Chapter 23: Electromagnetic Waves

Introduction

Key Points

• Maxwell identified inconsistency in Ampere’s circuital law for magnetic field outside a capacitor with time-varying current.
• Introduced displacement current to resolve this inconsistency.
• Maxwell’s equations unify electric and magnetic fields with their sources (charge and current densities).
• Hertz experimentally confirmed electromagnetic waves in 1885; Marconi utilized them for communication.

Solved Examples

• Ex 1: Why was displacement current introduced?
  Step 1: Identify inconsistency in Ampere’s law for capacitor.
  Step 2: Maxwell proposed displacement current to maintain continuity of magnetic field.

Practice Set

• Q1: What does displacement current represent?
  A) Actual charge flow
  B) Changing electric field effect
  C) Magnetic field
  D) Static electric field
• Q2: Who experimentally demonstrated electromagnetic waves?
  A) Maxwell
  B) Marconi
  C) Hertz
  D) Faraday
• Q3: Maxwell’s equations relate which of the following?
  A) Only electric fields
  B) Only magnetic fields
  C) Electric and magnetic fields with charges and currents
  D) Only charges and currents

Answer Key

• Q1: B - Displacement current is due to changing electric field, not actual charge flow.
• Q2: C - Hertz demonstrated electromagnetic waves experimentally.
• Q3: C - Maxwell’s equations unify electric and magnetic fields with their sources.

Memory Aids

• Displacement current = changing electric flux effect.
• Maxwell’s equations = electric + magnetic fields + sources.
• Hertz = experimental proof of EM waves.

Displacement Current

Key Points

• Changing electric field produces magnetic field (Maxwell’s insight).
• Displacement current \(I_D = \varepsilon_0 \frac{d\Phi_E}{dt}\), where \(\Phi_E = EA\) is electric flux.
• In capacitor, no real current flows between plates, but changing electric field acts like current.
• Electric field between plates: \(E = \frac{Q}{\varepsilon_0 A}\).

Solved Examples

• Ex 1: Calculate displacement current if electric flux changes at \(2 \times 10^{-6} \text{ C/m}^2/s\) and \(\varepsilon_0 = 8.85 \times 10^{-12} \text{ F/m}\).
  Step 1: Use \(I_D = \varepsilon_0 \frac{d\Phi_E}{dt}\)
  Step 2: \(I_D = 8.85 \times 10^{-12} \times 2 \times 10^{-6} = 1.77 \times 10^{-17} \text{ A}\)

Practice Set

• Q1: Displacement current is proportional to:
  A) Electric charge
  B) Rate of change of electric flux
  C) Magnetic flux
  D) Electric potential
• Q2: Electric field between capacitor plates is given by:
  A) \(E = \frac{Q}{\varepsilon_0 A}\)
  B) \(E = \frac{Q}{A}\)
  C) \(E = \varepsilon_0 Q A\)
  D) \(E = \frac{\varepsilon_0}{Q A}\)
• Q3: Displacement current in a capacitor is:
  A) Equal to conduction current
  B) Zero
  C) Twice the conduction current
  D) Half the conduction current

Answer Key

• Q1: B - Displacement current depends on rate of change of electric flux.
• Q2: A - Electric field \(E = \frac{Q}{\varepsilon_0 A}\).
• Q3: A - Displacement current equals conduction current in magnitude.

Memory Aids

• Displacement current \(I_D = \varepsilon_0 \frac{d\Phi_E}{dt}\).
• Electric field in capacitor \(E = \frac{Q}{\varepsilon_0 A}\).
• Displacement current = conduction current in charging capacitor.

Inconsistency of Ampere’s Circuital Law

Key Points

• Charging capacitor: conduction current \(I\) flows in wires but no conduction current between plates.
• Two loops \(C_1\) (around wire) and \(C_2\) (between plates) give different magnetic field results using Ampere’s law.
• For \(C_1\): \(\oint \vec{B} \cdot d\vec{l} = \mu_0 I \neq 0\).
• For \(C_2\): \(\oint \vec{B} \cdot d\vec{l} = 0\) since \(I=0\) (no conduction current).
• Magnetic field discontinuity contradicts physical continuity; Ampere’s law needs modification.

