NEET UG Physics · Chapter 22

Dual Nature of Radiation and Matter

Photoelectric effect, Einstein’s equation, photons and de Broglie waves.

NCERT Class 12NEET: 1–2 Qs per paper 11 min read

1. Chapter at a Glance

Light sometimes behaves like a wave (interference, diffraction) and sometimes like a stream of particles called photons (photoelectric effect). Matter shows the same dual behaviour: moving electrons have a wavelength. This chapter covers electron emission, the photoelectric effect and Einstein's equation, the photon picture, and de Broglie's matter waves. It is short but very scoring: NEET asks two questions from it almost every year.

What NEET asks most
  • Photon energy E = hν = hc/λ (use 1240 eV nm).
  • Einstein's photoelectric equation; stopping potential and threshold wavelength.
  • Graphs: photocurrent vs voltage, stopping potential vs frequency.
  • Effect of intensity and frequency on photoemission.
  • de Broglie wavelength of electrons, protons and α-particles.

2. Electron Emission

  • Free electrons in a metal need a minimum energy, the work function φ₀, to escape from its surface. It is measured in eV.
  • They can gain this energy through heating (thermionic emission), a strong electric field (field emission), or light (photoelectric emission).
  • Caesium has a low work function (~2.1 eV), so even visible light can eject electrons from it. Platinum has a high work function (~5.6 eV).

3. Photoelectric Effect: Experimental Facts

  • For each metal there is a threshold frequency ν₀. Below ν₀, no electrons are emitted, however intense the light.
  • Emission is instantaneous (within about 10⁻⁹ s), even for very dim light.
  • The photocurrent (number of electrons per second) is proportional to the intensity of the light.
  • The maximum kinetic energy of the electrons depends on the frequency, not on the intensity.
  • The stopping potential V₀ is the negative plate voltage that just stops the fastest electrons: eV₀ = Kmax.

The wave theory of light cannot explain the threshold frequency, the instant emission or the independence of Kmax from intensity. The photon theory explains all of them.

4. Einstein's Photoelectric Equation

hν = φ₀ + Kmax    Kmax = eV₀ = h(ν − ν₀)    φ₀ = hν₀ = hc/λ₀
  • One photon gives all its energy to one electron. Part of it is used to escape (φ₀), and the rest becomes kinetic energy.
  • A graph of V₀ against ν is a straight line with slope h/e and x-intercept ν₀. The slope is the same for all metals.
  • Photocurrent against collector voltage: for the same frequency but higher intensity, the saturation current is larger and the stopping potential is the same. For higher frequency at the same intensity, the stopping potential is more negative and the saturation current is the same.
ν₀(A)ν₀(B) metal Ametal B νV₀ slope = h/e (same for both)
V₀ against ν for two metals: parallel lines with the same slope h/e. Metal B has the larger work function, so its threshold frequency is higher.

5. The Photon

E = hν = hc/λ    E (eV) = 1240/λ (nm)    p = h/λ = E/c
  • h = 6.63 × 10⁻³⁴ J s. A photon has zero rest mass and always travels at c in vacuum.
  • Photons are electrically neutral, so electric and magnetic fields do not deflect them.
  • Intensity depends on the number of photons per second: number per second n = P/(hν).
  • In photon–electron collisions, energy and momentum are conserved, but the number of photons need not be.

6. de Broglie Matter Waves

λ = h/p = h/mv = h/√(2mK)    electron accelerated through V: λ = 1.227/√V nm
  • Every moving particle has a wavelength. For everyday objects it is far too small to detect; for electrons it is comparable to atomic spacing.
  • For a charge q accelerated from rest through V: λ = h/√(2mqV).
  • Heavier particles with the same kinetic energy have shorter wavelengths.
  • The wave nature of electrons is used in the electron microscope, which has a much better resolution than an optical microscope.
  • The Davisson–Germer experiment confirmed electron diffraction.

