Class 12 Physics Electrons Notes

Class 12 Physics Electrons Notes
Unit 5
Modern Physics
Class 12 Physics • Chapter 20

Electrons

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NEB/CDC syllabus scope: Chapter 20 is a 4-teaching-hour Modern Physics chapter. It covers Millikan’s oil-drop experiment and quantization of charge; motion and deflection of an electron beam in electric and magnetic fields with mathematical expressions; J.J. Thomson’s experiment for determining the electron’s specific charge; and related numerical problems.

1. Electron: Basic Idea and Properties

Electron An electron is a fundamental negatively charged particle with charge −e and rest mass me.

Electrons are constituents of atoms and are also produced as free particles in devices such as discharge tubes and electron guns.

  • Charge = −1.602 × 10⁻¹⁹ C.
  • Mass = 9.11 × 10⁻³¹ kg.
  • Magnitude of specific charge e/m ≈ 1.76 × 10¹¹ C kg⁻¹.
  • Electric fields can change their speed and direction.
  • Magnetic fields can bend their path without changing their kinetic energy when the magnetic force is perpendicular to velocity.
Notation In many textbook derivations, e denotes the positive magnitude of the electron charge. The actual electron charge is −e.

2. Important Constants and Units

QuantitySymbolValue / unit
Elementary charge magnitudee1.602 × 10⁻¹⁹ C
Electron massm9.11 × 10⁻³¹ kg
Specific charge magnitudee/m≈ 1.76 × 10¹¹ C kg⁻¹
Electric fieldEN C⁻¹ or V m⁻¹
Magnetic flux densityBtesla (T)
Dynamic viscosityηPa·s

3. Millikan’s Oil-Drop Experiment

Purpose Millikan’s oil-drop experiment measured the elementary electronic charge and demonstrated the quantization of electric charge.

3.1 Principle

The experiment compares the forces on a tiny charged oil drop as it moves through air. The analysis uses Stokes’ law for the viscous drag on a small sphere moving slowly through a fluid.

Stokes drag: Fv = 6πηrv

3.2 Main Apparatus

  • Two horizontal parallel metal plates separated by distance d.
  • A small hole in the upper plate for admitting atomized oil droplets.
  • A high-voltage source to produce an electric field E = V/d.
  • Illumination system and microscope to observe individual droplets.
  • A method of ionizing the air/droplets so a drop can gain or lose electronic charge.
Millikan Oil-Drop Experiment upper plate lower plate oil drop Microscope observe drop Atomizer E qE effective weight Observe terminal motion with field OFF and ON. Measured charges occur as q = ne.

Diagram 1: Simplified Millikan apparatus and force directions

4. Oil Drop Falling under Gravity

Let an oil drop have:

  • radius r,
  • oil density ρ,
  • air density σ,
  • terminal downward speed v1.

4.1 Weight, Upthrust and Viscous Drag

Weight = (4/3)πr³ρg Upthrust = (4/3)πr³σg

Hence the effective downward weight is:

Weff = (4/3)πr³(ρ − σ)g

At terminal speed:

(4/3)πr³(ρ − σ)g = 6πηrv₁

Therefore:

r = √[9ηv₁ / {2(ρ − σ)g}]
Why terminal velocity matters At terminal speed, acceleration is zero, so the forces balance and the drop radius can be calculated from its measured speed.
Falling Drop at Terminal Speed drop W upthrust viscous drag At terminal speed: weight = upthrust + drag

Diagram 2: Force balance on a falling oil drop

5. Charged Oil Drop Rising in an Electric Field

Suppose the field is adjusted so a negatively charged drop rises upward with terminal speed v2. The upward electric force is qE.

At terminal upward speed:

qE = effective weight + downward viscous drag qE = (4/3)πr³(ρ − σ)g + 6πηrv₂

From the falling case:

(4/3)πr³(ρ − σ)g = 6πηrv₁

Hence:

qE = 6πηr(v₁ + v₂) q = 6πηr(v₁ + v₂)/E

Since E = V/d:

q = 6πηrd(v₁ + v₂)/V

6. Oil Drop Held Stationary

If the electric field is adjusted until the charged oil drop is stationary, viscous drag is zero.

qE = effective weight
qE = (4/3)πr³(ρ − σ)g q = [4πr³(ρ − σ)g]/(3E)

With E = V/d:

q = 4πr³(ρ − σ)gd/(3V)
Common simplification Some short textbook treatments write qE = mg. A more complete force balance includes air buoyancy, giving the effective weight proportional to (ρ − σ).

