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

What Dual Nature of Radiation and Matter is about

Weightage: Dual Nature of Radiation and Matter is part of the unit Dual Nature of Radiation and Matter, which carries 7 of the 70 marks in the ISC Physics Paper 1 (Theory). Questions on it can appear in any section of the paper.Source: CISCE ISC Physics syllabus, examination year 2027, distribution of marks

Key concepts

Photons: energy and momentum

Planck proposed that radiation is emitted and absorbed in packets (quanta); Einstein extended this to light itself travelling as photons. Each photon has energy $E=hf=\frac{hc}{\lambda }$ (h = $6.63\times 10^{-34}$ J s), momentum $p=\frac{E}{c}=\frac{h}{\lambda }$, travels at c and has zero rest mass. A more intense beam of the same frequency carries more photons per second, not more energetic photons. Useful shortcut: $E(eV)=\frac{1240}{\lambda (nm)}$. Because photons carry momentum, light exerts a tiny force on surfaces (radiation pressure): an absorbing surface receives force $F=P/c$ from a beam of power P.

Worked example

For light of wavelength 500 nm, find the energy (in J and eV) and momentum of one photon. How many photons per second does a 2 mW laser of this wavelength emit?

Try it first, then show the solutionSolution

1. $E=\frac{hc}{\lambda }=\frac{6.63\times 10^{-34}\times 3\times 10^8}{500\times 10^{-9}}=3.98\times 10^{-19}J=2.49eV$
2. $p=\frac{h}{\lambda }=\frac{6.63\times 10^{-34}}{5\times 10^{-7}}=1.33\times 10^{-27}kgm{s}^{-1}$
3. $N=\frac{P}{E}=\frac{2\times 10^{-3}}{3.98\times 10^{-19}}=5.0\times 10^{15}$ per second

Answer: $3.98\times 10^{-19}J$ (2.49 eV); $1.33\times 10^{-27}kgm{s}^{-1}$; $5.0\times 10^{15}$ photons per second

Photoelectric effect: Hertz's and Lenard's observations

Hertz noticed that sparks jumped more easily when the electrodes were lit with ultraviolet light; Lenard showed that UV falling on a metal plate in an evacuated tube ejects electrons that make a current flow. In a photocell, a collector at positive potential draws the emitted electrons; making it negative slows them, and the stopping potential $V_s$ is the reverse potential that just stops the fastest ones: $e{V}_s=K_{max}$. Experimental facts: (1) for each metal there is a threshold frequency $f_0$, the minimum frequency below which no electrons are emitted, whatever the intensity; (2) $K_{max}$ (and $V_s$) increases linearly with frequency and does not depend on intensity; (3) the photocurrent (number of electrons per second) is proportional to intensity; (4) emission is practically instantaneous ($<10^{-9}$ s). The wave theory predicts that any frequency should work given enough intensity, that $K_{max}$ should rise with intensity and that dim light should cause a delay: all contradicted. Only a particle (quantum) picture explains the effect.

Worked example

In a photocell experiment, the intensity of the incident light is doubled, keeping its frequency fixed. What happens to (a) the saturation current and (b) the stopping potential? Then the frequency is increased at fixed intensity: what happens to each?

Try it first, then show the solutionSolution

1. (a) Double intensity means twice as many photons per second, so twice as many electrons: saturation current doubles. (b) Each photon still has energy hf, so $K_{max}$ and $V_s$ are unchanged.
2. Higher frequency: each photon has more energy, so $V_s$ increases; at fixed intensity there are fewer photons per second, so the saturation current stays about the same or falls slightly.

Answer: Intensity: current doubles, $V_s$ unchanged. Frequency: $V_s$ rises; current does not rise.

Einstein's photoelectric equation, threshold frequency and work function

An electron absorbs a single photon. Part of the energy, the work function $W_0$ (the minimum energy needed to free an electron from the surface), is used to escape; the rest appears as kinetic energy: $K_{max}=hf-W_0=e{V}_s$. The threshold frequency is the minimum frequency for emission: $h{f}_0=W_0$, so $f_0=W_0/h$ and the threshold wavelength ${\lambda }_0=\frac{hc}{W_0}$ (the maximum wavelength). Below $f_0$ no single photon has enough energy, so intensity does not help. Metals with low work functions (caesium, potassium, sodium) respond to visible light; most others need ultraviolet.

