PhysicsCBSE Class 12 · Chapter 11

Dual Nature of Radiation and Matter Class 12: Notes and Solved Numericals

Updated 14 October 2026 · Fact-checked

Dual Nature of Radiation and Matter is Chapter 11 of CBSE Class 12 Physics. It covers electron emission, the photoelectric effect with Hertz's and Lenard's observations and its experimental laws, why the wave theory fails, Einstein's photoelectric equation and the photon picture of light, and matter waves with the de Broglie relation $\lambda = \frac{h}{p}$.

What Dual Nature of Radiation and Matter is about

By 1900 light was firmly established as an electromagnetic wave. Yet when light falls on a metal and knocks out electrons (the photoelectric effect), the results make no sense for a wave: the energy of the electrons depends on the light's frequency, not its intensity; below a threshold frequency no electrons come out at all, however bright the light; and emission starts instantly. Einstein explained this in 1905 by treating light as a stream of photons, each of energy $h ν$ . One photon gives all its energy to one electron, which spends at least the work function $ϕ_{0}$ to escape; hence $K_{ma x} = h ν - ϕ_{0}$ .

If light, a wave, can behave like particles, de Broglie asked whether particles can behave like waves. His answer, $λ = \frac{h}{p}$ , gives every moving particle a wavelength, too small to notice for everyday objects but comparable to atomic spacings for electrons. So both radiation and matter show a dual nature.

The Davisson-Germer experiment, Heisenberg's uncertainty principle and the derivation of the wavelength of an accelerated electron are not in the 2026-27 syllabus. In our question bank this chapter appears in all eight board papers from 2023 to 2026, mainly as 1-2 mark questions on the photoelectric graphs, Einstein's equation and de Broglie wavelength, plus 4-mark case studies.

Weightage: This chapter is part of Units VII and VIII, Dual Nature of Radiation and Matter; Atoms and Nuclei, which carry 12 of the 70 theory marks together with Atoms and Nuclei.Source: CBSE Physics syllabus 2026-27, Units VII and VIII Dual Nature of Radiation and Matter; Atoms and Nuclei

In real board papers: asked in 8 of the 8 board papers we hold (2023 to 2026), on average 4.9 marks per paper. See the questions

Key concepts

Electron emission and work function

NCERT section 11.2

Free electrons in a metal are held inside it by the attraction of the positive ions. The minimum energy needed to pull an electron out of the surface is the work function $ϕ_{0}$ ; it depends on the metal and on the state of its surface, and is usually given in electron volts: $1 e V = 1.6 \times 1 0^{- 19} J$ (the energy an electron gains through a potential difference of 1 V).

Electrons can be given this energy in three ways:

Thermionic emission: heating the metal.
Field emission: a very strong electric field ( $1 0^{8} V m^{- 1}$ order) pulls electrons out.
Photoelectric emission: light of suitable frequency falls on the metal.

Worked example

The work function of a metal is 2.5 eV. Express it in joules, and find the longest wavelength of light that can eject electrons from it. ( $h c \approx 1240 e V nm$ ) [2 marks]

Try it first, then show the solutionSolution

$ϕ_{0} = 2.5 \times 1.6 \times 1 0^{- 19} = 4 \times 1 0^{- 19} J$
Threshold: $h ν_{0} = ϕ_{0}$ , so $λ_{0} = \frac{h c}{ϕ _{0}} = \frac{1240 e V nm}{2.5 e V}$
$λ_{0} = 496 nm$

Answer: $4 \times 1 0^{- 19} J$ ; about 496 nm

Photoelectric effect: Hertz's and Lenard's observations

NCERT section 11.3

Hertz (1887): while studying electromagnetic waves, he noticed that sparks jumped more easily across a gap when the electrodes were lit by ultraviolet light: the light was helping charged particles leave the metal.

Hallwachs and Lenard (1886 to 1902): in an evacuated tube with two electrodes, ultraviolet light on the emitter plate made a current flow in the circuit, and the current stopped when the light was cut off. Lenard concluded that the light ejected negative particles (electrons) from the emitter. Hallwachs found that a negatively charged zinc plate lost its charge under UV, a neutral plate became positive, and a positive plate became more positive.

They also found that for each metal there is a threshold frequency: below it no emission happens, however intense the light. Alkali metals (lithium, sodium, potassium, caesium) respond even to visible light, while metals such as zinc, cadmium and magnesium need ultraviolet. The emitted electrons are called photoelectrons and the current a photocurrent.

Worked example

A neutral zinc plate is lit with ultraviolet light. What charge does it acquire and why? What happens if it was negatively charged to begin with? [2 marks]

Try it first, then show the solutionSolution

UV light ejects electrons from the zinc, so the neutral plate loses negative charge and becomes positively charged.
A negatively charged plate loses its extra electrons first, so its negative charge decreases (it discharges).

Answer: It becomes positive; a negative plate gradually loses its negative charge.

Experimental study of the photoelectric effect

NCERT section 11.4

Light falls on an emitter C in an evacuated tube; electrons are collected by A, whose potential relative to C can be varied.

1. Effect of intensity: for a given frequency (above threshold) and fixed accelerating potential, the photocurrent is directly proportional to the intensity: more photons, more electrons.
2. Effect of potential: as A is made more positive the current rises to a saturation value (all emitted electrons are collected). Making A negative reduces the current; at a certain negative potential $- V_{0}$ , the stopping (cut-off) potential, even the fastest electrons are stopped and the current becomes zero: $e V_{0} = K_{ma x}$ . For a given frequency, $V_{0}$ is the same for all intensities: the maximum kinetic energy does not depend on intensity.
3. Effect of frequency: for the same intensity, a higher frequency gives a larger stopping potential (same saturation current). $V_{0}$ varies linearly with ν, and the line meets the ν axis at the threshold frequency $ν_{0}$ , below which there is no emission.
4. Emission is instantaneous (within about $1 0^{- 9} s$ ), even for very faint light.

Worked example

Sketch and explain how the photocurrent varies with collector potential for (a) two intensities $I_{1} > I_{2}$ at the same frequency, (b) two frequencies $ν_{1} > ν_{2}$ at the same intensity. [3 marks]

Try it first, then show the solutionSolution

(a) Both curves start from the same stopping potential $- V_{0}$ on the negative axis and rise to saturation; the $I_{1}$ curve saturates higher. Intensity changes the number of electrons, not their maximum energy.
(b) Both curves reach the same saturation current, but the $ν_{1}$ curve starts at a more negative stopping potential ( $V_{01} > V_{02}$ ). Higher frequency gives more energetic electrons.
In both, the current is non-zero at zero potential (electrons leave with some kinetic energy).

Answer: (a) Same $V_{0}$ , higher saturation for higher intensity. (b) Same saturation, larger stopping potential for higher frequency.