7. Quantization of Electric Charge

Repeating the oil-drop measurement for many droplets gave charges that were integral multiples of a smallest value.

q = ne

where n = 0, 1, 2, 3, … and:

e = 1.602 × 10⁻¹⁹ C
Quantization of charge Electric charge on an isolated body occurs in discrete integral multiples of the elementary charge.
Quantization of Electric Charge −2e −e 0 +e +2e q = ne Millikan’s measurements revealed discrete charge steps.

Diagram 3: Allowed charge values in units of e

8. Motion of an Electron in a Uniform Electric Field

For a particle of charge q in electric field E:

F = qE

For an electron, q = −e, so the force is opposite to E. Its acceleration magnitude is:

a = eE/m

8.1 Motion Parallel to the Field

If the electron’s velocity is parallel or antiparallel to E, the motion remains along a straight line but its speed changes because the electric force does work.

8.2 Motion Perpendicular to the Field

If the electron enters horizontally while E is vertical, the horizontal velocity is uniform while the vertical motion has constant acceleration. The path becomes parabolic.

9. Deflection of an Electron Beam in an Electric Field

Let an electron enter between parallel plates with horizontal speed u. Take the magnitude of vertical acceleration as:

a = eE/m

Horizontal displacement:

x = ut

Vertical deflection magnitude:

y = ½at² = ½(eE/m)t²

Since t = x/u:

y = eEx²/(2mu²)

If plate potential difference is V and separation is d:

E = V/d
y = eVx²/(2mdu²)

This has the form y = kx², so the electron beam follows a parabolic path inside the uniform transverse electric field.

9.1 Deflection at the End of Plates

If plate length is l:

yl = eEl²/(2mu²)

9.2 Angle of Emergence

Vertical velocity at exit:

vy = at = (eE/m)(l/u)

Horizontal velocity remains u, therefore:

tanθ = vy/u = eEl/(mu²)
Electron Beam in a Transverse Electric Field + − E u deflection Uniform transverse E-field → constant transverse acceleration → parabola

Diagram 4: Parabolic electron deflection between charged plates

10. Motion of an Electron in a Magnetic Field

The magnetic part of the Lorentz force has magnitude:

F = evB sinθ

where θ is the angle between velocity v and magnetic field B.

10.1 Important Cases

ConditionMagnetic forceMotion
v ∥ B0Straight line
v ⟂ BevBCircular path
v at an oblique angle to Bacts on perpendicular velocity componentHelical path
Magnetic force does no work For a charged particle moving in a magnetic field, F is perpendicular to the instantaneous velocity. Thus the field changes the direction of motion but not the kinetic energy in the ideal magnetic-only case.

11. Circular Motion of an Electron in a Magnetic Field

For v perpendicular to B, magnetic force provides the centripetal force:

evB = mv²/r

Therefore:

r = mv/(eB)

11.1 Angular Frequency

Since v = ωr:

ω = eB/m

11.2 Frequency and Time Period

f = eB/(2πm) T = 2πm/(eB)

In the non-relativistic model, f and T do not depend on the electron speed or radius.

Electron Perpendicular to a Uniform Magnetic Field ×× ×× ×× ×× ×× ×× ×× ×× ×× v FB evB = mv²/r → r = mv/(eB)

Diagram 5: Magnetic force supplies the centripetal force

12. Electron Accelerated through a Potential Difference

If an electron starts from rest and is accelerated through potential difference V0, the gain in kinetic energy equals the electrical work done:

eV₀ = ½mv²

Thus:

v = √(2eV₀/m)

This relation is valid when the speed is low enough that classical, non-relativistic mechanics is adequate.

12.1 Magnetic Radius after Acceleration

Substitute v into r = mv/(eB):

r = (1/B)√(2mV₀/e)

or:

e/m = 2V₀/(B²r²)

13. Crossed Electric and Magnetic Fields

Suppose E and B are perpendicular to each other and arranged so their forces on the electron are opposite.

