CHAPTER 20: ELECTRONS
Comprehensive study of subatomic charge carrier discovery, quantization of charge, parabolic & circular trajectories in electric and magnetic fields, and specific charge measurement ($e/m$).
Millikan’s Oil Drop Experiment & Charge Quantization
Performed by **Robert A. Millikan** in 1909, this landmark experiment directly measured the fundamental unit of electric charge ($e$) and proved that electric charge is **quantized** (exists only in discrete integer multiples of a fundamental unit $e$).
Experimental Setup Details
- Atomizer: Produces a fine spray of tiny oil drops into the upper chamber. Friction during spraying imparts a static charge onto drops.
- Parallel Metal Plates ($A$ & $B$): Highly polished brass plates separated by distance $d$, connected to a high-voltage DC supply ($0-10\text{ kV}$) to create a uniform electric field $E = V/d$.
- X-ray Beam / Ionization Source: Ionizes air molecules between the plates. The falling oil droplets capture these free electrons or ions, acquiring negative charge $q$.
- Viewing Microscope: Fitted with a calibrated micrometer eyepiece scale to measure terminal velocities of droplets.
- Illumination Lamp: Provides intense light at right angles to light up droplets (scattered Light effect / Tyndall effect).
Figure 20.1: Millikan's Oil Drop Experimental Setup (Line Schematic)
Comprehensive Mathematical Derivation
Case 1: Motion under gravity alone (Electric Field $E = 0$)
When an oil droplet of radius $r$ and density $\rho$ falls through air of density $\sigma$ and viscosity $\eta$ under gravity alone, it quickly reaches a constant downward terminal velocity $v_{1}$.
$W = m g = \frac{4}{3} \pi r^3 \rho g$
$U = \text{Vol} \times \sigma g = \frac{4}{3} \pi r^3 \sigma g$
$F_v = 6 \pi \eta r v_{1}$
At terminal equilibrium (Downward forces = Upward forces):
Solving Eq. (1) for droplet radius $r$ gives:
Case 2: Motion under Uniform Electric Field $E$
Method A: Upward Motion with Terminal Speed $v_{2}$
Apply a potential difference $V$ across plates $d$ apart so top plate $A$ is positive. The upward electric force $F_e = qE = q \frac{V}{d}$ pulls the negatively charged droplet upward with terminal velocity $v_{2}$. Since motion is upward, viscous drag $F_v' = 6 \pi \eta r v_{2}$ acts downward.
From Equation (1), we know $W - U = \frac{4}{3}\pi r^3(\rho - \sigma)g = 6\pi \eta r v_{1}$. Substituting this:
Method B: Alternative Stationary Drop Method ($v_{2} = 0$)
If the electric field is precisely adjusted so the droplet remains perfectly stationary in mid-air ($v_{2} = 0$, viscous force $F_v' = 0$):
Substituting $r = \sqrt{\frac{9 \eta v_{1}}{2(\rho-\sigma)g}}$ into Eq. (2):
Principle of Quantization of Charge
Millikan repeated the experiment thousands of times for various droplets and with different degrees of ionization. He discovered that every measured charge $q$ was always an integer multiple of a fundamental unit charge $e$:
- The minimum basic charge step observed was $e = 1.602 \times 10^{-19}\,\mathrm{C}$.
- Electric charge cannot exist in arbitrary fractional amounts (like $0.5e$ or $1.7e$) under free conditions.
- This conclusively proved the granular/particle nature of electric charge.
In highly precise calculations, when the drop radius $r$ is comparable to the mean free path of air molecules, Stokes' law requires air viscosity correction $\eta' = \eta / (1 + b/pr)$. However, for NEB Grade XII theory and numericals, standard Stokes' law $\eta$ without correction is used unless explicitly specified in the question.
Motion of Electrons in Electric and Magnetic Fields
A Motion of Electron in a Uniform Electric Field (Parabolic Path)
Consider an electron of mass $m$ and charge $-e$ entering perpendicularly into a uniform electric field $E$ directed vertically downwards (between two parallel plates of length $L$ and separation $d$) with initial horizontal velocity $v_x = v_0$.