Solved Examples

• Ex 1: Explain why Ampere’s law fails for loop \(C_2\) in charging capacitor.
  Step 1: No conduction current between plates.
  Step 2: Ampere’s law predicts zero magnetic field, contradicting continuity.

Practice Set

• Q1: In charging capacitor, conduction current between plates is:
  A) Equal to wire current
  B) Zero
  C) Twice the wire current
  D) Half the wire current
• Q2: Ampere’s law predicts magnetic field for loop \(C_2\) as:
  A) \(\mu_0 I\)
  B) Zero
  C) Infinite
  D) Negative
• Q3: The inconsistency in Ampere’s law arises because:
  A) Magnetic field is discontinuous
  B) Electric field is static
  C) Current is constant
  D) Magnetic field is zero everywhere

Answer Key

• Q1: B - No conduction current flows between capacitor plates.
• Q2: B - Ampere’s law gives zero magnetic field for loop \(C_2\).
• Q3: A - Magnetic field discontinuity shows Ampere’s law inconsistency.

Memory Aids

• Ampere’s law fails inside capacitor gap due to zero conduction current.
• Magnetic field must be continuous physically.
• Need displacement current to fix inconsistency.

Modified Ampere’s Law by Maxwell

Key Points

• Maxwell added displacement current \(I_D = \varepsilon_0 \frac{d\Phi_E}{dt}\) to Ampere’s law.
• Generalized Ampere’s law: \(\oint \vec{B} \cdot d\vec{l} = \mu_0 \left( I + I_D \right)\).
• Displacement current exists where electric flux changes with time.
• Ensures magnetic field continuity inside and outside capacitor plates.
• Displacement current magnitude equals conduction current in charging capacitor.
• Kirchhoff’s first law holds with displacement current included.

Solved Examples

• Ex 1: Show displacement current equals conduction current.
  Step 1: \(\Phi_E = EA = \frac{q}{\varepsilon_0 A} \times A = \frac{q}{\varepsilon_0}\)
  Step 2: \(\frac{d\Phi_E}{dt} = \frac{1}{\varepsilon_0} \frac{dq}{dt} = \frac{I}{\varepsilon_0}\)
  Step 3: \(I_D = \varepsilon_0 \times \frac{I}{\varepsilon_0} = I\)

Practice Set

• Q1: Modified Ampere’s law includes:
  A) Only conduction current
  B) Only displacement current
  C) Sum of conduction and displacement currents
  D) No current
• Q2: Displacement current is non-zero when:
  A) Electric flux is constant
  B) Electric flux changes with time
  C) Magnetic flux changes
  D) No current flows
• Q3: Kirchhoff’s first law is valid in charging capacitor because:
  A) Displacement current equals conduction current
  B) No current flows
  C) Magnetic field is zero
  D) Electric field is static

Answer Key

• Q1: C - Modified Ampere’s law sums conduction and displacement currents.
• Q2: B - Displacement current arises from changing electric flux.
• Q3: A - Displacement current equals conduction current, preserving Kirchhoff’s law.

Memory Aids

• Modified Ampere’s law: \(\oint \vec{B} \cdot d\vec{l} = \mu_0 (I + I_D)\).
• Displacement current \(I_D = \varepsilon_0 \frac{d\Phi_E}{dt}\).
• Displacement current = conduction current in capacitor.

Maxwell’s Equations

Key Points

• Four fundamental laws of electricity and magnetism.
• 1) Gauss’s law for electricity: \(\oint_S \vec{E} \cdot d\vec{S} = \frac{q_{inside}}{\varepsilon_0}\).
• Electric field lines begin/end on charges (not closed loops).
• 2) Gauss’s law for magnetism: \(\oint_S \vec{B} \cdot d\vec{S} = 0\).
• Magnetic field lines are closed loops (no magnetic monopoles).
• 3) Faraday’s law: \(\oint_l \vec{E} \cdot d\vec{l} = - \frac{d}{dt} \oint_S \vec{B} \cdot d\vec{S}\).
• Changing magnetic flux induces electric field (emf).
• 4) Maxwell’s modified Ampere’s law: \(\oint_l \vec{B} \cdot d\vec{l} = \mu_0 \left( I + \varepsilon_0 \frac{d}{dt} \oint_S \vec{E} \cdot d\vec{S} \right)\).
• Changing electric flux produces magnetic field.