Why the wave theory fails

NCERT section 11.5

In the wave picture, light energy is spread continuously over the wavefront and absorbed gradually. That predicts:

the maximum kinetic energy of electrons should increase with intensity: observed, it does not;
emission should happen at any frequency if the light is intense enough: observed, there is a threshold frequency;
with faint light, an electron should need a long time (hours) to gather enough energy: observed, emission is instantaneous.

All three predictions fail. The photon picture explains each one.

Worked example

Explain why very bright red light may eject no electrons from a metal while faint violet light ejects some. [2 marks]

Try it first, then show the solutionSolution

Each electron absorbs one photon. A red photon has energy hν that may be less than the work function, so no electron can escape however many red photons arrive.
A violet photon has higher frequency and energy; if $h ν > ϕ_{0}$ , each absorbed photon can free an electron, even if only a few photons arrive.

Answer: Emission depends on the energy of each photon (frequency), not on how many arrive (intensity).

Einstein's photoelectric equation

NCERT section 11.6

Einstein proposed that light energy comes in quanta (photons) of energy $h ν$ , where $h = 6.63 \times 1 0^{- 34} J s$ is Planck's constant. An electron absorbs one whole photon; part of the energy ( $ϕ_{0}$ ) is used to escape, and the rest is its kinetic energy. The most energetic electrons (from the surface) have

$K_{ma x} = h ν - ϕ_{0}$ , and since $K_{ma x} = e V_{0}$ : $V_{0} = \frac{h}{e} ν - \frac{ϕ _{0}}{e}$

This explains every observation:

$K_{ma x}$ depends linearly on ν and not on intensity.
Emission needs $h ν \geq ϕ_{0}$ , so there is a threshold $ν_{0} = \frac{ϕ _{0}}{h}$ .
Intensity sets the number of photons, so it sets the current.
Absorption of one photon is a single event, so emission is instant.

The graph of $V_{0}$ against ν is a straight line with slope $\frac{h}{e}$ (the same for every metal) and ν-intercept $ν_{0}$ . Millikan's measurements of this slope gave a value of h agreeing with Planck's.

Worked example

Light of wavelength 400 nm falls on a metal of work function 2.0 eV. Find (a) the energy of each photon, (b) the maximum kinetic energy of the photoelectrons, (c) the stopping potential, (d) the threshold wavelength. ( $h c \approx 1240 e V nm$ ) [3 marks]

Try it first, then show the solutionSolution

(a) $E = \frac{1240}{400} = 3.1 e V$
(b) $K_{ma x} = 3.1 - 2.0 = 1.1 e V$
(c) $V_{0} = \frac{K _{ma x}}{e} = 1.1 V$
(d) $λ_{0} = \frac{1240}{2.0} = 620 nm$

Answer: (a) 3.1 eV (b) 1.1 eV (c) 1.1 V (d) 620 nm

The photon: particle nature of light

NCERT section 11.7

In its interaction with matter, light behaves as if made of particles called photons:

Each photon of frequency ν has energy $E = h ν = \frac{h c}{λ}$ and momentum $p = \frac{h ν}{c} = \frac{h}{λ}$ .
All photons of a given frequency have the same energy and momentum, whatever the intensity. More intense light of the same frequency simply means more photons per second through a given area.
Photons travel at c in vacuum, are electrically neutral (not deflected by fields) and have zero rest mass.
In a photon-particle collision, total energy and momentum are conserved, but photons can be absorbed or created, so their number may change.

A source of power P emitting light of frequency ν sends out $n = \frac{P}{h ν}$ photons per second.

Worked example

A laser of power 10 mW emits light of wavelength 600 nm. Find the energy and momentum of each photon and the number of photons emitted per second. ( $h = 6.63 \times 1 0^{- 34} J s$ ) [3 marks]

Try it first, then show the solutionSolution

$E = \frac{h c}{λ} = \frac{6.63 \times 1 0 ^{- 34} \times 3 \times 1 0 ^{8}}{600 \times 1 0 ^{- 9}} = 3.315 \times 1 0^{- 19} J$
$p = \frac{h}{λ} = \frac{6.63 \times 1 0 ^{- 34}}{600 \times 1 0 ^{- 9}} \approx 1.1 \times 1 0^{- 27} k g m s^{- 1}$
$n = \frac{P}{E} = \frac{10 \times 1 0 ^{- 3}}{3.315 \times 1 0 ^{- 19}} \approx 3.0 \times 1 0^{16}$ photons per second

Answer: $3.3 \times 1 0^{- 19} J$ ; $1.1 \times 1 0^{- 27} k g m s^{- 1}$ ; about $3.0 \times 1 0^{16}$ per second

Wave nature of matter: the de Broglie relation

NCERT section 11.8

De Broglie suggested that moving particles have a wave character, with wavelength

$λ = \frac{h}{p} = \frac{h}{m v}$

The same relation holds for photons. Since $K = \frac{p ^{2}}{2 m}$ , it can also be written $λ = \frac{h}{2 m K}$ .

For everyday objects (large m), λ is far too small to detect, so they show no wave behaviour.
For electrons, λ is comparable to the spacing of atoms in crystals, so electron diffraction can be observed (the basis of the electron microscope).
For an electron accelerated from rest through a potential difference V, $K = e V$ , giving $λ = \frac{h}{2 m e V} \approx \frac{1.227}{V} nm$ . (Use this as a formula; its derivation is not in the 2026-27 syllabus.)
For particles with the same kinetic energy, λ ∝ $\frac{1}{m}$ : the lighter particle has the longer wavelength. With the same momentum, they have the same λ.

Worked example

Find the de Broglie wavelength of (a) an electron accelerated through 100 V, (b) a 60 g ball moving at 20 m s⁻¹. Why can we observe the wave nature of one but not the other? [2 marks]

Try it first, then show the solutionSolution

(a) $λ = \frac{1.227}{100} nm = 0.123 nm$
(b) $λ = \frac{h}{m v} = \frac{6.63 \times 1 0 ^{- 34}}{0.06 \times 20} \approx 5.5 \times 1 0^{- 34} m$
0.123 nm is comparable to atomic spacings, so electron waves can be diffracted by crystals; $1 0^{- 34} m$ is far smaller than anything, so the ball shows no wave effects.

Answer: (a) about 0.123 nm (b) about $5.5 \times 1 0^{- 34} m$ ; only the electron's wavelength is large enough to detect.

Physics in real life: Dual Nature of Radiation and Matter

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Photon energy and threshold

Suppose a light-operated door switch uses a cathode coated with a metal of work function 1.9 eV, and the designer must choose between a red lamp (700 nm) and a green lamp (530 nm).

Which lamp will work? Find the maximum kinetic energy of the photoelectrons for it.