Figure 20.2: Parabolic Path in Uniform Electric Field
Acceleration $a_x = 0$, velocity $v_x = v_0$ (constant).
$$x = v_x t \implies t = \frac{x}{v_x}$$
Electric force $F_y = eE$ upwards. Acceleration $a_y = \frac{eE}{m}$.
$$y = \frac{1}{2} a_y t^2 = \frac{1}{2} \left(\frac{eE}{m}\right) \left(\frac{x}{v_x}\right)^2$$
- Horizontal speed: $v_x = v_0$
- Vertical speed at exit ($x = L$): $v_y = a_y t = \frac{e E L}{m v_x}$
- Resultant velocity magnitude: $v = \sqrt{v_x^2 + v_y^2}$
- Deflection angle: $\tan \theta = \frac{v_y}{v_x} = \frac{e E L}{m v_x^2}$
If electron is accelerated from rest through potential difference $V_{acc}$:
$$\mathrm{KE} = e V_{acc} = \frac{1}{2} m v^2 \implies v = \sqrt{\frac{2 e V_{acc}}{m}}$$
B Motion of Electron in a Uniform Magnetic Field (Circular & Helical Paths)
The magnetic Lorentz force on a moving electron with velocity $\mathbf{v}$ in magnetic field $\mathbf{B}$ is given by:
Case 1: Perpendicular Entry ($\theta = 90^\circ$)
Magnetic force is perpendicular to velocity at all points, providing the required **centripetal force** for a circular path of radius $r$.
• Time Period ($T$): $T = \frac{2\pi r}{v} = \frac{2\pi m}{e B}$
• Cyclotron Frequency ($f$): $f = \frac{1}{T} = \frac{e B}{2\pi m}$
Case 2: Oblique Entry ($\theta \neq 0^\circ, 90^\circ, 180^\circ$)
Resolve velocity into components parallel and perpendicular to $\mathbf{B}$:
- $v_\parallel = v \cos\theta$ (Unchanged by $\mathbf{B}$, causes linear displacement along field)
- $v_\perp = v \sin\theta$ (Provides centripetal force, causes circular motion)
Resulting trajectory is a HELIX:
Radius: $$r = \frac{m v_\perp}{e B} = \frac{m v \sin\theta}{e B}$$
Pitch ($P$ = distance advanced along field in one full turn $T$):
$$P = v_\parallel \times T = (v \cos\theta) \left( \frac{2\pi m}{e B} \right) = \frac{2\pi m v \cos\theta}{e B}$$
Figure 20.3: Circular Path ($\theta=90^\circ$) vs Helical Trajectory ($\theta \neq 90^\circ$)
C Crossed Electric and Magnetic Fields (Velocity Selector / Filter)
When an electric field $\mathbf{E}$ (directed vertically downward) and a magnetic field $\mathbf{B}$ (directed perpendicularly into the page) are applied simultaneously in the same region, they exert opposite vertical forces on a moving electron:
Acts UPWARD towards the positive electric plate: $F_e = e E$.
By Fleming's Left Hand Rule, acts DOWNWARD: $F_m = e v B$.
Only electrons possessing exactly this velocity $v = E/B$ pass straight through without bending, regardless of their charge or mass!
J.J. Thomson’s Experiment: Determination of Specific Charge ($e/m$)
In 1897, **Sir J.J. Thomson** conducted his famous cathode ray tube experiment, determining the specific charge ($e/m$) of cathode rays. This led to the fundamental discovery of the **electron** as a universal constituent particle of all matter.
Figure 20.4: J.J. Thomson Cathode Ray Tube Apparatus Schematic
Step-by-Step Specific Charge Derivation
Both electric field $E$ and magnetic field $B$ are switched on simultaneously and adjusted until the spot returns to its original undeflected position $P_1$.
$$e E = e v B \implies v = \frac{E}{B}$$
Switch off magnetic field $B$, keeping $E$ on. The spot shifts to $P_2$ by vertical distance $y$.
Inside plates of length $L$, vertical acceleration is $a_y = \frac{eE}{m}$ for time $t = \frac{L}{v}$. Vertical velocity at exit of plates: $v_y = a_y t = \left(\frac{eE}{m}\right) \left(\frac{L}{v}\right)$.
If screen is at distance $D$ from the center of electric plates, total screen deflection $y$ is:
$$y = \frac{e E L D}{m v^2}$$
Substitute $v = E/B$ into $y = \frac{e E L D}{m v^2}$:
If electrons are accelerated by anode voltage $V_{acc}$ before entering fields: $\frac{1}{2} m v^2 = e V_{acc} \implies v^2 = \frac{2 e V_{acc}}{m}$. Substituting $v = E/B$:
- Thomson found $e/m$ was identical regardless of the gas in the cathode tube or the metal used for cathode electrodes.
