Cathode rays | Thomson's electron | Millikan's experiment
Discovery
🔬 Cathode Rays & Electron
William Crookes discovered cathode rays (1870). J.J. Thomson (1897) identified them as electrons — negatively charged particles. Measured e/m = 1.76 × 10¹¹ C/kg. Thomson won Nobel Prize (1906).
Millikan
💧 Oil-Drop Experiment (1913)
R.A. Millikan measured precise charge on electron: e = 1.602 × 10⁻¹⁹ C. Showed electric charge is quantised (always integer multiples of e). Won Nobel Prize 1923.
Key Values
📐 Important Constants
Charge of electron: e = 1.6 × 10⁻¹⁹ C Mass of electron: m = 9.1 × 10⁻³¹ kg 1 eV = 1.6 × 10⁻¹⁹ J Planck's constant: h = 6.63 × 10⁻³⁴ J·s
🎯 Practice MCQs — Section 11.1
Q1. The specific charge (e/m) of an electron is: Easy
A) 1.6 × 10¹⁹ C/kg
B) 1.76 × 10¹¹ C/kg
C) 9.1 × 10⁻³¹ C/kg
D) 6.63 × 10⁻³⁴ C/kg
The specific charge e/m of electron = 1.76 × 10¹¹ C/kg, as measured by J.J. Thomson. This value is independent of the cathode material or gas in the discharge tube, confirming the universality of electrons. Option B.
Q2. 1 electron volt (eV) in joules is: Easy
A) 1.6 × 10⁻¹⁷ J
B) 1.6 × 10⁻¹⁹ J
C) 9.1 × 10⁻³¹ J
D) 6.63 × 10⁻³⁴ J
1 eV = energy gained by electron accelerated through 1 V = e × V = 1.6×10⁻¹⁹ × 1 = 1.6×10⁻¹⁹ J. Option B.
Q3. Millikan's oil-drop experiment established that: Easy
A) Light is a wave
B) Electrons have dual nature
C) Electric charge is quantised
D) Electrons are positively charged
Millikan found charge on oil drops is always an integer multiple of 1.6×10⁻¹⁹ C, establishing that electric charge is quantised (comes in discrete units of e). Option C.
⚡
11.2 Electron Emission & Work Function
Work function | Thermionic | Field | Photoelectric emission
Work Function
🔋 Work Function (φ₀)
Minimum energy required to pull an electron OUT of a metal surface. Denoted by φ₀, measured in eV. Different for each metal. Electrons are held inside by ion attraction.
Types
🔌 Types of Electron Emission
Thermionic: By heating the metal Field emission: By strong electric field (~10⁸ V/m) Photoelectric: By light of suitable frequency → produces photoelectrons
Values
📊 Work Functions of Metals
Caesium: 2.14 eV (lowest — sensitive to visible light) Sodium: 2.75 eV Zinc: 4.31 eV Copper: 4.65 eV Silver: 4.74 eV Platinum: 5.65 eV (highest)
Work Function
φ₀ = minimum energy for electron to escape from metal surface (in eV)
Energy Unit
1 eV = 1.6 × 10⁻¹⁹ J (energy gained by electron through 1 volt potential difference)
🎯 Practice MCQs — Section 11.2
Q4. Which type of electron emission uses heat energy? Easy
A) Photoelectric emission
B) Field emission
C) Thermionic emission
D) Secondary emission
Thermionic emission uses thermal (heat) energy to give electrons sufficient energy to overcome the work function and escape from the metal surface. Used in vacuum tubes and CRT displays. Option C.
Q5. The work function of caesium is 2.14 eV. In joules this equals: Moderate
A) 2.14 × 10⁻¹⁹ J
B) 3.42 × 10⁻¹⁹ J
C) 2.14 × 10⁻³⁴ J
D) 1.6 × 10⁻¹⁹ J
φ₀ = 2.14 eV × 1.6×10⁻¹⁹ J/eV = 3.424 × 10⁻¹⁹ J ≈ 3.42 × 10⁻¹⁹ J. Option B.