For zero net deflection:

electric force = magnetic force eE = evB
v = E/B

If the electric field is produced by plates with potential difference V and separation d:

E = V/d v = V/(Bd)
Velocity selector Crossed fields allow particles of speed v = E/B to pass undeflected.
Crossed-Field Velocity Selector electron beam, v electric force magnetic force × × × × B into page No deflection: eE = evB → v = E/B

Diagram 6: Opposing electric and magnetic forces select one speed

14. J.J. Thomson’s Experiment

Purpose J.J. Thomson’s cathode-ray experiment determined the electron’s specific charge e/m and established important properties of the electron.

14.1 Main Arrangement

  • Evacuated discharge tube.
  • Cathode and anode produce a fine electron beam.
  • Parallel plates provide a transverse electric field.
  • A magnetic field is applied perpendicular to both the electron velocity and electric field.
  • A fluorescent screen indicates beam position.

14.2 Principle

The electric and magnetic fields are adjusted so their deflections cancel. The electron beam then travels undeflected through the crossed-field region.

J.J. Thomson Specific-Charge Experiment Electron gun cathode → anode Crossed E and B region + − E × × × × undeflected beam Screen Tune E and B until electric and magnetic deflections cancel. Then v = E/B.

Diagram 7: Conceptual Thomson crossed-field arrangement

15. Determination of Specific Charge e/m

15.1 Step 1: Find Electron Speed

At zero deflection in crossed electric and magnetic fields:

eE = evB
v = E/B

15.2 Step 2: Use Accelerating Potential

If the electron beam was accelerated from rest through potential V0:

eV₀ = ½mv²

Hence:

e/m = v²/(2V₀)

Using v = E/B:

e/m = E²/(2V₀B²)

And if E = V/d:

e/m = V²/(2V₀B²d²)

15.3 Alternative Radius Method

If the electron speed v is known and a magnetic field bends the beam into a circle of radius r:

evB = mv²/r
e/m = v/(Br)

With v = E/B:

e/m = E/(B²r)
Expected experimental result The magnitude of the electron’s specific charge is approximately: e/m ≈ 1.76 × 10¹¹ C kg⁻¹
Thomson e/m Calculation Flow Crossed fields v = E/B Acceleration eV₀ = ½mv² Combine e/m = E²/(2V₀B²) If E = V/d → e/m = V²/(2V₀B²d²)

Diagram 8: Main equations used in Thomson’s specific-charge experiment

16. Millikan and Thomson Together

Thomson measured e/m, while Millikan measured e. Combining the two results allows the electron mass to be obtained:

m = e/(e/m)

Using e ≈ 1.602 × 10⁻¹⁹ C and e/m ≈ 1.76 × 10¹¹ C kg⁻¹ gives:

m ≈ 9.11 × 10⁻³¹ kg
ExperimentMain quantity measuredImportance
J.J. ThomsonSpecific charge e/mEstablished universal negatively charged cathode-ray particles and their charge-to-mass ratio
MillikanElementary charge eMeasured charge magnitude and demonstrated charge quantization
Combined resultsElectron mass mm = e/(e/m)

17. Worked Numericals

Example 1: Quantization of Charge

A measured oil-drop charge is 4.806 × 10⁻¹⁹ C. How many elementary charges are present?

n = q/e n = (4.806 × 10⁻¹⁹)/(1.602 × 10⁻¹⁹) = 3

Answer: q = 3e.

Example 2: Electric Acceleration

Find the magnitude of acceleration of an electron in E = 2.0 × 10⁴ V m⁻¹.

a = eE/m a = (1.602 × 10⁻¹⁹ × 2.0 × 10⁴)/(9.11 × 10⁻³¹) a ≈ 3.52 × 10¹⁵ m s⁻²
Example 3: Magnetic Radius

An electron moves at 3.0 × 10⁶ m s⁻¹ perpendicular to B = 0.020 T.

r = mv/(eB) r = (9.11 × 10⁻³¹ × 3.0 × 10⁶)/(1.602 × 10⁻¹⁹ × 0.020) r ≈ 8.53 × 10⁻⁴ m
Example 4: Velocity Selector

E = 3.0 × 10⁴ V m⁻¹ and B = 2.0 × 10⁻³ T.

v = E/B v = (3.0 × 10⁴)/(2.0 × 10⁻³) v = 1.5 × 10⁷ m s⁻¹
Example 5: Speed after Accelerating Potential

An electron is accelerated from rest through 100 V.

v = √(2eV₀/m) v = √[2(1.602 × 10⁻¹⁹)(100)/(9.11 × 10⁻³¹)] v ≈ 5.93 × 10⁶ m s⁻¹
Example 6: Specific Charge from Crossed Fields