7. Formula Master Table

QuantityFormula
Photon energyE = hν = hc/λ = 1240/λ(nm) eV
Photon momentump = h/λ
Photons per secondn = P/(hν)
Einstein's equationKmax = hν − φ₀
Stopping potentialeV₀ = Kmax
Work functionφ₀ = hν₀ = hc/λ₀
de Broglie wavelengthλ = h/mv = h/√(2mK)
Accelerated chargeλ = h/√(2mqV)
Electronλ = 1.227/√V nm

8. Master Summary Table

If you increase…PhotocurrentKmax / V₀NEET Note
Intensity (ν fixed, ν > ν₀)IncreasesNo changeMore photons, same energy each
Frequency (intensity fixed)About the sameIncreases linearlyKmax not ∝ ν (offset φ₀)
Work function—DecreasesThreshold ν₀ rises
Frequency below ν₀Zero—No emission at any intensity
Accelerating voltage (electron)——λ ∝ 1/√V

9. Mind Map

Dual Nature
Emission
  • Work function
  • Thermionic, field, photo
Photoelectric effect
  • Threshold ν₀
  • Stopping potential
  • Graphs
Photon
  • E = hν
  • p = h/λ
  • n = P/hν
Matter waves
  • λ = h/p
  • 1.227/√V nm
  • Electron microscope

10. Tips and Tricks

Shortcuts
  • E (eV) = 1240/λ (nm) saves time; e.g. 620 nm ⇒ 2 eV.
  • Kmax in eV equals V₀ in volts numerically.
  • For the same V: λ ∝ 1/√(mq). For the same K: λ ∝ 1/√m. For the same speed: λ ∝ 1/m.
  • λp : λα through the same potential = 2√2 : 1.
Common NEET traps
  • Saying brighter light gives faster electrons (it gives more electrons).
  • Thinking doubling ν doubles Kmax (it more than doubles, because of φ₀).
  • Using 1.227/√V for protons (it is only for electrons).
  • Forgetting that the V₀–ν slope is the same for every metal.

11. Important One-liners

  • The photoelectric effect supports the particle nature of light.
  • Photoemission is instantaneous.
  • The saturation photocurrent depends on intensity; the stopping potential depends on frequency.
  • The slope of the V₀–ν graph is h/e.
  • A photon has energy and momentum but zero rest mass.
  • The de Broglie wavelength of a particle at rest is infinite (undefined).
  • Electron microscopes use the wave nature of electrons.
  • Alkali metals like caesium emit electrons even with visible light.

12. Solved Numericals: Average to Advanced

Five average and five advanced numericals. Take h = 6.63 × 10⁻³⁴ J s, c = 3 × 10⁸ m s⁻¹, e = 1.6 × 10⁻¹⁹ C and hc = 1240 eV nm. Click to reveal each answer and working.

Average level

N1Average

Find the energy of a photon of violet light of wavelength 400 nm, in eV and in joules.

Show answer & solution

Concept: E = 1240/λ(nm) eV.

  1. E = 1240/400 = 3.1 eV.
  2. E = 3.1 × 1.6 × 10⁻¹⁹ ≈ 4.97 × 10⁻¹⁹ J.

Answer: 3.1 eV; about 4.97 × 10⁻¹⁹ J

N2Average

Ultraviolet light of wavelength 310 nm falls on a metal of work function 2.8 eV. Find the maximum kinetic energy of the photoelectrons and the stopping potential.

Show answer & solution

Concept: Kmax = hν − φ₀.

  1. Photon energy = 1240/310 = 4.0 eV.
  2. Kmax = 4.0 − 2.8 = 1.2 eV.
  3. V₀ = Kmax/e = 1.2 V.

Answer: 1.2 eV; 1.2 V

N3Average

The work function of a metal is 2.0 eV. Find its threshold wavelength. Will orange light of 600 nm eject electrons from it?

Show answer & solution

Concept: λ₀ = hc/φ₀.