Try it first, then show the solutionSolution

Red: $E = \frac{1240}{700} \approx 1.77 e V < 1.9 e V$ : no emission
Green: $E = \frac{1240}{530} \approx 2.34 e V > 1.9 e V$ : emission occurs
$K_{ma x} = 2.34 - 1.9 = 0.44 e V$

Answer: Only the green lamp works, giving electrons of up to about 0.44 eV.

Memory hook: Colour decides if it works; brightness decides how much.

Number of photons

Imagine a small 1 W bulb that emits all its energy as yellow light of wavelength 600 nm.

Suppose all 1 W goes into 600 nm light. How many photons does it emit per second?

Try it first, then show the solutionSolution

$E = \frac{h c}{λ} = 3.315 \times 1 0^{- 19} J$ per photon
$n = \frac{1}{3.315 \times 1 0 ^{- 19}} \approx 3.0 \times 1 0^{18}$ per second
Such huge numbers are why light looks continuous to us.

Answer: About $3 \times 1 0^{18}$ photons every second.

Memory hook: Light is grainy, but the grains are unimaginably many.

de Broglie wavelength

Suppose an electron microscope accelerates electrons through 10 000 V to examine fine structures.

Find the de Broglie wavelength of the electrons and compare it with the wavelength of visible light (about 500 nm).

Try it first, then show the solutionSolution

$λ \approx \frac{1.227}{10 000} nm = \frac{1.227}{100} nm \approx 0.0123 nm$
Ratio: $\frac{500}{0.0123} \approx 4 \times 1 0^{4}$
Shorter wavelength means finer details can be resolved, which is why electron microscopes see much smaller things than light microscopes.

Answer: About 0.012 nm, roughly 40 000 times shorter than visible light.

Memory hook: Faster electrons, shorter waves, sharper pictures.

Topics in this chapter

Dual Nature of Radiation and Matter formula sheet

Electron volt: $1 e V = 1.6 \times 1 0^{- 19} J$
Photon energy and momentum: $E = h ν = \frac{h c}{λ}, p = \frac{h}{λ}$; $h c \approx 1240 e V nm$ ; $h = 6.63 \times 1 0^{- 34} J s$ .
Number of photons per second: $n = \frac{P}{h ν}$
Einstein's photoelectric equation: $K_{ma x} = h ν - ϕ_{0} = e V_{0}$
Threshold frequency and wavelength: $ν_{0} = \frac{ϕ _{0}}{h}, λ_{0} = \frac{h c}{ϕ _{0}}$
Stopping potential against frequency: $V_{0} = \frac{h}{e} ν - \frac{ϕ _{0}}{e}$; Slope h/e is the same for all metals.
Maximum speed of photoelectrons: $\frac{1}{2} m v_{ma x}^{2} = e V_{0}$
de Broglie wavelength: $λ = \frac{h}{p} = \frac{h}{m v} = \frac{h}{2 m K}$
Electron accelerated through V (formula only): $λ = \frac{h}{2 m e V} \approx \frac{1.227}{V} nm$

Dual Nature of Radiation and Matter: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

Experimental set-up for the photoelectric effect

Describing the experimental study (3-5 marks).

Draw

Draw an evacuated glass tube with a quartz window.
Inside, draw a photosensitive plate C (emitter) and a collector plate A.
Show light falling through the window onto C.
Connect A and C through a microammeter, a battery and a potential divider (with a commutator) so that A's potential can be made positive or negative; add a voltmeter across A and C.

Label

Evacuated tube
Quartz window
Emitter C
Collector A
Incident light
Microammeter
Battery with potential divider
Voltmeter

Watch out: Quartz is used because it lets ultraviolet light through; say so if asked why.

Photoelectric graphs

Almost every 2-3 mark question on this chapter.

Draw

Current against collector potential: curves rising from a common point $- V_{0}$ on the negative axis to flat saturation levels (different heights for different intensities).
Same axes for two frequencies: curves starting at $- V_{01}$ and $- V_{02}$ and reaching the same saturation level.
Stopping potential against frequency: straight lines starting at $ν_{0}$ on the frequency axis, parallel for different metals.

Label

Saturation current
Stopping potential $- V_{0}$
$ν_{0}$ (threshold)
Slope = h/e

Watch out: Label which curve is for higher intensity or higher frequency.

How the board asks Dual Nature of Radiation and Matter

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Photoelectric graphs	Current against potential for different intensities or frequencies; stopping potential against frequency for two metals; current against intensity.	Intensity changes saturation current only; frequency changes stopping potential only. $V_{0}$ -ν lines for different metals are parallel (slope h/e) with different intercepts.
Einstein's equation numericals	Find $K_{ma x}$ , $V_{0}$ , $v_{ma x}$ , work function or threshold from given data; two frequencies give two stopping potentials.	Work in eV with $h c = 1240 e V nm$ where possible; $e V_{0} = h ν - ϕ_{0}$ ; subtract two equations to eliminate $ϕ_{0}$ .
Photon counting	Number of photons from a lamp or laser; momentum or energy of a photon.	$n = P λ / h c$ ; keep power in watts and λ in metres.
Wave theory versus photon theory	Assertion-reason and 2-3 mark explanations of why intensity does not affect $K_{ma x}$ or why there is a threshold.	State the observation, the wave prediction and the photon explanation in one line each.
de Broglie wavelength	λ of an electron, proton, alpha particle or ball; ratios for the same K, the same V or the same momentum.	$λ = h / p$ ; for the same K, $λ \propto 1/ m$ ; for the same V, $λ \propto 1/ m q$ .

Tricks to remember Dual Nature of Radiation and Matter

Short memory aids. Say each one aloud once, then use it on a question the same day.

Intensity = how many; frequency = how hard
Intensity sets HOW MANY photons (current); frequency sets HOW HARD each one hits (kinetic energy, stopping potential).
Use it when: Every graph and assertion-reason question.
1240 shortcut
Photon energy in eV = 1240 ÷ wavelength in nm.
Use it when: Fast Einstein-equation numericals.
Work, then play
A photon's energy pays the exit fee (work function) first; whatever is left is the electron's kinetic energy.
Use it when: Writing $K_{ma x} = h ν - ϕ_{0}$ .
Same slope for all
$V_{0}$ -ν graphs for every metal are parallel lines of slope h/e; only the intercept (threshold) changes.
Use it when: Comparing metals on one graph.
1.227 over root V
Electron through V volts: λ ≈ 1.227/√V nm.
Use it when: Quick electron wavelengths.

Dual Nature of Radiation and Matter: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Photon energy · 1 mark · MCQBoard level

The energy of a photon of wavelength 620 nm is about:
(a) 1 eV
(b) 2 eV
(c) 3 eV
(d) 4 eV

Try it first, then show the solutionSolution

$E = \frac{1240}{620} = 2 e V$

Answer: (b) 2 eV

Example 2: Intensity and kinetic energy · 1 mark · Assertion-reasonBoard level

Assertion (A): Doubling the intensity of light falling on a photosensitive surface doubles the photocurrent but does not change the stopping potential.
Reason (R): Intensity determines the number of photons, while frequency determines the energy of each photon.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.