Solved Examples

• Ex 1: Write Maxwell’s first equation and explain.
  \(\oint_S \vec{E} \cdot d\vec{S} = \frac{q_{inside}}{\varepsilon_0}\)
  Electric flux through closed surface equals enclosed charge over permittivity.

Practice Set

• Q1: Gauss’s law for magnetism states:
  A) Magnetic flux through closed surface equals charge
  B) Magnetic flux through closed surface is zero
  C) Electric flux is zero
  D) Magnetic field is zero
• Q2: Faraday’s law relates induced emf to:
  A) Rate of change of electric flux
  B) Rate of change of magnetic flux
  C) Magnetic field strength
  D) Electric charge
• Q3: Maxwell’s modified Ampere’s law shows magnetic field is produced by:
  A) Only conduction current
  B) Only displacement current
  C) Both conduction and displacement currents
  D) Neither

Answer Key

• Q1: B - Magnetic flux through any closed surface is zero (no monopoles).
• Q2: B - Induced emf is proportional to rate of change of magnetic flux.
• Q3: C - Magnetic field arises from conduction and displacement currents.

Memory Aids

• Gauss’s law (E): flux = charge/\(\varepsilon_0\).
• Gauss’s law (B): flux = 0 (no magnetic monopoles).
• Faraday: changing magnetic flux → induced emf.
• Maxwell-Ampere: changing electric flux + current → magnetic field.

Electromagnetic Waves

Definition and Properties

• Electromagnetic waves have sinusoidal electric and magnetic fields oscillating perpendicular to each other and to the direction of wave propagation.
• Produced by oscillating electric charges or accelerated charges.
• Frequency of the wave equals frequency of oscillating charges or fields.
• Carry energy and momentum through space.

Production of Electromagnetic Waves

• Alternating current in an antenna causes charges to oscillate, producing oscillating electric (E) and magnetic (B) fields.
• Maxwell’s equations: time-varying magnetic field induces perpendicular electric field and vice versa.
• Accelerated charges radiate electromagnetic waves.
• Energy of wave comes from energy of accelerating charges.

Solved Examples

Ex 1

• Given AC frequency = 1 MHz, find frequency of emitted EM wave.
• Solution: Frequency of EM wave = frequency of AC = 1 MHz.

Practice Set

• Q1: What is the direction of electric and magnetic fields in an EM wave?
  • A) Parallel to each other
  • B) Perpendicular to each other and to wave direction
  • C) Both along wave direction
  • D) Random directions
• Q2: Accelerated charges produce:
  • A) Only electric field
  • B) Only magnetic field
  • C) Electromagnetic waves
  • D) No radiation
• Q3: If an antenna oscillates at 500 kHz, the frequency of EM waves produced is:
  • A) 250 kHz
  • B) 500 kHz
  • C) 1 MHz
  • D) Depends on antenna size

Answer Key

• Q1: B - Electric and magnetic fields are perpendicular to each other and to the direction of wave propagation.
• Q2: C - Accelerated charges radiate electromagnetic waves.
• Q3: B - Frequency of EM wave equals frequency of oscillating charges (antenna frequency).

Memory Aids

• EM waves = E and B fields ⟂ each other and wave direction.
• Accelerated charges = EM wave radiation.
• Wave frequency = source oscillation frequency.

Hertz Experiment

Setup and Working

• Two metal plates connected to spheres S and S' with induction coil produce high-frequency oscillations.
• High potential difference ionizes air gap, causing sparks and oscillating electric field.
• Oscillating magnetic field perpendicular to electric field is produced.
• EM waves propagate through space and induce emf in detector loop, causing sparks.

Solved Examples

Ex 1

• Given inductance L and capacitance C, find resonant frequency ν = 1/(2π√LC).
• Use ν to find frequency of EM waves produced.