  1. λ₀ = 1240/2.0 = 620 nm.
  2. 600 nm < 620 nm, so its photons (2.07 eV) exceed φ₀: yes, electrons are emitted.

Answer: 620 nm; yes

N4Average

Find the de Broglie wavelength of an electron accelerated from rest through 400 V.

Show answer & solution

Concept: λ = 1.227/√V nm (electrons only).

  1. √400 = 20.
  2. λ = 1.227/20 ≈ 0.0613 nm.

Answer: about 0.061 nm

N5Average

A 10 mW laser emits light of wavelength 632.8 nm. How many photons does it emit each second?

Show answer & solution

Concept: n = P/E.

  1. E = 1240/632.8 ≈ 1.96 eV ≈ 3.14 × 10⁻¹⁹ J.
  2. n = 0.01/3.14 × 10⁻¹⁹ ≈ 3.2 × 10¹⁶ per second.

Answer: about 3.2 × 10¹⁶ photons per second

Advanced level

N6Advanced

For a certain metal, the stopping potential is 1.1 V for light of 400 nm and 2.13 V for light of 300 nm. Find the work function and threshold wavelength, and estimate Planck's constant from the data.

Show answer & solution

Concept: eV₀ = hν − φ₀ for each wavelength.

  1. For 400 nm: E = 3.1 eV ⇒ φ₀ = 3.1 − 1.1 = 2.0 eV. (Check with 300 nm: 4.13 − 2.0 = 2.13 ✓.)
  2. λ₀ = 1240/2.0 = 620 nm.
  3. ν₁ = 7.5 × 10¹⁴ Hz, ν₂ = 1.0 × 10¹⁵ Hz; h = e(ΔV₀)/(Δν) = 1.6 × 10⁻¹⁹ × 1.03/2.5 × 10¹⁴ ≈ 6.6 × 10⁻³⁴ J s.

Answer: 2.0 eV; 620 nm; h ≈ 6.6 × 10⁻³⁴ J s

N7Advanced

A proton and an α-particle are accelerated from rest through the same potential difference. Find the ratio of their de Broglie wavelengths.

Show answer & solution

Concept: λ = h/√(2mqV) ∝ 1/√(mq).

  1. mα = 4mp, qα = 2e.
  2. λp/λα = √(mαqα/mpqp) = √(4 × 2) = 2√2.

Answer: λp : λα = 2√2 : 1

N8Advanced

An electron and a photon both have a wavelength of 1 nm. Find the energy of each (me = 9.11 × 10⁻³¹ kg).

Show answer & solution

Concept: photon E = hc/λ; electron K = p²/2m with p = h/λ.

  1. Photon: E = 1240/1 = 1240 eV.
  2. Electron: p = 6.63 × 10⁻³⁴/10⁻⁹ = 6.63 × 10⁻²⁵ kg m s⁻¹.
  3. K = (6.63 × 10⁻²⁵)²/(2 × 9.11 × 10⁻³¹) ≈ 2.41 × 10⁻¹⁹ J ≈ 1.51 eV.
  4. For the same wavelength, the photon carries far more energy.

Answer: photon 1240 eV; electron about 1.5 eV

N9Advanced

Light of photon energy 3 eV ejects electrons with Kmax = 1 eV from a metal. If the frequency of the light is doubled, what is the new Kmax? Is it double the old value?

Show answer & solution

Concept: Kmax = hν − φ₀; φ₀ stays fixed.

  1. φ₀ = 3 − 1 = 2 eV.
  2. Doubled frequency ⇒ photon energy 6 eV ⇒ Kmax = 6 − 2 = 4 eV.
  3. 4 eV is four times, not twice, the old value, because of the fixed work function.

Answer: 4 eV; no, it becomes four times

N10Advanced

A 5 mW beam of 400 nm light falls on a photocell. If only 1% of the photons eject an electron, find the photocurrent.

Show answer & solution

Concept: photons per second n = P/E; current I = (electrons per second) × e.