Try it first, then show the solutionSolution

Twice the photons give twice the electrons (current), each with the same maximum energy hν − φ₀ (same $V_{0}$ ). R explains A.

Answer: (a)

Example 3: Work function from two experiments · 2 marks · Very short answerHarder board level

When light of wavelength 300 nm falls on a metal, the stopping potential is 1.6 V. Find the work function and the threshold wavelength. ( $h c = 1240 e V nm$ ) [2 marks]

Try it first, then show the solutionSolution

$E = \frac{1240}{300} \approx 4.13 e V$
$ϕ_{0} = E - e V_{0} = 4.13 - 1.6 = 2.53 e V$
$λ_{0} = \frac{1240}{2.53} \approx 490 nm$

Answer: About 2.53 eV; about 490 nm

Example 4: Photoelectric graphs explained · 3 marks · Short answerHarder board level

Draw the graph of stopping potential against frequency of incident light for two metals A and B with work functions $ϕ_{A} < ϕ_{B}$ . (a) Why are the lines parallel? (b) Which line has the larger intercept on the frequency axis? (c) What does the intercept on the potential axis (extended) give? [3 marks]

Try it first, then show the solutionSolution

$V_{0} = \frac{h}{e} ν - \frac{ϕ _{0}}{e}$ : straight lines.
(a) The slope h/e is a universal constant, the same for every metal, so the lines are parallel.
(b) Intercept $ν_{0} = ϕ_{0} / h$ , so B (larger work function) has the larger threshold frequency.
(c) Extending to ν = 0 gives $V_{0} = - ϕ_{0} / e$ , i.e. the work function in eV with a minus sign.

Answer: Parallel lines (slope h/e); B has the larger ν-intercept; the extended V-intercept is $- ϕ_{0} / e$ .

Example 5: Full photoelectric problem · 5 marks · Long answerHarder board level

(a) State three experimental observations of the photoelectric effect that the wave theory cannot explain. (b) Explain them with Einstein's equation. (c) Light of wavelength 400 nm and power 2 mW falls on a metal of work function 2.0 eV. Find the stopping potential and the maximum number of photoelectrons emitted per second, assuming every photon ejects one electron. ( $h c = 1240 e V nm$ , $1 e V = 1.6 \times 1 0^{- 19} J$ ) [5 marks]

Try it first, then show the solutionSolution

(a) $K_{ma x}$ independent of intensity; a threshold frequency exists; emission is instantaneous.
(b) $K_{ma x} = h ν - ϕ_{0}$ : each electron gets one photon's energy hν (set by frequency); below $ν_{0} = ϕ_{0} / h$ no photon has enough energy; one photon-electron interaction is immediate.
(c) $E = \frac{1240}{400} = 3.1 e V$ ; $V_{0} = 3.1 - 2.0 = 1.1 V$ .
Photon energy in joules: $3.1 \times 1.6 \times 1 0^{- 19} = 4.96 \times 1 0^{- 19} J$ .
Photons per second $= \frac{2 \times 1 0 ^{- 3}}{4.96 \times 1 0 ^{- 19}} \approx 4.0 \times 1 0^{15}$ , so at most $4.0 \times 1 0^{15}$ electrons per second.

Answer: $V_{0} = 1.1 V$ ; at most about $4 \times 1 0^{15}$ electrons per second.

Example 6: Ratio of de Broglie wavelengths · 2 marks · Very short answerHarder board level

A proton and an alpha particle are accelerated from rest through the same potential difference. Find the ratio of their de Broglie wavelengths. ( $m_{α} = 4 m_{p}$ , $q_{α} = 2 e$ ) [2 marks]

Try it first, then show the solutionSolution

K = qV, so $λ = \frac{h}{2 m q V} \propto \frac{1}{m q}$
$\frac{λ _{p}}{λ _{α}} = \frac{m _{α} q _{α}}{m _{p} q _{p}} = \frac{4 \times 2}{1 \times 1} = 8 = 22$

Answer: $λ_{p} : λ_{α} = 22 : 1$

Example 7: Two frequencies, one metal · 3 marks · Short answerHOTS / competency

When a metal is lit with light of frequency $8 \times 1 0^{14} H z$ , the stopping potential is 1.3 V; with $10 \times 1 0^{14} H z$ it is 2.1 V. Find Planck's constant and the work function from these data. ( $e = 1.6 \times 1 0^{- 19} C$ ) [3 marks]

Try it first, then show the solutionSolution

$e V_{1} = h ν_{1} - ϕ_{0}$ and $e V_{2} = h ν_{2} - ϕ_{0}$ ; subtract: $e (V_{2} - V_{1}) = h (ν_{2} - ν_{1})$
$h = \frac{1.6 \times 1 0 ^{- 19} \times 0.8}{2 \times 1 0 ^{14}} = 6.4 \times 1 0^{- 34} J s$
$ϕ_{0} = h ν_{1} - e V_{1} = 6.4 \times 1 0^{- 34} \times 8 \times 1 0^{14} - 1.6 \times 1 0^{- 19} \times 1.3 = 5.12 \times 1 0^{- 19} - 2.08 \times 1 0^{- 19} = 3.04 \times 1 0^{- 19} J$
$ϕ_{0} = \frac{3.04 \times 1 0 ^{- 19}}{1.6 \times 1 0 ^{- 19}} = 1.9 e V$

Answer: $h \approx 6.4 \times 1 0^{- 34} J s$ ; $ϕ_{0} = 1.9 e V$

Example 8: Same kinetic energy, different particles · 2 marks · Very short answerHOTS / competency

An electron, a proton and an alpha particle all have the same kinetic energy. Arrange them in order of increasing de Broglie wavelength, and find $λ_{e} / λ_{p}$ . (Take $m_{p} = 1836 m_{e}$ , $m_{α} = 4 m_{p}$ .) [2 marks]

Try it first, then show the solutionSolution

Same K: $λ = \frac{h}{2 m K} \propto \frac{1}{m}$ , so the heaviest has the shortest wavelength.
Order: alpha < proton < electron.
$\frac{λ _{e}}{λ _{p}} = \frac{m _{p}}{m _{e}} = 1836 \approx 42.8$

Answer: α < p < e; $λ_{e} / λ_{p} \approx 43$

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 1 + 1 marks.

Testing a photocell

Suppose a student shines light of different colours on a photocell whose cathode has a work function of 2.1 eV. She records the stopping potential and the saturation current for each colour, keeping the intensity the same. Take $h c = 1240 e V nm$ .