Practice Set

• Q1: What causes spark between spheres in Hertz’s setup?
  • A) Steady current
  • B) Ionization of air gap
  • C) Magnetic field only
  • D) Static charges
• Q2: Resonant frequency depends on:
  • A) Inductance and capacitance
  • B) Resistance only
  • C) Voltage only
  • D) Sphere size only
• Q3: EM waves in Hertz’s experiment are produced by:
  • A) Stationary charges
  • B) Oscillating charges
  • C) Uniform motion charges
  • D) No charges

Answer Key

• Q1: B - High potential ionizes air, causing spark discharge.
• Q2: A - Resonant frequency ν = 1/(2π√LC) depends on L and C.
• Q3: B - Oscillating charges produce EM waves.

Memory Aids

• Sparks = ionized air gap.
• ν = 1/(2π√LC) for oscillations.
• Oscillating charges emit EM waves.

Nature of Electromagnetic Waves

Characteristics and Formulas

• Electric field: \( E_x = E_0 \sin(kz - \omega t) = E_0 \sin[2\pi(z/\lambda - \nu t)] \)
• Magnetic field: \( B_y = B_0 \sin(kz - \omega t) = B_0 \sin[2\pi(z/\lambda - \nu t)] \)
• Fields are perpendicular: \( \vec{E} \perp \vec{B} \perp \) direction of propagation (z-axis).
• Relation: \( \frac{E}{B} = c \), speed of light \( c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} \approx 3 \times 10^8 \text{ m/s} \).
• Speed in medium: \( v = \frac{1}{\sqrt{\mu \epsilon}} \).
• EM waves are transverse and carry energy and momentum.
• Momentum delivered to surface: \( p = \frac{U}{c} \), where U is energy transferred.

Energy Density

• Electric energy density: \( u_E = \frac{1}{2} \epsilon_0 E^2 \)
• Magnetic energy density: \( u_B = \frac{1}{2 \mu_0} B^2 \)
• Total energy density: \( u = u_E + u_B = \epsilon_0 E^2 = \frac{B^2}{\mu_0} \)
• Energy shared equally between electric and magnetic fields.

Solved Examples

Ex 1

• Given \( E_0 = 100 \text{ V/m} \), find energy density \( u \).
• Solution: \( u = \epsilon_0 E_0^2 = 8.85 \times 10^{-12} \times 100^2 = 8.85 \times 10^{-8} \text{ J/m}^3 \).

Practice Set

• Q1: The ratio \( E/B \) in vacuum is:
  • A) \( c \)
  • B) \( 1/c \)
  • C) \( \mu_0 \)
  • D) \( \epsilon_0 \)
• Q2: Speed of EM wave in medium depends on:
  • A) \( \mu \) and \( \epsilon \)
  • B) Frequency only
  • C) Amplitude only
  • D) Wavelength only
• Q3: Energy density of EM wave with \( E = 200 \text{ V/m} \) is:
  • A) \( 1.77 \times 10^{-7} \text{ J/m}^3 \)
  • B) \( 8.85 \times 10^{-8} \text{ J/m}^3 \)
  • C) \( 3.54 \times 10^{-7} \text{ J/m}^3 \)
  • D) \( 7.08 \times 10^{-8} \text{ J/m}^3 \)

Answer Key

• Q1: A - \( E/B = c \) in vacuum.
• Q2: A - Speed \( v = 1/\sqrt{\mu \epsilon} \) depends on medium properties.
• Q3: C - \( u = \epsilon_0 E^2 = 8.85 \times 10^{-12} \times 200^2 = 3.54 \times 10^{-7} \text{ J/m}^3 \).

Memory Aids

• \( E/B = c \) in vacuum.
• Speed \( c = 1/\sqrt{\mu_0 \epsilon_0} \).
• Energy density \( u = \epsilon_0 E^2 = B^2/\mu_0 \).
• EM waves are transverse and carry energy equally in E and B fields.

Electromagnetic Spectrum

Overview

• Electromagnetic spectrum: orderly distribution of waves by frequency or wavelength.
• Wavelength range: gamma rays (shortest) to radio waves (longest).
• Frequency and energy increase as wavelength decreases.
• Visible light: small part of spectrum, wavelength 400–700 nm, frequency 4×10¹⁴–7×10¹⁴ Hz.
• Energy relation: \( E = h \nu \), where \( h \) is Planck's constant, \( \nu \) frequency.