Worked example

Light of wavelength 400 nm falls on a metal of work function 2.0 eV. Find the maximum kinetic energy of the photoelectrons, the stopping potential, and the threshold frequency and wavelength. (Use hc = 1240 eV nm.)

Try it first, then show the solutionSolution

1. Photon energy $=\frac{1240}{400}=3.1eV$; $K_{max}=3.1-2.0=1.1eV$, so $V_s=1.1V$
2. $f_0=\frac{W_0}{h}=\frac{2.0\times 1.6\times 10^{-19}}{6.63\times 10^{-34}}=4.83\times 10^{14}Hz$
3. ${\lambda }_0=\frac{1240}{2.0}=620nm$

Answer: 1.1 eV; 1.1 V; $4.83\times 10^{14}$ Hz; 620 nm

Stopping potential against frequency: finding Planck's constant

From $e{V}_s=hf-W_0$: $V_s=\frac{h}{e}f-\frac{W_0}{e}$. A graph of $V_s$ (y-axis) against f (x-axis) is a straight line: its slope is h/e (the same for all metals), it cuts the f-axis at the threshold frequency $f_0$, and its extension cuts the $V_s$-axis at $-W_0/e$. So $h=e\times slope$. Different metals give parallel lines shifted along the f-axis. (The slope is not h itself.) On a graph of photocurrent against collector potential, curves for different intensities have the same stopping potential but different saturation currents; curves for different frequencies have different stopping potentials.

Worked example

A metal gives stopping potentials of 0.50 V for light of $6.0\times 10^{14}$ Hz and 1.33 V for $8.0\times 10^{14}$ Hz. Find Planck's constant, the threshold frequency and the work function.

Try it first, then show the solutionSolution

1. Slope $=\frac{1.33-0.50}{8.0\times 10^{14}-6.0\times 10^{14}}=4.15\times 10^{-15}Vs$; $h=e\times slope=1.6\times 10^{-19}\times 4.15\times 10^{-15}=6.64\times 10^{-34}Js$
2. $V_s=0$ at $f_0=6.0\times 10^{14}-\frac{0.50}{4.15\times 10^{-15}}=4.8\times 10^{14}Hz$
3. $W_0=h{f}_0=3.18\times 10^{-19}J\approx 2.0eV$

Answer: $h\approx 6.64\times 10^{-34}$ J s; $f_0\approx 4.8\times 10^{14}$ Hz; $W_0\approx 2.0$ eV

de Broglie hypothesis: matter waves

Wave nature of radiation shows in interference and diffraction; particle nature in the photoelectric effect. De Broglie argued that, by symmetry, moving particles should have a wavelength $\lambda =\frac{h}{p}=\frac{h}{mv}$. In terms of kinetic energy K, $p=\sqrt{2mK}$, so $\lambda =\frac{h}{\sqrt{2mK}}$; for a charge q accelerated from rest through V, $\lambda =\frac{h}{\sqrt{2mqV}}$; for an electron $\lambda =\frac{1.227}{\sqrt{V}}$ nm. Hence $\lambda \propto \frac{1}{\sqrt{K}}$: four times the kinetic energy halves the wavelength. Everyday objects have absurdly small wavelengths (a cricket ball's is about $10^{-34}$ m), so their wave nature is never noticed; electrons have wavelengths similar to atomic spacings (about 0.1 nm), so crystals can diffract them.

Worked example

Find the de Broglie wavelength of (a) an electron accelerated from rest through 100 V and (b) a 0.1 kg ball moving at $10m{s}^{-1}$.

Try it first, then show the solutionSolution

1. (a) $\lambda =\frac{h}{\sqrt{2meV}}=\frac{6.63\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 1.6\times 10^{-19}\times 100}}=1.23\times 10^{-10}m$
2. (b) $\lambda =\frac{h}{mv}=\frac{6.63\times 10^{-34}}{0.1\times 10}=6.63\times 10^{-34}m$

Answer: (a) 0.123 nm (b) $6.6\times 10^{-34}$ m: far too small to detect

Comparing particles; electron diffraction and the Davisson-Germer experiment

For equal kinetic energy, $\lambda \propto \frac{1}{\sqrt{m}}$, so an electron has a much longer wavelength than a proton. For equal wavelength, the momenta are equal and $K=\frac{p^2}{2m}$, so the lighter particle has more kinetic energy. Davisson-Germer experiment (qualitative): a beam of electrons from a heated filament, accelerated through about 54 V, was directed at a nickel crystal, and the number of electrons scattered at different angles was measured with a movable detector. A sharp peak in scattered intensity appeared at about 50°, just as for X-rays diffracted by a crystal. The wavelength obtained from the diffraction agreed with de Broglie's $\frac{h}{p}$ (about 0.165 nm), confirming that electrons behave as waves. Electron diffraction (rings when electrons pass through thin metal foil) is a phenomenon showing the wave nature of particles; it is used in the electron microscope.