(i) · 1 mark
Will light of wavelength 650 nm produce a photocurrent?
Answer
Answer: No
$E = 1240/650 \approx 1.91 e V < 2.1 e V$
(ii) · 1 mark
Find the stopping potential for light of wavelength 400 nm.
Answer
Answer: 1.0 V
$E = 3.1 e V$ ; $V_{0} = 3.1 - 2.1 = 1.0 V$
(iii) · 1 mark
If the intensity of the 400 nm light is doubled, what happens to the stopping potential and the saturation current?
Answer
Answer: Stopping potential unchanged; saturation current doubles.
Intensity changes the number of photons, not their energy.
(iv) · 1 mark
What is the threshold wavelength of the cathode?
Answer
Answer: About 590 nm
$λ_{0} = 1240/2.1 \approx 590 nm$

Waves of particles

Suppose a lab compares the de Broglie wavelengths of an electron accelerated through 400 V and of a neutron moving at $2 \times 1 0^{3} m s^{- 1}$ . Take $h = 6.63 \times 1 0^{- 34} J s$ and the neutron mass as $1.67 \times 1 0^{- 27} k g$ .

(i) · 1 mark
Find the electron's wavelength.
Answer
Answer: About 0.061 nm
$λ \approx 1.227/ 400 = 1.227/20 nm$
(ii) · 1 mark
Find the neutron's wavelength.
Answer
Answer: About 0.20 nm
$λ = \frac{6.63 \times 1 0 ^{- 34}}{1.67 \times 1 0 ^{- 27} \times 2 \times 1 0 ^{3}} \approx 1.99 \times 1 0^{- 10} m$
(iii) · 1 mark
If the electron's accelerating voltage is made four times larger, what is its new wavelength?
Answer
Answer: Half, about 0.031 nm
$λ \propto 1/ V$
(iv) · 1 mark
Why are such particles useful for studying crystals?
Answer
Answer: Their wavelengths are comparable to atomic spacings in crystals, so they are diffracted by them.
Diffraction needs wavelengths comparable to the spacing of the 'slits'.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

The work function of a metal is 2.0 eV. Light of wavelength 400 nm falls on it. Find the stopping potential. ( $h c = 1240 e V nm$ ) [2 marks] (2 marks)

Step	Marks
$E = \frac{1240}{400} = 3.1 e V$	1
$e V_{0} = 3.1 - 2.0 = 1.1 e V$ , so $V_{0} = 1.1 V$	1

Write Einstein's photoelectric equation and use it to explain (a) the existence of a threshold frequency, (b) why the maximum kinetic energy of photoelectrons is independent of intensity, (c) why emission is instantaneous. [3 marks] (3 marks)

Step	Marks
$K_{ma x} = h ν - ϕ_{0}$ ; (a) $K_{ma x} \geq 0$ needs $h ν \geq ϕ_{0}$ , i.e. $ν \geq ν_{0} = ϕ_{0} / h$	1
(b) Each electron absorbs one photon, whose energy hν depends only on frequency; intensity changes the number of photons, not their energy	1
(c) Absorption of a photon by an electron is a single, immediate event; no time is needed to collect energy	1

(a) Draw graphs of photocurrent against collector potential for two frequencies at the same intensity, and of stopping potential against frequency for two metals. (b) Explain the features using Einstein's equation. (c) Light of frequency $1.0 \times 1 0^{15} H z$ falls on a metal of work function 2.0 eV. Find $K_{ma x}$ in eV. (h = $6.63 \times 1 0^{- 34}$ J s) [5 marks] (5 marks)

Step	Marks
(a) I-V graph: same saturation current, larger stopping potential for higher frequency	1
$V_{0}$ -ν graph: two parallel straight lines with different ν-intercepts	1
(b) $V_{0} = \frac{h}{e} ν - \frac{ϕ _{0}}{e}$ : slope h/e same for all metals; intercept $ν_{0} = ϕ_{0} / h$ depends on the metal	1
(c) $h ν = \frac{6.63 \times 1 0 ^{- 34} \times 1 0 ^{15}}{1.6 \times 1 0 ^{- 19}} \approx 4.14 e V$	1
$K_{ma x} = 4.14 - 2.0 = 2.14 e V$	1

Where students lose marks in Dual Nature of Radiation and Matter

The slips examiners see most often, and what to write instead.

Mistake: Saying stopping potential increases with intensity
Thinking brighter means more energetic.
Do this: Stopping potential depends only on frequency (and the metal). Intensity changes only the saturation current.
Usually costs: 1 mark in MCQs and graph questions.
Mistake: Mixing eV and joules
Substituting φ₀ in eV into a formula in joules.
Wrong working
$K = 6.63 \times 1 0^{- 34} \times 1 0^{15} - 2.0$
Right working
$K = \frac{6.63 \times 1 0 ^{- 34} \times 1 0 ^{15}}{1.6 \times 1 0 ^{- 19}} - 2.0 = 4.14 - 2.0 = 2.14 e V$
Do this: Convert every energy to the same unit first.
Usually costs: 1-2 marks.
Mistake: Using $λ = h / m v$ with the wrong mass or speed for photons
Treating photons like ordinary particles.
Do this: For a photon use $p = h / λ$ or $E / c$ ; it has zero rest mass.
Usually costs: 1 mark.
Mistake: Drawing $V_{0}$ -ν lines through the origin
Forgetting the threshold.
Do this: The line meets the ν axis at $ν_{0} > 0$ ; extended, it meets the $V_{0}$ axis at $- ϕ_{0} / e$ .
Usually costs: 1 mark for the graph.
Mistake: Writing about Davisson-Germer or the uncertainty principle
Older syllabi included them.
Do this: Not in the 2026-27 syllabus; keep matter-wave answers to $λ = h / p$ and simple applications.
Usually costs: Time.

NCERT exercise guide: Dual Nature of Radiation and Matter

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Exercises 11.1 to 11.4 (photons)

What it tests: Energy and momentum of photons, maximum kinetic energy and threshold for caesium, maximum electron speed from cut-off voltage, and number of photons from a laser.

How to approach it: $E = h c / λ$ , $p = h / λ$ ; $K_{ma x} = h ν - ϕ_{0}$ ; $\frac{1}{2} m v^{2} = e V_{0}$ ; $n = P / E$ .

Same kind as Exercise 11.3

The stopping potential in a photoelectric experiment is 1.1 V. Find the maximum kinetic energy and speed of the photoelectrons. ( $m = 9.1 \times 1 0^{- 31} k g$ )

Show the solutionSolution

$K_{ma x} = e V_{0} = 1.1 e V = 1.76 \times 1 0^{- 19} J$
$v_{ma x} = \frac{2 K}{m} = \frac{2 \times 1.76 \times 1 0 ^{- 19}}{9.1 \times 1 0 ^{- 31}} = 3.87 \times 1 0^{11}$
$v_{ma x} \approx 6.2 \times 1 0^{5} m s^{- 1}$

Answer: 1.1 eV ( $1.76 \times 1 0^{- 19} J$ ); about $6.2 \times 1 0^{5} m s^{- 1}$

Exercises 11.5 to 11.9 (Einstein's equation)

What it tests: Planck's constant from the slope of a $V_{0}$ -ν graph, threshold and emission at given frequencies, whether a metal emits under given light, and work function from cut-off voltage.