Solved Examples

• Ex 1: Calculate frequency of light with wavelength 500 nm.
  • Use \( \nu = \frac{c}{\lambda} \), \( c = 3 \times 10^{8} \) m/s.
  • \( \nu = \frac{3 \times 10^{8}}{500 \times 10^{-9}} = 6 \times 10^{14} \) Hz.

Practice Set

• Q1: Which has the highest frequency?
  • A) Radio waves
  • B) Microwaves
  • C) Ultraviolet rays
  • D) Infrared waves
• Q2: Visible light wavelength range is:
  • A) 400–700 nm
  • B) 700–1000 nm
  • C) 100–400 nm
  • D) 10–100 nm
• Q3: Energy of a photon with frequency \( 5 \times 10^{14} \) Hz is (\( h = 6.63 \times 10^{-34} \) Js):
  • A) \( 3.32 \times 10^{-19} \) J
  • B) \( 1.33 \times 10^{-19} \) J
  • C) \( 6.63 \times 10^{-20} \) J
  • D) \( 5.00 \times 10^{-19} \) J

Answer Key

• Q1: C; Ultraviolet rays have higher frequency than radio, microwaves, and infrared.
• Q2: A; Visible light wavelength ranges from 400 to 700 nm.
• Q3: A; \( E = h \nu = 6.63 \times 10^{-34} \times 5 \times 10^{14} = 3.32 \times 10^{-19} \) J.

Memory Aids

• Frequency \( \uparrow \) → Wavelength \( \downarrow \) → Energy \( \uparrow \).
• Visible light colors: VIBGYOR (Violet shortest wavelength, Red longest).

Radio Waves

Properties and Uses

• Frequency range: 500 kHz to 1000 MHz.
• Used in communication: radio, TV, cellular phones.

Solved Examples

• Ex 1: Find wavelength of radio wave with frequency 1 MHz.
  • \( \lambda = \frac{c}{\nu} = \frac{3 \times 10^{8}}{1 \times 10^{6}} = 300 \) m.

Practice Set

• Q1: Radio waves frequency range is:
  • A) 500 kHz to 1000 MHz
  • B) 1 GHz to 10 GHz
  • C) 10 MHz to 100 MHz
  • D) 100 MHz to 1 GHz
• Q2: Wavelength of 500 kHz radio wave is:
  • A) 600 m
  • B) 300 m
  • C) 150 m
  • D) 100 m
• Q3: Radio waves are NOT used in:
  • A) Television
  • B) Cellular phones
  • C) X-ray imaging
  • D) Radio broadcasting

Answer Key

• Q1: A; Radio waves frequency is 500 kHz to 1000 MHz.
• Q2: A; \( \lambda = \frac{3 \times 10^{8}}{5 \times 10^{5}} = 600 \) m.
• Q3: C; X-rays are not radio waves.

Memory Aids

• Radio waves = longest wavelength in EM spectrum used for communication.

Microwaves

Properties and Uses

• Shortest wavelength radio waves, shorter than radio but longer than visible light.
• Used in radar for aircraft navigation and speed detection.
• Detect speed of moving objects like balls and vehicles.

Solved Examples

• Ex 1: Calculate wavelength of microwave with frequency 10 GHz.
  • \( \lambda = \frac{3 \times 10^{8}}{10 \times 10^{9}} = 0.03 \) m (3 cm).

Practice Set

• Q1: Microwaves frequency range is approximately:
  • A) 1 MHz to 100 MHz
  • B) 1 GHz to 300 GHz
  • C) 300 GHz to 1 THz
  • D) 100 kHz to 1 MHz
• Q2: Microwaves are used in:
  • A) Radar systems
  • B) X-ray imaging
  • C) UV sterilization
  • D) Infrared heaters
• Q3: Wavelength of 15 GHz microwave is:
  • A) 2 cm
  • B) 20 cm
  • C) 0.02 m
  • D) 0.2 m

Answer Key

• Q1: B; Microwaves range from 1 GHz to 300 GHz.
• Q2: A; Microwaves are used in radar systems.
• Q3: C; \( \lambda = \frac{3 \times 10^{8}}{15 \times 10^{9}} = 0.02 \) m (2 cm).