- This proved that cathode rays consist of identical negatively charged universal subatomic particles — electrons.
- The extremely high $e/m$ value (compared to $H^+$ ion $e/m \approx 9.58 \times 10^7\mathrm{C\,kg^{-1}}$) proved electrons have negligible mass ($\approx 1/1836$ of hydrogen atom).
6 Solved Standard Textbook Numerical Examples
Fully worked step-by-step solutions with exact physical constants verified.
Problem: In a Millikan oil drop experiment, an oil drop of density $875\mathrm{kg\,m^{-3}}$ is held stationary between two horizontal parallel plates separated by $1.2\,\mathrm{cm}$ when a potential difference of $2400\,\mathrm{V}$ is applied. If the radius of the drop is $1.5 \times 10^{-6}\text{ m}$, calculate: (a) Mass of the drop (neglecting air buoyancy), (b) Charge $q$ on the drop, and (c) Number of excess electrons on the drop.
Solution:
Given: Density $\rho = 875\mathrm{kg\,m^{-3}}$, Distance $d = 1.2\,\mathrm{cm} = 0.012\text{ m}$, Voltage $V = 2400\,\mathrm{V}$, Radius $r = 1.5 \times 10^{-6}\text{ m}$, $g = 9.8\mathrm{m\,s^{-2}}$, $e = 1.602 \times 10^{-19}\,\mathrm{C}$.
(a) Mass of droplet ($m$):
$$m = \frac{4}{3} \pi r^3 \rho = \frac{4}{3} \times 3.1416 \times (1.5 \times 10^{-6})^3 \times 875 = 1.237 \times 10^{-14}\text{ kg}$$
(b) Charge on droplet ($q$):
For stationary drop: $q E = m g \implies q \left(\frac{V}{d}\right) = m g \implies q = \frac{m g d}{V}$
$$q = \frac{(1.237 \times 10^{-14}) \times 9.8 \times 0.012}{2400} = 6.06 \times 10^{-19}\,\mathrm{C}$$
(c) Number of excess electrons ($n$):
$$n = \frac{q}{e} = \frac{6.06 \times 10^{-19}}{1.602 \times 10^{-19}} \approx 3.78 \approx 4 \text{ electrons}$$
Problem: An electron accelerated through a potential difference of $1000\,\mathrm{V}$ enters perpendicularly into a uniform electric field of $2000\text{ V/m}$ between two horizontal plates of length $4\,\mathrm{cm}$. Find the vertical deflection $y$ of the electron as it emerges from the plates. ($e/m = 1.76 \times 10^{11}\mathrm{C\,kg^{-1}}$).
Solution:
Given: $V_{acc} = 1000\,\mathrm{V}$, $E = 2000\text{ V/m}$, Plate length $L = 0.04\text{ m}$, $e/m = 1.76 \times 10^{11}\mathrm{C\,kg^{-1}}$.
Step 1: Horizontal velocity $v_x$:
$$\frac{1}{2} m v_x^2 = e V_{acc} \implies v_x^2 = \frac{2 e V_{acc}}{m} = 2 \times (1.76 \times 10^{11}) \times 1000 = 3.52 \times 10^{14}\mathrm{m^2\,s^{-2}}$$
$$v_x = 1.876 \times 10^7\mathrm{m\,s^{-1}}$$
Step 2: Deflection $y$:
$$y = \frac{e E L^2}{2 m v_x^2} = \frac{1}{2} \left(\frac{e}{m}\right) \frac{E L^2}{v_x^2} = \frac{1}{2} \times (1.76 \times 10^{11}) \times \frac{2000 \times (0.04)^2}{3.52 \times 10^{14}}$$
$$y = 0.0008\text{ m} = 0.8\,\mathrm{mm}$$
Problem: An electron is accelerated from rest through $500\,\mathrm{V}$ and then enters a uniform magnetic field $B = 2.0 \times 10^{-3}\text{ T}$ perpendicular to its direction of motion. Calculate: (a) Speed of the electron, (b) Radius of the circular path, and (c) Time period of revolution. ($m_e = 9.11 \times 10^{-31}\text{ kg}$, $e = 1.6 \times 10^{-19}\,\mathrm{C}$).