Worked example

An electron and a proton have the same kinetic energy. Find the ratio of their de Broglie wavelengths. If instead they had the same wavelength, what would be the ratio of their kinetic energies?

Try it first, then show the solutionSolution

1. Same K: $\frac{{\lambda }_e}{{\lambda }_p}=\sqrt{\frac{m_p}{m_e}}=\sqrt{1835}=42.8$ (the electron's is about 43 times longer)
2. Same λ means same p: $\frac{K_e}{K_p}=\frac{m_p}{m_e}=1835$

Answer: ${\lambda }_e:{\lambda }_p\approx 43:1$; $K_e:K_p\approx 1835:1$

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.

Topics in this chapter

• Photoelectric Effect and Einstein's Equation
• Stopping Potential and the Frequency Graph
• De Broglie Wavelength and Matter Waves

Dual Nature of Radiation and Matter formula sheet

Photon energy

$E=hf=\frac{hc}{\lambda },E(eV)=\frac{1240}{\lambda (nm)}$

h = $6.63\times 10^{-34}$ J s; 1 eV = $1.6\times 10^{-19}$ J.

Photon momentum

$p=\frac{E}{c}=\frac{hf}{c}=\frac{h}{\lambda }$

Force of an absorbed beam F = P/c.

Number of photons per second

$N=\frac{P}{hf}=\frac{P\lambda }{hc}$

P = power of the beam.

Einstein's photoelectric equation

$K_{max}=hf-W_0=e{V}_s$

One photon frees one electron.

Threshold frequency and wavelength

$W_0=h{f}_0=\frac{hc}{{\lambda }_0}$

No emission if $f<f_0$ ($\lambda >{\lambda }_0$).

Stopping potential graph

$V_s=\frac{h}{e}f-\frac{W_0}{e}$

Slope h/e; f-intercept $f_0$.

de Broglie wavelength

$\lambda =\frac{h}{p}=\frac{h}{mv}=\frac{h}{\sqrt{2mK}}=\frac{h}{\sqrt{2mqV}}$

λ ∝ 1/√K.

Electron through V volts

$\lambda =\frac{1.227}{\sqrt{V}}nm$

Non-relativistic.

Dual Nature of Radiation and Matter: diagrams to practise

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

How the board asks Dual Nature of Radiation and Matter

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

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.

• 1240 rule

  Photon energy in eV = 1240 ÷ wavelength in nm.

  Use it when: Fast photoelectric numericals.

• Intensity counts heads, frequency counts energy

  Intensity → number of photons → current. Frequency → energy per photon → $V_s$.

  Use it when: Effect-of-change questions.

• Slope is h over e

  On the $V_s$-f line, 'h is hiding over e': slope = h/e, never h alone.

  Use it when: Graph questions.

• Threshold means minimum

  Threshold frequency = minimum frequency; work function = minimum energy. Miss 'minimum' and miss the mark.

  Use it when: Definitions.

• de Broglie: h over p

  λ = h/p; with energy, put p = √(2mK). Quadruple K → half λ.

  Use it when: Ratio MCQs.

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.

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 5-mark question split into 1 + 1 + 2 + 1 marks.

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.

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 changes with intensity

  CISCE examiners' comments, 2024: asked how stopping potential changes when the intensity of UV is increased, candidates said it increases or decreases; they also confused frequency with wavelength when choosing the metal with the lowest threshold frequency.

  Wrong working

  More intense light gives a larger stopping potential.

  Right working

  $V_s$ is unchanged: it depends on f and $W_0$ only. The metal with the smallest work function has the lowest $f_0$.

  ✓Do this: Write $e{V}_s=hf-W_0$ before answering.

  Usually costs: 1 mark each.

• ✗Mistake: Dropping 'minimum' from the definitions

  CISCE examiners' comments, 2025: candidates missed the key word 'minimum' in defining threshold frequency and work function, confused it with a maximum cut-off, or wrote 'photons' instead of electrons being emitted.