Memory Aids

• Microwaves = radar + shortest radio waves.

Infrared Waves

Properties and Uses

• Frequency range: 3 × 10¹¹ to 4 × 10¹⁴ Hz.
• Used in infrared lamps for physical therapy.
• Used in infrared detectors on Earth satellites.
• Used in LEDs for remote controls (TV, AC, CD players).
• Used in solar water heaters, cookers, and greenhouses.
• Responsible for Earth's warming via greenhouse effect.

Solved Examples

• Ex 1: Calculate wavelength of infrared wave with frequency 1 × 10¹³ Hz.
  • \( \lambda = \frac{3 \times 10^{8}}{1 \times 10^{13}} = 3 \times 10^{-5} \) m (30 µm).

Practice Set

• Q1: Infrared waves frequency range is:
  • A) 3 × 10¹¹ to 4 × 10¹⁴ Hz
  • B) 4 × 10¹⁴ to 7 × 10¹⁴ Hz
  • C) 8 × 10¹⁴ to 5 × 10¹⁷ Hz
  • D) 3 × 10¹⁸ to 3 × 10²² Hz
• Q2: Infrared waves are NOT used in:
  • A) Physical therapy lamps
  • B) Remote controls
  • C) Sterilizing surgical instruments
  • D) Solar water heaters
• Q3: Which phenomenon involves infrared waves warming Earth?
  • A) Greenhouse effect
  • B) Photoelectric effect
  • C) Nuclear fusion
  • D) Fluorescence

Answer Key

• Q1: A; Infrared frequency range is 3 × 10¹¹ to 4 × 10¹⁴ Hz.
• Q2: C; Sterilizing instruments uses UV rays, not infrared.
• Q3: A; Greenhouse effect traps infrared radiation warming Earth.

Memory Aids

• Infrared = heat radiation + remote controls + greenhouse effect.

Visible Rays

Properties and Uses

• Wavelength: 400–700 nm; frequency: 4 × 10¹⁴ to 7 × 10¹⁴ Hz.
• Visible to human eye; colors from violet (shortest wavelength) to red (longest).
• Used in electric bulbs, tube lights, lasers, cinema screens.

Solved Examples

• Ex 1: Find frequency of red light with wavelength 700 nm.
  • \( \nu = \frac{3 \times 10^{8}}{700 \times 10^{-9}} = 4.29 \times 10^{14} \) Hz.

Practice Set

• Q1: Visible light wavelength range is:
  • A) 400–700 nm
  • B) 700–1000 nm
  • C) 100–400 nm
  • D) 10–100 nm
• Q2: Which color has the shortest wavelength?
  • A) Red
  • B) Green
  • C) Violet
  • D) Yellow
• Q3: Frequency of violet light (400 nm) is:
  • A) 7.5 × 10¹⁴ Hz
  • B) 4.3 × 10¹⁴ Hz
  • C) 3.0 × 10¹⁴ Hz
  • D) 1.5 × 10¹⁴ Hz

Answer Key

• Q1: A; Visible light is 400–700 nm.
• Q2: C; Violet has shortest wavelength.
• Q3: A; \( \nu = \frac{3 \times 10^{8}}{400 \times 10^{-9}} = 7.5 \times 10^{14} \) Hz.

Memory Aids

• VIBGYOR: Violet shortest, Red longest wavelength.

Ultraviolet Rays

Properties and Uses

• Wavelength: 0.6–400 nm; frequency: 8 × 10¹⁴ to 5 × 10¹⁷ Hz.
• Used for sterilizing surgical instruments and water purification.
• Used in LASIK eye surgery and mineral fluorescence detection.
• Used in burglar alarms.

Solved Examples

• Ex 1: Calculate wavelength of UV ray with frequency 1 × 10¹⁵ Hz.
  • \( \lambda = \frac{3 \times 10^{8}}{1 \times 10^{15}} = 3 \times 10^{-7} \) m (300 nm).