  1. E = 3.1 eV = 4.96 × 10⁻¹⁹ J; n = 5 × 10⁻³/4.96 × 10⁻¹⁹ ≈ 1.0 × 10¹⁶ s⁻¹.
  2. Electrons per second = 1% of this ≈ 1.0 × 10¹⁴ s⁻¹.
  3. I = 1.0 × 10¹⁴ × 1.6 × 10⁻¹⁹ ≈ 1.6 × 10⁻⁵ A = 16 μA.

Answer: about 16 μA

13. NEET-pattern Questions with Detailed Answers

These questions are original and based on concepts that NEET UG asks again and again.

Q1Easy

The maximum kinetic energy of photoelectrons depends on:

  1. The intensity of light
  2. The frequency of light
  3. The time of exposure
  4. The area of the metal
Show solution

Kmax = hν − φ₀ depends only on the frequency (and on the metal). Intensity changes the number of electrons, not their maximum energy.

Answer: (B)

Q2Easy

The slope of the graph of stopping potential against frequency is:

  1. h
  2. e/h
  3. h/e
  4. φ₀/e
Show solution

V₀ = (h/e)ν − φ₀/e, so the slope is h/e. φ₀/e is the size of the intercept on the V₀-axis.

Answer: (C)

Q3Medium

If the intensity of incident light is doubled (frequency above threshold kept the same), the stopping potential:

  1. Doubles
  2. Halves
  3. Stays the same
  4. Becomes four times
Show solution

V₀ depends only on the frequency. Doubling the intensity doubles the saturation current but does not change V₀.

Answer: (C)

Q4Medium

An electron, a proton and an α-particle have the same kinetic energy. The one with the longest de Broglie wavelength is:

  1. Electron
  2. Proton
  3. α-particle
  4. All equal
Show solution

λ = h/√(2mK) ∝ 1/√m when K is the same. The electron is the lightest, so it has the longest wavelength.

Answer: (A)

Q5Medium

The energy of a photon of wavelength 620 nm is about:

  1. 1 eV
  2. 2 eV
  3. 3 eV
  4. 4 eV
Show solution

E = 1240/620 = 2 eV. This red-orange light is at the low-energy end of the visible spectrum.

Answer: (B)

Q6Medium

If the accelerating voltage of an electron beam is increased four times, its de Broglie wavelength becomes:

  1. Four times
  2. Twice
  3. Half
  4. One-quarter
Show solution

λ ∝ 1/√V. Making V four times larger halves λ. (D) wrongly assumes λ ∝ 1/V.

Answer: (C)

Q7Hard

Light of frequency 1.5 times the threshold frequency falls on a metal. If the frequency is halved and the intensity is doubled, the photocurrent becomes:

  1. Double
  2. Half
  3. Four times
  4. Zero
Show solution

The new frequency is 0.75ν₀, which is below the threshold. No electrons are emitted, however intense the light, so the photocurrent is zero.

Answer: (D)

Q8Hard

Two metals A and B have work functions 2 eV and 4 eV. Which statement about their V₀–ν graphs is correct?

  1. They have different slopes
  2. They are parallel, with B's threshold frequency higher
  3. They intersect at the origin
  4. B's line lies above A's
Show solution

Both lines have slope h/e, so they are parallel. The x-intercept is ν₀ = φ₀/h, which is larger for B. B's line is shifted to the right, so it lies below A's line at any given frequency.

Answer: (B)

14. Quick Revision Summary

Remember before the exam
  • E = hν = 1240/λ(nm) eV; p = h/λ.
  • Kmax = hν − φ₀ = eV₀; φ₀ = hc/λ₀.
  • Intensity ⇒ current; frequency ⇒ Kmax; below ν₀ ⇒ nothing.
  • V₀–ν slope = h/e (same for all metals).
  • λ = h/p = h/√(2mK); electron: 1.227/√V nm.
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