Q6. Which metal has the lowest work function and is most sensitive to visible light? Easy
A) Platinum
B) Zinc
C) Caesium
D) Copper
Caesium has the lowest work function (2.14 eV) among common metals and is sensitive even to visible light. Alkali metals (Li, Na, K, Cs) have low work functions. Option C.
☀️
11.3 & 11.4 Photoelectric Effect — Experimental Study
Hertz discovery | Laws of photoelectric effect | Stopping potential | Saturation current
Discovery
🔴 Hertz's Discovery (1887)
Discovered photoelectric emission accidentally during EM wave experiments. UV light enhanced sparks. Hallwachs (1888) and Lenard (1886-1902) studied it in detail. UV → negatively charged particles emitted from zinc plate.
Key Terms
📋 Important Terms
Photocurrent: Current due to photoelectrons Saturation current: Max photocurrent when all electrons reach collector Stopping potential (V₀): Min retarding potential to stop all photoelectrons Threshold frequency (ν₀): Min freq for photoelectric emission
4 Laws
📜 Laws of Photoelectric Effect
1. Photocurrent ∝ intensity (for ν > ν₀) 2. Stopping potential independent of intensity 3. Below threshold frequency ν₀ → no emission (any intensity) 4. Emission is instantaneous (~10⁻⁹ s)
Stopping Potential
Kmax = eV₀ → V₀ = Kmax/e
V₀ depends on frequency of light, NOT on intensity. Same V₀ for different intensities at same frequency.
Key Relationships
Photocurrent ∝ Intensity (at fixed frequency above ν₀)
Saturation current ∝ Intensity
Stopping potential V₀ increases linearly with frequency ν
V₀ is independent of intensity
🖼️[NCERT Fig. 11.1: Experimental setup — emitter C, collector A, quartz window | Fig. 11.2: Photocurrent vs Intensity (linear) | Fig. 11.3: Photocurrent vs potential for I₁,I₂,I₃ (same V₀) | Fig. 11.4: Photocurrent vs potential for ν₁,ν₂,ν₃ | Fig. 11.5: V₀ vs ν (linear)]
Add photoelectric effect experimental graphs here
🎯 Practice MCQs — Sections 11.3 & 11.4
Q7. In a photoelectric experiment, if the intensity of incident light is doubled (at fixed frequency above threshold), the saturation current: Easy
A) Halves
B) Doubles
C) Remains same
D) Quadruples
Saturation current ∝ intensity (more photons → more photoelectrons per second). Doubling intensity doubles the number of photoelectrons, hence saturation current doubles. Option B.
Q8. The stopping potential in a photoelectric experiment is 1.5 V. The maximum kinetic energy of photoelectrons is: Easy
A) 1.5 J
B) 3.0 eV
C) 1.5 eV
D) 0.75 eV
Kmax = eV₀ = e × 1.5 = 1.5 eV. (NCERT Ex 11.3 — cut-off voltage 1.5V → Kmax = 1.5 eV). Option C.
Q9. If frequency of incident light in a photoelectric experiment is doubled (keeping intensity same), the stopping potential: Moderate
A) Doubles
B) Halves
C) Increases (but not exactly doubles)
D) Remains unchanged
V₀ = (h/e)ν − φ₀/e. So V₀ vs ν is linear but doesn't double when ν doubles (because of the work function constant term). Stopping potential increases with frequency, not simply doubles. Option C.
⚡
11.5 & 11.6 Wave Theory Failure & Einstein's Equation
1. Kmax should increase with intensity — but it doesn't ❌ 2. No threshold frequency should exist — but it does ❌ 3. Emission should be delayed (hours!) — but it's instant ❌ Wave theory cannot explain any of these!