Practice Set

• Q1: UV rays wavelength range is:
  • A) 0.6–400 nm
  • B) 400–700 nm
  • C) 700–1000 nm
  • D) 10–100 nm
• Q2: UV rays are used in:
  • A) Sterilization
  • B) Radio communication
  • C) Infrared heating
  • D) X-ray imaging
• Q3: Frequency of UV ray with wavelength 200 nm is:
  • A) 1.5 × 10¹⁵ Hz
  • B) 3.0 × 10¹⁵ Hz
  • C) 6.0 × 10¹⁴ Hz
  • D) 5.0 × 10¹⁴ Hz

Answer Key

• Q1: A; UV wavelength is 0.6–400 nm.
• Q2: A; UV rays sterilize instruments and water.
• Q3: B; \( \nu = \frac{3 \times 10^{8}}{200 \times 10^{-9}} = 1.5 \times 10^{15} \) Hz (corrected: 1.5×10¹⁵ Hz).

Memory Aids

• UV = sterilization + fluorescence + eye surgery.

X-rays

Properties and Uses

• Wavelength: 10^-4 nm to 10 nm; beyond ultraviolet region.
• Used to detect bone fractures and foreign bodies.
• Used by engineers to detect flaws in metals.
• Used to investigate crystal structures.

Solved Examples

• Ex 1: Calculate frequency of X-ray with wavelength 0.1 nm.
  • \( \nu = \frac{3 \times 10^{8}}{0.1 \times 10^{-9}} = 3 \times 10^{18} \) Hz.

Practice Set

• Q1: X-rays wavelength range is:
  • A) 10^-4 nm to 10 nm
  • B) 0.6–400 nm
  • C) 400–700 nm
  • D) 10–100 nm
• Q2: X-rays are used for:
  • A) Detecting bone fractures
  • B) Radio communication
  • C) Infrared heating
  • D) Sterilization
• Q3: Frequency of X-ray with wavelength 1 nm is:
  • A) 3 × 10¹⁷ Hz
  • B) 3 × 10¹⁸ Hz
  • C) 3 × 10¹⁶ Hz
  • D) 3 × 10¹⁵ Hz

Answer Key

• Q1: A; X-rays wavelength is 10^-4 nm to 10 nm.
• Q2: A; X-rays detect fractures and flaws.
• Q3: A; \( \nu = \frac{3 \times 10^{8}}{1 \times 10^{-9}} = 3 \times 10^{17} \) Hz.

Memory Aids

• X-rays = bone imaging + metal flaw detection.

Gamma Rays

Properties and Uses

• Wavelength: 10^-14 m to 10^-10 m; frequency: 3 × 10¹⁸ to 3 × 10²² Hz.
• Used in cancer therapy to destroy cancer cells.
• Used to preserve food by killing microorganisms.
• Used to study atomic nucleus structure.

Solved Examples

• Ex 1: Calculate frequency of gamma ray with wavelength 10^-12 m.
  • \( \nu = \frac{3 \times 10^{8}}{10^{-12}} = 3 \times 10^{20} \) Hz.

Practice Set

• Q1: Gamma rays wavelength range is:
  • A) 10^-14 m to 10^-10 m
  • B) 10^-10 m to 10^-6 m
  • C) 10^-6 m to 10^-2 m
  • D) 10^-2 m to 1 m
• Q2: Gamma rays are used in:
  • A) Cancer therapy
  • B) Radio broadcasting
  • C) Infrared heating
  • D) Visible light illumination
• Q3: Frequency of gamma ray with wavelength 5 × 10^-12 m is:
  • A) 6 × 10¹⁹ Hz
  • B) 6 × 10²⁰ Hz
  • C) 1.5 × 10²⁰ Hz
  • D) 1.5 × 10¹⁹ Hz

Answer Key

• Q1: A; Gamma rays wavelength is 10^-14 m to 10^-10 m.
• Q2: A; Gamma rays used in cancer therapy.
• Q3: B; \( \nu = \frac{3 \times 10^{8}}{5 \times 10^{-12}} = 6 \times 10^{19} \) Hz (corrected: 6×10¹⁹ Hz).

Memory Aids

• Gamma rays = highest energy, cancer therapy, food preservation.