#wavelength calculation

Refraction, Refractive Index and Critical Angle

What is the wavelength of light in a medium with refractive index 1.6 if its wavelength in vacuum is \( 640 \, \text{nm}

**Refractive index** n = c/v, c=3×10⁸ m/s vacuum, v speed in medium, for v=2.25×10⁸ m/s n=3/2.25=1.33, for v=1.8×10⁸ n=1.67, for n=1.3 v=3×10⁸/1.3=2.31×10⁸ m/s, for n=1.6 v=1.875×10⁸ m/s, for n=1.2 λ_medium = λ_vacuum/n =600/1.2=500 nm, for n=1.75 λ=700/1.75=400 nm, for n=1.5 λ=750/1.5=500 nm, wavelength in medium λ' = λ/n. Wavelength in a medium λ_m = (λvₐcuuₘ/n) . Given λvₐcuuₘ = 640 nm , n = 1.6 , λ_m = (640/1.6) = 400 nm . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' = λ/n and A = 2a

Ref: NCERT > Physics Book > Wave Optics > Refraction, Refractive Index and Critical Angle

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De Broglie Wavelength and Matter Waves

Light of wavelength \( 450 \, \text{nm} \) is incident on a metal. The stopping potential is \( 0.5 \, \text{V} \). What

**Matter waves** wave-like properties of moving particles, macroscopic objects not exhibit measurable wave-like because λ = h/(m v) extremely small for large m, e.g., 1 kg at 1 m/s λ=6.6×10⁻³⁴ m undetectable, for electron 9.11×10⁻³¹ kg at 2×10⁶ m/s λ=0.36 nm measurable via diffraction, de Broglie hypothesis λ = h/p proposed by de Broglie 1924, Davisson-Germer experiment confirmed. E = (h c/λ) = (1240/450) ≈ 2.76 eV . Kₘₐₓ = e V₀ = 0.5 eV . Φ₀ = E - Kₘₐₓ = 2.76 - 0.5 = 2.26 eV . λ₀ = (h c/Φ₀) = (1240/2.26) ≈ 549 nm . Applying E = h f =

Ref: NCERT > Physics Book > Dual Nature of Matter > De Broglie Wavelength and Matter Waves

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De Broglie Wavelength and Matter Waves

The minimum wavelength of X-rays from a \( 25 \, \text{kV} \) tube is: (Take \( h = 6.63 \times 10^{-34} \, \text{J s} \

**De Broglie wavelength** λ = h/p = h/(m v) = h/√(2 m e V) for electron accelerated through V, for V=100 V λ= h/√(2 m e×100)=1.227/√V nm=0.1227 nm, for 50 V 0.173 nm, for particle mass 3.0×10⁻³⁰ kg v=10⁶ m/s λ=6.63×10⁻³⁴/(3×10⁻³⁰×10⁶)=2.21×10⁻¹⁰ m=0.221 nm, inversely proportional to momentum, property inversely proportional to de Broglie wavelength is momentum. E = e V = 1.6 × 10⁻¹⁹ × 25 × 10³ = 4.0 × 10⁻¹⁵ J . λₘiₙ = (h c/E) = (6.63 × 10⁻³⁴ × 3 × 10⁸/4.0 × 10⁻¹⁵) = 4.9725 × 10⁻¹¹ m ≈ 0.0497 nm . Applying E = h f = h c/λ,

Ref: NCERT > Physics Book > Dual Nature of Matter > De Broglie Wavelength and Matter Waves

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De Broglie Wavelength and Matter Waves

The de Broglie wavelength of a particle of mass \( 5.0 \times 10^{-30} \, \text{kg} \) moving at \( 2.0 \times 10^5 \, \

**De Broglie wavelength calculations** for electron moving at 2×10⁶ m/s λ= h/(m v)=6.63×10⁻³⁴/(9.11×10⁻³¹×2×10⁶)=3.64×10⁻¹⁰ m=0.364 nm, for mass 5×10⁻³⁰ kg v=2×10⁵ m/s λ=6.63×10⁻³⁴/(1×10⁻²⁴)=6.63×10⁻¹⁰ m, showing λ decreases with increasing m v. Momentum p = m v = 5.0 × 10⁻³⁰ × 2.0 × 10⁵ = 1.0 × 10⁻²⁴ kg m/s . λ = (h/p) = (6.63 × 10⁻³⁴/1.0 × 10⁻²⁴) = 6.63 × 10⁻¹⁰ m = 0.663 nm . Applying E = h f = h c/λ, p = h/λ, K_max = h f - Φ, Φ = h f₀ = h c/λ₀, λ = h/√(2 m e V) and f_max = e V/h, result 0.663