Solution:
(a) Speed $v$: $v = \sqrt{\frac{2 e V}{m}} = \sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 500}{9.11 \times 10^{-31}}} = 1.326 \times 10^7\mathrm{m\,s^{-1}}$.
(b) Radius $r$: $r = \frac{m v}{e B} = \frac{9.11 \times 10^{-31} \times 1.326 \times 10^7}{1.6 \times 10^{-19} \times 2.0 \times 10^{-3}} = 0.0377\text{ m} = 3.77\,\mathrm{cm}$.
(c) Period $T$: $T = \frac{2\pi m}{e B} = \frac{2 \times 3.1416 \times 9.11 \times 10^{-31}}{1.6 \times 10^{-19} \times 2.0 \times 10^{-3}} = 1.788 \times 10^{-8}\,\mathrm{s} = 17.88\text{ ns}$.
Problem: In a velocity selector, an electric field $E = 1.5 \times 10^4\text{ V/m}$ and a magnetic field $B = 0.03\text{ T}$ act perpendicularly to each other and to the electron beam. (a) What velocity will allow electrons to pass through undeflected? (b) If the magnetic field is turned off, what will be the initial acceleration of the electron?
Solution:
(a) Undeflected velocity $v$: $v = \frac{E}{B} = \frac{1.5 \times 10^4}{0.03} = 5.0 \times 10^5\mathrm{m\,s^{-1}}$.
(b) Acceleration with $E$ field alone:
$$a = \frac{e E}{m} = (1.76 \times 10^{11}) \times 1.5 \times 10^4 = 2.64 \times 10^{15}\mathrm{m\,s^{-2}}$$
Problem: In a J.J. Thomson $e/m$ experiment, electric plates are $5\,\mathrm{cm}$ long ($L$) with potential difference $150\,\mathrm{V}$ across plates separated by $1.5\,\mathrm{cm}$. When a magnetic field $B = 1.2 \times 10^{-3}\text{ T}$ is applied simultaneously, the electron beam spot is undeflected. When magnetic field is turned off, the beam spot on screen (located $20\,\mathrm{cm}$ from plate center) shifts by $1.18\,\mathrm{cm}$. Calculate specific charge $e/m$.
Solution:
Given: $L = 0.05\text{ m}$, $V_p = 150\,\mathrm{V}$, $d = 0.015\text{ m}$, $E = V_p/d = 150/0.015 = 10,000\text{ V/m}$, $B = 1.2 \times 10^{-3}\text{ T}$, $D = 0.20\text{ m}$, $y = 0.0118\text{ m}$.
Formula: $\frac{e}{m} = \frac{y E}{B^2 L D}$
$$\frac{e}{m} = \frac{0.0118 \times 10000}{(1.2 \times 10^{-3})^2 \times 0.05 \times 0.20} = \frac{118}{1.44 \times 10^{-8} \times 0.01} = 1.763 \times 10^{11}\mathrm{C\,kg^{-1}}$$
Problem: An electron with velocity $v = 4.0 \times 10^6\mathrm{m\,s^{-1}}$ enters a uniform magnetic field $B = 0.02\text{ T}$ at an angle $\theta = 30^\circ$ to the field lines. Calculate: (a) Radius of helical path, and (b) Pitch of the helix.
Solution:
Given: $v = 4 \times 10^6\mathrm{m\,s^{-1}}$, $B = 0.02\text{ T}$, $\theta = 30^\circ$, $e/m = 1.76 \times 10^{11}\mathrm{C\,kg^{-1}}$.
Velocity components: $v_\perp = v \sin 30^\circ = 2.0 \times 10^6\mathrm{m\,s^{-1}}$, $v_\parallel = v \cos 30^\circ = 3.464 \times 10^6\mathrm{m\,s^{-1}}$.
(a) Radius $r$: $r = \frac{m v_\perp}{e B} = \frac{2.0 \times 10^6}{(1.76 \times 10^{11}) \times 0.02} = 5.68 \times 10^{-4}\text{ m} = 0.568\,\mathrm{mm}$.
(b) Period $T$: $T = \frac{2\pi m}{e B} = \frac{2\times 3.1416}{1.76 \times 10^{11} \times 0.02} = 1.785 \times 10^{-9}\,\mathrm{s}$.
