to \infty

). (3) Uncertainty is fundamental, not a measurement flaw. (4) Use

ℏ = h /2 π

, not h, in

Δ x Δ p \geq ℏ/2

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