E = 4.0 × 10⁴ V m⁻¹, B = 2.0 × 10⁻³ T and accelerating potential V₀ = 1140 V.

e/m = E²/(2V₀B²) e/m = (4.0 × 10⁴)² / [2(1140)(2.0 × 10⁻³)²] e/m ≈ 1.75 × 10¹¹ C kg⁻¹
Example 7: Electron Frequency in Magnetic Field

B = 0.050 T.

f = eB/(2πm) f = (1.602 × 10⁻¹⁹ × 0.050)/(2π × 9.11 × 10⁻³¹) f ≈ 1.40 × 10⁹ Hz

18. Complete Formula Sheet

TopicFormula
Stokes dragF = 6πηrv
Effective drop weightWeff = (4/3)πr³(ρ − σ)g
Oil-drop radiusr = √[9ηv₁/{2(ρ − σ)g}]
Rising-drop chargeq = 6πηr(v₁ + v₂)/E
Stationary-drop chargeq = 4πr³(ρ − σ)g/(3E)
Charge quantizationq = ne
Electric forceF = qE
Electron acceleration magnitudea = eE/m
Electric-field pathy = eEx²/(2mu²)
Exit angletanθ = eEl/(mu²)
Magnetic forceF = evB sinθ
Circular radiusr = mv/(eB)
Angular frequencyω = eB/m
Frequencyf = eB/(2πm)
Time periodT = 2πm/(eB)
Acceleration by potentialeV₀ = ½mv²
Electron speedv = √(2eV₀/m)
Crossed fieldsv = E/B
Thomson specific chargee/m = E²/(2V₀B²)
With E = V/de/m = V²/(2V₀B²d²)
Magnetic-radius methode/m = v/(Br)

19. Common Exam Mistakes

  • Writing the electron charge as +e. The electron charge is −e; e is only its magnitude.
  • Using Stokes drag as 6πηr²v. Correct: 6πηrv.
  • Ignoring air upthrust in the full Millikan derivation.
  • Using oil density ρ alone instead of the effective density difference (ρ − σ).
  • Forgetting that terminal velocity means net force is zero.
  • Writing measured oil-drop charge as any arbitrary value instead of q = ne.
  • Forgetting E = V/d for parallel plates.
  • Using y ∝ x in electric deflection. The transverse electric-field path is parabolic: y ∝ x².
  • Forgetting that the electron deflects opposite to the electric-field direction.
  • Using magnetic force evB for every angle. General form: evB sinθ.
  • Claiming a magnetic field changes the electron’s kinetic energy in pure magnetic deflection.
  • Forgetting that v ∥ B gives zero magnetic force.
  • Writing r = eB/(mv). Correct: r = mv/(eB).
  • Writing T proportional to v. In the non-relativistic result, T = 2πm/(eB).
  • Confusing v = E/B with e/m. v = E/B is the undeflected speed in crossed fields.
  • Using eV = mv² instead of eV = ½mv².
  • Confusing Thomson’s e/m measurement with Millikan’s e measurement.
  • Forgetting the square on E and B in e/m = E²/(2V₀B²).
  • Mixing the accelerating potential V₀ with the deflecting-plate potential V.
  • Using the relativistic regime with simple classical formulas without checking speed.

20. Important Exam Questions

Short-Answer Questions

  1. Define an electron and state its charge and mass.
  2. Define specific charge.
  3. State the purpose of Millikan’s oil-drop experiment.
  4. State Stokes’ law used in the oil-drop experiment.
  5. Why is a non-volatile oil suitable for Millikan’s experiment?
  6. What is terminal velocity?
  7. Derive the effective weight of an oil drop in air.
  8. Derive the expression for the radius of a falling oil drop.
  9. Derive the expression for charge on a rising oil drop.
  10. Write the charge expression when the oil drop is stationary.
  11. What is meant by quantization of charge?
  12. How does Millikan’s experiment prove q = ne?
  13. Write the force on an electron in an electric field.
  14. Derive the acceleration of an electron in an electric field.
  15. Show that the path of an electron in a transverse uniform electric field is parabolic.
  16. Derive the deflection of an electron at the end of parallel plates.
  17. Derive the angle of emergence from an electric field.
  18. Write the magnetic force on a moving electron.
  19. What happens when an electron moves parallel to B?
  20. Derive r = mv/(eB).
  21. Derive the frequency and time period of an electron in a magnetic field.
  22. Derive the speed acquired by an electron through potential V₀.
  23. What is a velocity selector?
  24. Derive v = E/B for crossed fields.
  25. State the purpose and principle of J.J. Thomson’s experiment.
  26. Derive e/m = E²/(2V₀B²).
  27. How can Millikan and Thomson results be combined to find electron mass?