  Wrong working

  Threshold frequency is the frequency at which photons are emitted.

  Right working

  Threshold frequency is the minimum frequency of incident light that can eject (photo)electrons from the metal.

  ✓Do this: Underline 'minimum' and 'electrons' in your definitions.

  Usually costs: 1 mark each.

• ✗Mistake: Drawing a curve or calling the slope h

  CISCE examiners' comments, 2025: for the $V_s$-f graph, axes were missing or wrongly labelled, a curve was drawn, and candidates said the slope itself is Planck's constant.

  Wrong working

  A curve through the origin; h = slope.

  Right working

  A straight line starting at $f_0$ on the f-axis; h = e × slope.

  ✓Do this: Derive $V_s=(h/e)f-W_0/e$ next to the graph.

  Usually costs: 1-2 marks.

• ✗Mistake: Wrong de Broglie relation in terms of kinetic energy

  CISCE examiners' comments, 2024 and 2025: candidates chose $\frac{2h}{\sqrt{2mE}}$ instead of $\frac{h}{\sqrt{2mE}}$, and could not say how λ changes when K is made four times.

  Wrong working

  K four times, so λ becomes four times smaller.

  Right working

  $\lambda =\frac{h}{\sqrt{2mK}}\propto K^{-1/2}$: four times K halves λ.

  ✓Do this: Derive from p = √(2mK) each time.

  Usually costs: 1 mark.

• ✗Mistake: Naming an experiment instead of a phenomenon

  CISCE examiners' comments, 2025: asked for a phenomenon in which particles behave as waves, candidates wrote 'de Broglie waves' or 'Davisson-Germer experiment', or gave a particle-nature effect.

  Wrong working

  Davisson-Germer experiment.

  Right working

  Electron diffraction (by a crystal or thin foil).

  ✓Do this: A phenomenon is something that happens (diffraction); an experiment is how it was shown.

  Usually costs: 1 mark.

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.

1. Q1 · 1 mark · MCQBoard level

   The momentum of a photon of wavelength λ is:
   (a) hλ
   (b) h/λ
   (c) hc/λ
   (d) λ/h

   Hint

   p = E/c.

   Answer

   (b) h/λ

2. Q2 · 1 mark · MCQBoard level

   The maximum kinetic energy of photoelectrons depends on:
   (a) intensity of light
   (b) frequency of light
   (c) time of exposure
   (d) area of the surface

   Hint

   Einstein's equation.

   Answer

   (b)

3. Q3 · 1 mark · MCQBoard level

   The de Broglie wavelength of an electron accelerated through V volts is approximately:
   (a) $\frac{1.227}{V}$ nm
   (b) $\frac{1.227}{\sqrt{V}}$ nm
   (c) $1.227\sqrt{V}$ nm
   (d) $\frac{12.27}{V^2}$ nm

   Hint

   λ ∝ 1/√V.

   Answer

   (b)

4. Q4 · 1 mark · Very short answerBoard level

   Define work function of a metal.

   Hint

   Minimum energy.

   Answer

   The minimum energy required to free an electron from the surface of the metal.

5. Q5 · 2 marks · Short answerBoard level

   Find the energy (in eV) and momentum of a photon of wavelength 500 nm.

   Hint

   E = hc/λ; p = h/λ.

   Answer

   2.49 eV ($3.98\times 10^{-19}$ J); $1.33\times 10^{-27}kgm{s}^{-1}$

6. Q6 · 2 marks · Short answerHarder board level

   Find the de Broglie wavelength of an electron accelerated through 100 V.

   Hint

   $\lambda =h/\sqrt{2meV}$.

   Answer

   $1.23\times 10^{-10}$ m (0.123 nm)

7. Q7 · 2 marks · Short answerHarder board level

   Draw graphs of photocurrent against collector potential for (a) two intensities at the same frequency, (b) two frequencies at the same intensity. Explain the differences.

   Hint

   Saturation current versus stopping potential.

   Answer

   (a) Same stopping potential, higher saturation current for higher intensity (more photons). (b) Different stopping potentials (more negative for higher f, since K_max is larger); similar saturation currents.

8. Q8 · 2 marks · Short answerHOTS / competency

   A 0.1 kg ball moves at $10m{s}^{-1}$. Find its de Broglie wavelength and explain why its wave nature is never observed.

   Hint

   Compare with sizes of any aperture.