Einstein 1905
💡 Einstein's Photon Concept
Light consists of discrete energy packets called photons. Energy of one photon = hν. One photon knocks out one electron. Intensity = number of photons/sec (not energy per photon). Nobel Prize 1921.
E = hν = hc/λ | p = hν/c = h/λ
h = 6.626 × 10⁻³⁴ J·s (Planck's constant)
🖼️[Graph: V₀ vs ν — straight line with slope h/e, x-intercept at ν₀ for different metals (Metal A and B with different threshold frequencies)]
Add Einstein's photoelectric equation verification graph here
🎯 Practice MCQs — Sections 11.5 & 11.6
Q10. The work function of caesium is 2.14 eV. The threshold frequency is: (h = 6.63×10⁻³⁴ J·s) Moderate
A) 3.0 × 10¹⁴ Hz
B) 5.16 × 10¹⁴ Hz
C) 8.2 × 10¹⁴ Hz
D) 1.0 × 10¹⁵ Hz
ν₀ = φ₀/h = (2.14 × 1.6×10⁻¹⁹)/(6.63×10⁻³⁴) = (3.424×10⁻¹⁹)/(6.63×10⁻³⁴) = 5.16 × 10¹⁴ Hz. (NCERT Ex 11.2a) Option B.
Q11. The threshold frequency for a metal is 3.3×10¹⁴ Hz. Light of 8.2×10¹⁴ Hz is incident. The stopping potential is: (h = 6.63×10⁻³⁴ J·s) Hard
A) 1.5 V
B) 2.0 V
C) 3.3 V
D) 5.0 V
eV₀ = h(ν−ν₀) = 6.63×10⁻³⁴ × (8.2−3.3)×10¹⁴ = 6.63×10⁻³⁴ × 4.9×10¹⁴ = 3.249×10⁻¹⁹ J. V₀ = 3.249×10⁻¹⁹/(1.6×10⁻¹⁹) ≈ 2.0 V. (NCERT Ex 11.6) Option B.
Q12. In photoelectric effect, the slope of V₀ vs frequency (ν) graph gives: Easy
A) Work function φ₀
B) Planck's constant h
C) h/e (Planck's constant / charge)
D) Charge e
From Einstein's equation: V₀ = (h/e)ν − φ₀/e. This is y = mx + c form. Slope = h/e. Knowing e, Planck's constant h can be determined. This was Millikan's method. Option C.
🔆
11.7 Particle Nature of Light — The Photon
Photon properties | Energy | Momentum | Compton effect
Photon
🔆 What is a Photon?
Discrete quantum of electromagnetic radiation. Has both energy and momentum — treated as a particle. Compton (1924) confirmed photon momentum by X-ray scattering from electrons.
Properties
📋 Photon Properties
✅ Energy: E = hν = hc/λ ✅ Momentum: p = hν/c = h/λ ✅ Speed = c (speed of light) ✅ Electrically neutral ✅ Not deflected by E or B fields ✅ Energy and momentum conserved in collisions
Key Point
⚡ Photon Energy vs Intensity
All photons of same frequency have same energy (hν), regardless of intensity. Intensity only determines number of photons per second, not energy per photon. Increasing intensity → more electrons (more photons), not faster electrons.
Photon Energy and Momentum
E = hν = hc/λ
p = hν/c = h/λ (momentum of photon)
Energy of Photon (Laser Problem)
Power P = N × E = N × hν → N = P/(hν) (photons per second)
🎯 Practice MCQs — Section 11.7
Q13. A laser of frequency 6.0×10¹⁴ Hz has power 2×10⁻³ W. Number of photons emitted per second is: (h = 6.63×10⁻³⁴ J·s) Moderate
A) 2 × 10¹² /s
B) 1.0 × 10¹⁴ /s
C) 5.0 × 10¹⁵ /s
D) 3.0 × 10¹⁸ /s
E = hν = 6.63×10⁻³⁴ × 6×10¹⁴ = 3.978×10⁻¹⁹ J. N = P/E = 2×10⁻³/(3.978×10⁻¹⁹) ≈ 5×10¹⁵ photons/s. (NCERT Ex 11.1) Option C.