Long Questions / Derivations

  1. Describe Millikan’s oil-drop experiment with labelled diagram and derive the electronic charge formula.
  2. Explain how Millikan’s experiment establishes charge quantization.
  3. Derive the equation of the path of an electron beam in a uniform transverse electric field.
  4. Derive the radius, angular frequency, frequency and time period of an electron moving perpendicular to a magnetic field.
  5. Explain crossed electric and magnetic fields and derive the velocity-selector condition.
  6. Describe J.J. Thomson’s experiment with suitable diagram and derive the specific charge of an electron.
  7. Compare the effects of electric and magnetic fields on an electron beam.

Numerical Practice

  1. Find n from a measured charge q = ne.
  2. Calculate oil-drop radius from terminal speed and viscosity.
  3. Calculate electronic charge from Millikan data.
  4. Find electron acceleration in a given electric field.
  5. Calculate electric-field deflection between parallel plates.
  6. Find magnetic radius from v and B.
  7. Calculate cyclotron frequency/time period for an electron.
  8. Find electron speed after acceleration through a given potential.
  9. Find undeflected speed in crossed fields.
  10. Calculate e/m from E, B and V₀.
Exam Strategy This is a compact 4-hour chapter. Prioritize four derivations: Millikan oil-drop charge → parabolic electric deflection → magnetic circular motion → Thomson e/m. Memorize the distinction: Millikan measured e; Thomson measured e/m.

21. One-Minute Revision

  • Unit 5, Modern Physics, Chapter 20: Electrons.
  • Electron charge = −1.602 × 10⁻¹⁹ C.
  • Electron mass = 9.11 × 10⁻³¹ kg.
  • Specific charge magnitude ≈ 1.76 × 10¹¹ C kg⁻¹.
  • Millikan’s experiment uses tiny oil drops and Stokes drag.
  • Stokes drag = 6πηrv.
  • Effective drop weight = (4/3)πr³(ρ − σ)g.
  • Falling terminal-speed radius: r = √[9ηv₁/{2(ρ − σ)g}].
  • Rising-drop charge: q = 6πηr(v₁ + v₂)/E.
  • Charge is quantized: q = ne.
  • Electric force on charge: F = qE.
  • Electron acceleration magnitude in E: a = eE/m.
  • Transverse electric field gives a parabolic path.
  • y = eEx²/(2mu²).
  • Magnetic force magnitude = evB sinθ.
  • For v ⟂ B, electron moves in a circle.
  • r = mv/(eB).
  • ω = eB/m.
  • f = eB/(2πm).
  • T = 2πm/(eB).
  • Acceleration through potential: eV₀ = ½mv².
  • v = √(2eV₀/m).
  • Crossed fields: undeflected speed v = E/B.
  • Thomson measured e/m.
  • e/m = E²/(2V₀B²).
  • Millikan measured e.
  • Electron mass follows from m = e/(e/m).

22. Diagram Practice

Students should practice these labelled diagrams for the NEB examination:

  1. Millikan oil-drop apparatus.
  2. Forces on a falling oil drop.
  3. Quantization-of-charge diagram.
  4. Electron deflection in a transverse electric field.
  5. Circular motion in a perpendicular magnetic field.
  6. Crossed E–B velocity selector.
  7. J.J. Thomson experiment.
  8. Thomson e/m derivation flow.
Source handling: The original Nepal eNotes PDF remains embedded above using the verified Google Drive file. The source page identifies this resource as Unit 5, Modern Physics, Chapter 20 – Electrons. The typed section follows the verified NEB/CDC Chapter 20 syllabus and is designed as a searchable, responsive study companion. Equations from the source material are presented here in a cleaned and internally consistent form, including buoyancy in the full Millikan force balance. Where the PDF viewer does not expose page text, the typed section is a syllabus-aligned reconstruction and is not claimed to be a word-for-word transcription.
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