How to approach it: Slope of $V_{0}$ -ν is h/e. Compare photon energy with the work function. $ϕ_{0} = h ν - e V_{0}$ .

Same kind as Exercise 11.6

The threshold frequency of a metal is $3.3 \times 1 0^{14} H z$ . Light of frequency $6.0 \times 1 0^{14} H z$ falls on it. Find the stopping potential. (h = $6.63 \times 1 0^{- 34}$ J s)

Show the solutionSolution

$e V_{0} = h (ν - ν_{0}) = 6.63 \times 1 0^{- 34} \times 2.7 \times 1 0^{14} \approx 1.79 \times 1 0^{- 19} J$
$V_{0} = \frac{1.79 \times 1 0 ^{- 19}}{1.6 \times 1 0 ^{- 19}} \approx 1.12 V$

Answer: About 1.1 V

Exercises 11.10 and 11.11 (matter waves)

What it tests: De Broglie wavelengths of a bullet, a ball and a dust particle, and showing that for an EM wave the de Broglie relation gives the wave's own wavelength.

How to approach it: $λ = h / m v$ ; for a photon $p = E / c = h ν / c$ , so $h / p = c / ν = λ$ .

Same kind as Exercise 11.10

Find the de Broglie wavelength of a proton moving at $2 \times 1 0^{5} m s^{- 1}$ . ( $m_{p} = 1.67 \times 1 0^{- 27} k g$ )

Show the solutionSolution

$λ = \frac{h}{m v} = \frac{6.63 \times 1 0 ^{- 34}}{1.67 \times 1 0 ^{- 27} \times 2 \times 1 0 ^{5}}$
$λ \approx 1.99 \times 1 0^{- 12} m$

Answer: About $2 \times 1 0^{- 12} m$

Dual Nature of Radiation and Matter: previous year questions

Asked in 8 of the 8 board papers we hold (2023 to 2026), on average 4.9 marks per paper; 24 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Dual Nature of Radiation and Matter

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

2026, set 55/1/1, Q14 · 1 mark · Assertion-reasonCompetency-based

Assertion (A) : If accelerated electrons are passed through a narrow slit, a diffraction pattern is observed.
Reason (R) : Electrons behave as both particles and waves.

(A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C)Assertion (A) is true, but Reason (R) is false.
(D)Both Assertion (A) and Reason (R) are false.

Try it first, then show our solutionSolution

Electrons have a de Broglie wavelength, so they diffract at a narrow slit. This wave nature (dual behaviour) is exactly what explains the assertion.
1 mark

Answer: Option (A)

Where the marks are

Both true, and wave–particle duality explains the diffraction (1 mark)

See it in the full 2026 board paper, set 55/1/1

2026, set 55/1/1, Q17 · 2 marks · Very short answer

In a photoelectric experiment, the emitter plate is irradiated with radiation of 200 nm. The photocurrent becomes zero when the collector plate potential is $- 0.80$ V. Calculate the work function (in eV) of the emitter.

Try it first, then show our solutionSolution

Photon energy: $E = \frac{h c}{λ} = \frac{1240 eV nm}{200 nm} = 6.2 eV$
1 mark
The photocurrent stops at a collector potential of $- 0.80$ V, so the stopping potential is $V_{0} = 0.80$ V. Hence $K_{ma x} = e V_{0} = 0.80$ eV.
½ mark
Einstein's photoelectric equation gives $ϕ_{0} = \frac{h c}{λ} - e V_{0} = 6.2 - 0.80 = 5.4 eV$
½ mark

Answer: Work function ≈ 5.4 eV

Where the marks are

Calculates photon energy of about 6.2 eV (1 mark)
Identifies the stopping potential as 0.80 V and uses Einstein's equation (½ mark)
Obtains a work function of 5.4 eV (½ mark)

See it in the full 2026 board paper, set 55/1/1

2026, set 55/1/1, Q30 · 4 marks · Case-basedCompetency-based

A researcher performs an experiment on photo-electric effect using two metals A and B with unknown work functions. She illuminates the surfaces of A and B with monochromatic radiation of various frequencies and records the value of corrosponding stopping potentials ( $V_{s}$ ). The graph shows the variation of stopping potential ( $V_{s}$ ) with the frequency of incident radiation ( $ν$ ) for metals A and B.
Answer the following questions :

(I)
From the graph, the work functions of A and B are (h is Planck’s constant and e value of charge on an electron)
1. (A) $ν_{1}$ and $ν_{2}$
2. (B) $V_{1}$ and $V_{2}$
3. (C) $h ν_{1}$ and $h ν_{2}$
4. (D) $\frac{h ν _{1}}{e}$ and $\frac{h ν _{2}}{e}$
1 mark
(II)
For radiation of frequency $ν > ν_{2}$ incident on the surfaces of A and B, the maximum kinetic energy of ejected electron is
1. (A)greater for metal A because it has a smaller work function.
2. (B)greater for metal B because it has a larger work function.
3. (C)greater for metal B because it has higher threshold frequency.
4. (D)the same for both metal A and metal B because it is independent of work functions of metals.
1 mark
(III)
If the intensity of the incident radiation for both metals A and B, is doubled keeping its frequency constant, then
1. (A)the slope of the parallel lines will increase.
2. (B)the slope of the parallel lines will decrease.
3. (C)the threshold frequencies for both A and B will decrease.
4. (D)the slope of the parallel lines will not change but more electrons will be emitted per second.
1 mark
(IV)
The threshold frequency for a metal surface is $ν_{0}$ . If the radiation of frequency $3 ν_{0}$ illuminates the surface, the maximum kinetic energy (KE) of photoelectrons is $E_{1}$ . If the frequency were increased to $6 ν_{0}$ , the maximum KE of the photoelectrons becomes $E_{2}$ . Then $(\frac{E _{1}}{E _{2}})$ equals
1. (A)1/3
2. (B)1/2
3. (C)2/5
4. (D)3/4
1 mark
OR
(IV)
Let m be the slope of the graph line for metal B. If e is the value of electron charge, then Planck’s constant ‘h’ is given by
1. (A)me
2. (B) $\frac{1}{m e}$
3. (C) $\frac{m}{e}$
4. (D) $\frac{e}{m}$
1 mark

Figure in the paper: Graph of stopping potential $V_{s}$ (vertical axis) versus frequency $ν$ (horizontal axis) with two parallel straight lines A and B. Line A cuts the frequency axis at $ν_{1}$ and line B at a larger $ν_{2}$ ( $ν_{1} < ν_{2}$ ). Extended dashed portions meet the negative $V_{s}$ axis at $V_{1}$ (for A) and $V_{2}$ (for B), with $V_{2}$ more negative. See it in the official paper.