   Answer

   $\lambda =h/mv=6.6\times 10^{-34}m$, about $10^{19}$ times smaller than an atomic nucleus; no obstacle or slit is small enough to diffract it, so it behaves as a pure particle.

9. Q9 · 3 marks · Short answerHarder board level

   A 2 mW source emits 500 nm light. Find the number of photons emitted per second. If one photon in 1000 ejects an electron from a photocathode, find the saturation photocurrent.

   Hint

   N = P/E; I = ne.

   Answer

   $N=5.0\times 10^{15}{s}^{-1}$; electrons $5.0\times 10^{12}{s}^{-1}$; $I=5.0\times 10^{12}\times 1.6\times 10^{-19}=8.0\times 10^{-7}A$ (0.8 µA).

10. Q10 · 3 marks · Short answerHarder board level

    An electron microscope accelerates electrons through 50 kV. Find their de Broglie wavelength (non-relativistic) and say why electron microscopes resolve finer detail than optical ones.

    Hint

    Resolving power depends on λ.

    Answer

    $\lambda =5.5\times 10^{-12}$ m. Resolving power is better for shorter wavelength; this is about $10^5$ times shorter than visible light, so much finer details can be distinguished.

11. Q11 · 3 marks · Short answerHarder board level

    Describe the Davisson-Germer experiment qualitatively and state its conclusion. Name one phenomenon that shows the wave nature of electrons.

    Hint

    Electron gun, nickel crystal, detector.

    Answer

    A beam of electrons accelerated through about 54 V struck a nickel crystal; a movable detector measured the scattered intensity at different angles and found a sharp maximum near 50°, like X-ray diffraction. The wavelength from the diffraction matched h/p, confirming de Broglie's hypothesis. Phenomenon: electron diffraction.

12. Q12 · 5 marks · Long answerHarder board level

    (a) Write Einstein's photoelectric equation and use it to explain the existence of a threshold frequency and the independence of $K_{max}$ from intensity. [3]
    (b) A metal has work function 2.0 eV. Find its threshold wavelength and the stopping potential for 400 nm light. [2]

    Hint

    hc = 1240 eV nm.

    Answer

    (a) $K_{max}=hf-W_0$. If $hf<W_0$ no single photon can free an electron, so there is a minimum frequency $f_0=W_0/h$. $K_{max}$ depends on the energy of one photon (hf), not on how many arrive, so it is independent of intensity. (b) ${\lambda }_0=620$ nm; $V_s=3.1-2.0=1.1$ V.

13. Q13 · 2 marks · Short answerHarder board level

    A 5 mW laser beam is completely absorbed by a surface. Find the force it exerts.

    Hint

    Momentum per second = P/c.

    Answer

    $F=P/c=5\times 10^{-3}/3\times 10^8=1.7\times 10^{-11}N$

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. 1.Einstein's photoelectric equation?Show answer

   $K_{max}=hf-W_0=e{V}_s$.

2. 2.Define stopping potential.Show answer

   The minimum negative (retarding) potential of the collector that just stops the most energetic photoelectrons.

3. 3.Slope of the $V_s$-f graph?Show answer

   h/e.

4. 4.Momentum of a photon?Show answer

   $p=h/\lambda =E/c$.

5. 5.de Broglie wavelength in terms of kinetic energy?Show answer

   $\lambda =h/\sqrt{2mK}$.

6. 6.Name a phenomenon showing the particle nature of light.Show answer

   The photoelectric effect.

7. 7.Name a phenomenon showing the wave nature of particles.Show answer

   Electron diffraction.

8. 8.Conclusion of the Davisson-Germer experiment?Show answer

   Electrons are diffracted by a crystal; their wavelength equals h/p, confirming matter waves.

Quick revision checklist

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

• ☐Can you calculate photon energy in eV with 1240/λ?
• ☐Can you list the four experimental facts of the photoelectric effect?
• ☐Can you explain each fact with Einstein's equation and say why wave theory fails?
• ☐Can you define threshold frequency and work function with the word 'minimum'?
• ☐Can you draw the $V_s$-f graph and get h from its slope?
• ☐Can you say how intensity and frequency affect current and stopping potential?
• ☐Can you write λ = h/p, h/√(2mK) and h/√(2mqV) and use them in ratios?
• ☐Can you describe the Davisson-Germer experiment and its conclusion?

Dual Nature of Radiation and Matter: frequently asked questions