(c) Pitch $P$: $P = v_\parallel \times T = (3.464 \times 10^6) \times (1.785 \times 10^{-9}) = 6.18 \times 10^{-3}\text{ m} = 6.18\,\mathrm{mm}$.
Interactive Physics Simulators
Adjust physical parameters in real-time to observe beam deflection, helical motion, and charge quantization.
Apparatus Controls
• Electron Speed $v$: 1.87 × 10⁷ m/s
• Electric Force $F_e$: --
• Magnetic Force $F_m$: --
• Estimated $e/m$: 1.76 × 10¹¹ C/kg
NEB Grade XII Question Bank
High-frequency Board Exam conceptual short questions, long derivations, and unsolved practice numericals.
Conceptual Short Answer Questions
Q1. Why are oil drops preferred over water drops in Millikan’s experiment?
Model AnswerQ2. An electron moving horizontally enters a uniform magnetic field directed vertically downwards. What is the shape of its trajectory?
Model AnswerQ3. Does the specific charge ($e/m$) of an electron depend on its velocity at relativistic speeds?
Model AnswerNEB Board Exam Long Answer Derivations [5 Marks]
Describe Millikan's oil drop experiment. Derive mathematical expressions for droplet radius $r$ and charge $q$ under downward free fall and upward electric field motion. Explain charge quantization.
Describe J.J. Thomson's experiment with a neat labeled diagram to determine $e/m$ of an electron. Explain velocity selection using crossed fields and derive the deflection formula.
Unsolved Numerical Practice Set (With Answers & Hints)
P1. An oil drop carrying 10 excess electrons is held stationary between plates separated by $1\,\mathrm{cm}$ at a potential difference of $1200\,\mathrm{V}$. Find the radius of the oil droplet. ($\rho_{oil} = 900\mathrm{kg\,m^{-3}}, g = 9.8\mathrm{m\,s^{-2}}$).
Ans: $r = 1.98 \times 10^{-6}\text{ m} = 1.98\ \mu\text{m}$. [Hint: $q E = m g \implies n e (V/d) = \frac{4}{3}\pi r^3 \rho g$].
P2. An electron beam passes undeflected through crossed fields $E = 3.2 \times 10^4\text{ V/m}$ and $B = 2.0 \times 10^{-3}\text{ T}$. If electric field is removed, calculate the radius of the circular path traversed by the electron in the magnetic field.
Ans: $r = 0.0455\text{ m} = 4.55\,\mathrm{cm}$. [Hint: Find $v = E/B = 1.6 \times 10^7\mathrm{m\,s^{-1}}$, then $r = mv / eB$].
P3. An electron with kinetic energy $2.0\text{ keV}$ enters a magnetic field $B = 0.1\text{ T}$ at an angle of $45^\circ$ to field lines. Calculate the helical path radius and pitch.
Ans: Radius $r = 1.07\,\mathrm{mm}$, Pitch $P = 6.72\,\mathrm{mm}$. [Hint: $\mathrm{KE} = 2000 \times 1.6 \times 10^{-19}\text{ J}$, $v_\perp = v \sin 45^\circ$, $v_\parallel = v \cos 45^\circ$].
Standard Physical Constants Reference (NEB Syllabus)
| Physical Quantity | Symbol | Standard SI Value |
|---|---|---|
| Elementary Charge | $e$ | $1.60217663 \times 10^{-19}\,\mathrm{C} \approx 1.602 \times 10^{-19}\,\mathrm{C}$ |
| Electron Rest Mass | $m_e$ | $9.109383 \times 10^{-31}\text{ kg} \approx 9.11 \times 10^{-31}\text{ kg}$ |
| Specific Charge of Electron | $e/m_e$ | $1.75882 \times 10^{11}\mathrm{C\,kg^{-1}} \approx 1.76 \times 10^{11}\mathrm{C\,kg^{-1}}$ |
| Dynamic Viscosity of Air ($20^\circ\text{C}$) | $\eta$ | $1.81 \times 10^{-5}\mathrm{N\,s\,m^{-2}} \text{ (or kg/m}\cdot\text{s)}$ |
| Acceleration due to Gravity | $g$ | $9.80665\mathrm{m\,s^{-2}} \approx 9.8\mathrm{m\,s^{-2}} \text{ or } 9.81\mathrm{m\,s^{-2}}$ |