Try it first, then show our solutionSolution

(I) Threshold frequencies are $ν_{1}$ (A) and $ν_{2}$ (B), and $ϕ_{0} = h ν_{0}$ . The work functions are therefore $h ν_{1}$ and $h ν_{2}$ . Option (C).
1 mark
(II) $K_{ma x} = h ν - ϕ_{0}$ . Since $ϕ_{A} < ϕ_{B}$ , $K_{ma x}$ is greater for A. Option (A).
1 mark
(III) Intensity affects only the number of photons. The slope $h / e$ and the threshold frequencies stay the same, but more electrons are emitted per second. Option (D).
1 mark
(IV)(a) $E_{1} = h (3 ν_{0}) - h ν_{0} = 2 h ν_{0}$ and $E_{2} = 6 h ν_{0} - h ν_{0} = 5 h ν_{0}$ , so $\frac{E _{1}}{E _{2}} = \frac{2}{5}$ . Option (C).
(IV)(b) $V_{s} = \frac{h}{e} ν - \frac{ϕ _{0}}{e}$ , so the slope is $m = \frac{h}{e}$ and $h = m e$ . Option (A).
1 mark

Answer: (I) C; (II) A; (III) D; (IV) C [or alternative: A]

Where the marks are

(I) hν₁ and hν₂, C (1 mark)
(II) Greater for A, A (1 mark)
(III) Slope unchanged, more electrons emitted, D (1 mark)
(IV) 2/5, C; or alternative h = me, A (1 mark)

See it in the full 2026 board paper, set 55/1/1

Every Dual Nature of Radiation and Matter question, by year

For 2027

Sample paper 2026-27: Q30 (4 marks, solved)

2026

Set 55/1/1: Q14 (1 mark, solved), Q17 (2 marks, solved), Q30 (4 marks, solved)

Set 55/5/1: Q7 (1 mark, solved), Q16 (1 mark, solved), Q27 (3 marks, solved)

Sample paper 2025-26: Q21(II) (2 marks, solved), Q30 (4 marks, solved)

2025

Set 55/1/1: Q9 (1 mark, solved), Q10 (1 mark, solved), Q26 (3 marks, solved)

Set 55/7/1: Q8 (1 mark, solved), Q16 (1 mark, solved), Q26 (3 marks, solved)

Sample paper 2024-25: Q16 (1 mark, solved), Q17 (2 marks, solved), Q30 (4 marks, solved)

2024

Set 55/1/1: Q11 (1 mark, solved), Q12 (1 mark, solved), Q13 (1 mark, solved), Q20 (2 marks, solved)

Set 55/5/1: Q8 (1 mark, solved), Q22 (3 marks, solved)

Sample paper 2023-24: Q3 (1 mark, solved), Q13 (1 mark, solved), Q18 (2 marks, solved)

2023

Set 55/1/1: Q18 (1 mark, solved), Q29 (3 marks, solved)

Set 55/5/1: Q7 (1 mark, solved), Q11 (1 mark, solved), Q20(a) (2 marks, solved), Q20(b) (2 marks, solved)