Q14. The momentum of a photon of wavelength 600 nm is: Moderate
A) 3.32 × 10⁻²⁷ kg·m/s
B) 1.1 × 10⁻²⁷ kg·m/s
C) 6.63 × 10⁻³⁴ kg·m/s
D) 5.0 × 10⁻¹⁹ kg·m/s
p = h/λ = 6.63×10⁻³⁴/(600×10⁻⁹) = 6.63×10⁻³⁴/6×10⁻⁷ = 1.105×10⁻²⁷ ≈ 1.1×10⁻²⁷ kg·m/s. Option B.
Q15. Photons are NOT deflected by: Easy
A) Gravitational fields
B) Electric and magnetic fields
C) Glass prisms
D) Mirrors
Photons are electrically neutral — they carry no charge and are therefore not deflected by electric or magnetic fields. They ARE affected by gravity (gravitational lensing) and by optical elements. Option B.
🌊
11.8 Wave Nature of Matter — de Broglie Hypothesis
Matter waves | de Broglie wavelength | Dual nature of particles
Hypothesis (1924)
🌊 de Broglie's Bold Idea
Louis de Broglie (1924): If radiation has dual (wave + particle) nature, then matter too should have wave-like nature. Moving particles have associated wavelength. Nobel Prize 1929.
Formula
📐 de Broglie Wavelength
λ = h/p = h/(mv) where p = momentum of particle, m = mass, v = speed. Smaller for heavier/faster particles. For macroscopic objects, λ is unmeasurably tiny → no observable wave effects.
Significance
⚛️ Why Electrons Show Waves
Electron: m = 9.1×10⁻³¹ kg (tiny) → λ is measurable (~0.1 nm, comparable to X-rays and crystal planes). Macroscopic ball: m = 0.1 kg → λ ≈ 10⁻³⁴ m (unmeasurable). Wave effects only in quantum/atomic domain.
de Broglie Relation
λ = h/p = h/(mv)
h = 6.63×10⁻³⁴ J·s, p = momentum (kg·m/s)
For Electron Accelerated through Potential V
KE = eV = ½mv² → p = mv = √(2meV)
λ = h/√(2meV)
de Broglie wavelength of a Photon
For photon: p = hν/c = h/λ → λ = h/p ✓ (consistent!)
🎯 Practice MCQs — Section 11.8
Q16. An electron moves at 5.4×10⁶ m/s. Its de Broglie wavelength is: (m = 9.11×10⁻³¹ kg, h = 6.63×10⁻³⁴ J·s) Moderate
A) 0.5 nm
B) 0.135 nm
C) 1.47 nm
D) 2.76 nm
p = mv = 9.11×10⁻³¹ × 5.4×10⁶ = 4.92×10⁻²⁴ kg·m/s. λ = h/p = 6.63×10⁻³⁴/4.92×10⁻²⁴ = 0.135 nm. (NCERT Ex 11.3a) Option B.
Q17. A ball of mass 150 g moves at 30 m/s. Its de Broglie wavelength is: Moderate
A) 0.135 nm
B) 2.76 × 10⁻³⁴ m
C) 1.47 × 10⁻³⁴ m
D) 6.63 × 10⁻³⁴ m
p = mv = 0.150 × 30 = 4.50 kg·m/s. λ = h/p = 6.63×10⁻³⁴/4.50 = 1.47×10⁻³⁴ m. (NCERT Ex 11.3b) This is ~10⁻¹⁹ times the size of a proton — completely unmeasurable! Option C.
Q18. If an electron is accelerated through a potential of 100 V, its de Broglie wavelength is approximately: (m=9.1×10⁻³¹ kg, e=1.6×10⁻¹⁹ C, h=6.63×10⁻³⁴) Hard