Practice questions on Dual Nature of Radiation and Matter

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The stopping potential in a photoelectric experiment depends on:
(a) intensity of light only
(b) frequency of light only
(c) both intensity and frequency
(d) neither
Hint
Einstein's equation.
Answer
(b) frequency of light only (for a given metal)
Q2 · 1 mark · MCQBoard level
The slope of the graph of stopping potential against frequency is:
(a) h
(b) e/h
(c) h/e
(d) $ϕ_{0} / e$
Hint
$V_{0} = (h / e) ν - ϕ_{0} / e$ .
Answer
(c) h/e
Q3 · 1 mark · Very short answerBoard level
Write the de Broglie relation and explain the symbols.
Hint
Wavelength and momentum.
Answer
$λ = h / p$ , where λ is the de Broglie wavelength, h Planck's constant and p = mv the momentum of the particle.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Photoelectric emission does not take place if the frequency of light is below the threshold frequency, however intense the light.
Reason (R): The energy of a photon depends on its frequency, and one electron absorbs only one photon.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
One photon, one electron.
Answer
(a)
Q5 · 2 marks · Very short answerHarder board level
Light of wavelength 500 nm falls on a metal of work function 1.9 eV. Find the maximum kinetic energy of the photoelectrons in eV and the stopping potential. ( $h c = 1240 e V nm$ )
Hint
Photon energy first.
Answer
$E = 2.48 e V$ ; $K_{ma x} = 0.58 e V$ ; $V_{0} = 0.58 V$ .
Q6 · 2 marks · Very short answerHarder board level
A 5.0 mW laser of wavelength 660 nm shines on a photosensitive surface of work function 1.8 eV. (a) Find the energy and momentum of each photon. (b) How many photons strike the surface per second? (c) If only one photon in $1 0^{4}$ ejects an electron, find the saturation photocurrent. ( $h = 6.63 \times 1 0^{- 34} J s$ )
Hint
$E = h c / λ$ ; photons per second $= P / E$ ; current = electrons per second × e.
Answer
(a) $E = \frac{6.63 \times 1 0 ^{- 34} \times 3 \times 1 0 ^{8}}{660 \times 1 0 ^{- 9}} \approx 3.0 \times 1 0^{- 19} J$ (about 1.88 eV, just above 1.8 eV, so emission occurs); $p = \frac{h}{λ} \approx 1.0 \times 1 0^{- 27} k g m s^{- 1}$ . (b) $n = \frac{5.0 \times 1 0 ^{- 3}}{3.0 \times 1 0 ^{- 19}} \approx 1.7 \times 1 0^{16}$ per second. (c) Electrons per second $\approx 1.7 \times 1 0^{12}$ , so $I \approx 1.7 \times 1 0^{12} \times 1.6 \times 1 0^{- 19} \approx 2.7 \times 1 0^{- 7} A$ .
Q7 · 2 marks · Very short answerHarder board level
Draw the graph of photocurrent against collector potential for light of the same frequency but different intensities $I_{1} > I_{2}$ . Explain why both curves meet the potential axis at the same point.
Hint
Stopping potential depends on frequency.
Answer
Both curves start at the same $- V_{0}$ and rise to different saturation currents (higher for $I_{1}$ ). They meet at the same point because the stopping potential depends only on the maximum kinetic energy, $h ν - ϕ_{0}$ , which is the same for both intensities.
Q8 · 2 marks · Very short answerHarder board level
An electron and an alpha particle have the same de Broglie wavelength. Find the ratio of (a) their speeds, (b) their kinetic energies. ( $m_{α} = 4 m_{p} = 4 \times 1836 m_{e}$ )
Hint
Same λ means the same momentum p = mv; then $K = \frac{p ^{2}}{2 m}$ .
Answer
Same λ, so the same momentum: $m_{e} v_{e} = m_{α} v_{α}$ . (a) $\frac{v _{e}}{v _{α}} = \frac{m _{α}}{m _{e}} = 4 \times 1836 = 7344$ . (b) $K = \frac{p ^{2}}{2 m}$ with p equal, so $\frac{K _{e}}{K _{α}} = \frac{m _{α}}{m _{e}} = 7344$ : the electron has far more kinetic energy.
Q9 · 3 marks · Short answerHarder board level
(a) Derive the relation between stopping potential and frequency from Einstein's equation and sketch the graph. (b) Stopping potentials of 0.60 V and 1.84 V are found for frequencies $6.0 \times 1 0^{14}$ and $9.0 \times 1 0^{14}$ Hz. Find h, the work function in eV and the threshold frequency. ( $e = 1.6 \times 1 0^{- 19} C$ )
Hint
Subtract the two equations.
Answer
(a) $e V_{0} = h ν - ϕ_{0} \Rightarrow V_{0} = \frac{h}{e} ν - \frac{ϕ _{0}}{e}$ : a straight line with slope h/e meeting the ν axis at $ν_{0}$ . (b) $h = \frac{e ( V _{2} - V _{1} )}{ν _{2} - ν _{1}} = \frac{1.6 \times 1 0 ^{- 19} \times 1.24}{3.0 \times 1 0 ^{14}} \approx 6.61 \times 1 0^{- 34} J s$ . $ϕ_{0} = h ν_{1} - e V_{1} = 3.97 \times 1 0^{- 19} - 0.96 \times 1 0^{- 19} \approx 3.01 \times 1 0^{- 19} J \approx 1.88 e V$ . $ν_{0} = \frac{ϕ _{0}}{h} \approx 4.55 \times 1 0^{14} H z$ .
Q10 · 3 marks · Short answerHOTS / competency
A photon and an electron have the same de Broglie wavelength of 1 nm. Compare their energies. ( $h = 6.63 \times 1 0^{- 34} J s$ , $m_{e} = 9.1 \times 1 0^{- 31} k g$ )
Hint
Same p; photon E = pc, electron $K = p^{2} /2 m$ .
Answer
$p = h / λ = 6.63 \times 1 0^{- 25} k g m s^{- 1}$ for both. Photon: $E = p c = 1.99 \times 1 0^{- 16} J$ (about 1240 eV). Electron: $K = \frac{p ^{2}}{2 m} = \frac{( 6.63 \times 1 0 ^{- 25} ) ^{2}}{2 \times 9.1 \times 1 0 ^{- 31}} \approx 2.4 \times 1 0^{- 19} J$ (about 1.5 eV). The photon has about 800 times more energy.
Q11 · 2 marks · Very short answerHOTS / competency
Why can a bright red light fail to eject electrons from a metal when a faint ultraviolet light succeeds? What would happen if the red light were made a thousand times brighter?
Hint
Energy per photon.
Answer
Each electron absorbs one photon. Red photons have energy below the work function, so none can free an electron; UV photons have more than enough. Making the red light brighter only sends more red photons, each still too weak, so there is still no emission.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Define work function and threshold frequency.Show answer
Work function: the minimum energy needed to remove an electron from a metal surface. Threshold frequency: the minimum frequency of light that can cause photoemission, $ν_{0} = ϕ_{0} / h$ .
2.State Einstein's photoelectric equation.Show answer
$K_{ma x} = h ν - ϕ_{0}$ .
3.What does the slope of a $V_{0}$ -ν graph give?Show answer
h/e.
4.How does photocurrent depend on intensity?Show answer
It is directly proportional to intensity (above threshold).
5.Energy of a 620 nm photon in eV?Show answer
$1240/620 = 2 e V$ .
6.Write the de Broglie relation.Show answer
$λ = h / p = h / m v$ .
7.An electron and a proton have the same kinetic energy. Which has the longer de Broglie wavelength?Show answer
The electron (λ ∝ 1/√m).
8.Give two properties of photons.Show answer
Energy hν and momentum h/λ; travel at c; uncharged; zero rest mass; number not conserved (any two).

Best order to study it

Start with electron emission, work function and the electron volt.
Learn Hertz's and Lenard's observations and the experimental study: intensity, potential and frequency effects.
Understand why the wave theory fails on each point.
Learn Einstein's equation, threshold frequency and the $V_{0}$ -ν graph; practise numericals in eV.
Study the photon: energy, momentum and photon counting.
Finish with matter waves: $λ = h / p$ , comparisons of particles and electron wavelengths.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you name three ways of electron emission?
Can you draw current-potential graphs for different intensities and frequencies?
Can you explain stopping potential and why it is independent of intensity?
Can you list three failures of the wave theory?
Can you derive the $V_{0}$ -ν relation and read h/e and $ν_{0}$ from a graph?
Can you solve Einstein-equation problems in eV using 1240 eV nm?
Can you count photons per second from a source of given power?
Can you find the de Broglie wavelength of an electron, proton and ball?
Can you compare wavelengths for the same kinetic energy or the same accelerating voltage?
Can you explain why everyday objects show no wave behaviour?

Dual Nature of Radiation and Matter: frequently asked questions

Is the Davisson-Germer experiment in the 2026-27 syllabus?

No. The Davisson-Germer experiment and Heisenberg's uncertainty principle are not in the 2026-27 syllabus.

Do I need to derive the de Broglie wavelength of an accelerated electron?

No, the derivation is not in the syllabus. You can still use \lambda = h/\sqrt{2meV} (about 1.227/√V nm) in numericals.

Why does the photocurrent not start at zero potential?

Photoelectrons leave the metal with some kinetic energy, so some of them reach the collector even with no accelerating potential; a negative (retarding) potential is needed to stop them all.

What is the most common type of board question here?

In our question bank, de Broglie wavelength and photoelectric effect questions are the most frequent, mostly as 1-mark MCQs or assertion-reason items and short numericals.

Does increasing the intensity of light increase the speed of photoelectrons?

No. It increases the number of photons and so the number of photoelectrons (the current), but each electron's maximum kinetic energy depends only on the frequency.

What is meant by the dual nature of radiation?

Light behaves as a wave in interference and diffraction, and as a stream of particles (photons) when it interacts with matter, as in the photoelectric effect.

Based on: CBSE Physics (042) syllabus 2026-27; NCERT Physics Part II, Textbook for Class XII, Chapter 11 (Reprint 2026-27)