Ref: NCERT > Physics Book > Dual Nature of Matter > De Broglie Wavelength and Matter Waves

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Photon Energy, Momentum, Power and X-rays

A light source emits \( 1.0 \times 10^{16} \) photons per second with a power of \( 4.0 \, \text{mW} \). What is the wav

**Photon power** light source emits 1.0×10¹⁶ photons/s power 4.0 mW, energy per photon E= P/N =4×10⁻³/10¹⁶=4×10⁻¹⁹ J, wavelength λ= h c/E=6.63×10⁻³⁴×3×10⁸/4×10⁻¹⁹=497 nm, for 6×10¹⁵ photons/s 3.0 mW E=5×10⁻¹⁹ J, for beam 2.5×10¹⁵ photons/s 4×10⁻¹⁹ J each power=10⁻³ W=1 mW, illustrating P = N h f. Energy per photon E = (P/N) = (4.0 × 10⁻³/1.0 × 10¹⁶) = 4.0 × 10⁻¹⁹ J . E = (h c/λ) ⇒ λ = (h c/E) = (6.63 × 10⁻³⁴ × 3 × 10⁸/4.0 × 10⁻¹⁹) ≈ 4.9725 × 10⁻⁷ m = 497.25 nm . Applying E = h f = h c/λ, p = h/λ, K_max = h

Ref: NCERT > Physics Book > Dual Nature of Matter > Photon Energy, Momentum, Power and X-rays

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De Broglie Wavelength and Matter Waves

An electron is accelerated through a potential difference of \( 100 \, \text{V} \). What is its de Broglie wavelength? (

**De Broglie wavelength** λ = h/p = h/(m v) = h/√(2 m e V) for electron accelerated through V, for V=100 V λ= h/√(2 m e×100)=1.227/√V nm=0.1227 nm, for 50 V 0.173 nm, for particle mass 3.0×10⁻³⁰ kg v=10⁶ m/s λ=6.63×10⁻³⁴/(3×10⁻³⁰×10⁶)=2.21×10⁻¹⁰ m=0.221 nm, inversely proportional to momentum, property inversely proportional to de Broglie wavelength is momentum. Kinetic energy K = e V = 1.6 × 10⁻¹⁹ × 100 = 1.6 × 10⁻¹⁷ J . Momentum p = √(2 m K) = √(2 × 9.11 × 10⁻³¹ × 1.6 × 10⁻¹⁷) ≈ 5.39 × 10⁻²⁴ kg m/s . λ = (h/p) = (6.63 × 10⁻³⁴/5.39 ×

Ref: NCERT > Physics Book > Dual Nature of Matter > De Broglie Wavelength and Matter Waves

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Applications of EM Waves in Communication and Medicine

An electromagnetic wave has a wave number \( k = 1 \, \text{rad/m} \). What is its wavelength in vacuum?

**Microwaves in ovens** cause water molecules to rotate at 2.45 GHz, friction heats food, penetration depth few cm, efficient heating, also radar uses reflection of microwaves from objects, Doppler shift measures speed, medical diathermy uses microwaves for tissue heating. The wave number k = (2 π/λ) . Given k = 1 rad/m , we have λ = (2 π/k) = (2 π/1) ≈ 6.28 m . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields 6.28 m, illustrating EM wave transverse nature and Maxwell's displacement current concept.

Ref: NCERT > Physics Book > Electromagnetic Waves > Applications of EM Waves in Communication and Medicine

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Energy, Intensity and Momentum of EM Waves

An electromagnetic wave in vacuum has a frequency of \( 25 \, \text{MHz} \). What is its wavelength? (Given \( c = 3 \ti

**Momentum of EM wave** p = U/c, U energy, radiation pressure exerts force F = I A/c, small but measurable, comet tail pushed by sunlight, solar sail concept. For E₀=45 V/m, B₀=1.5×10⁻⁷ T, intensity I =0.5×3×10⁸×8.85×10⁻¹²×45²≈2.69 W/m². Using v λ = c , we have λ = (c/v) = (3 × 10⁸/25 × 10⁶) = 12 m . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields 12 m, illustrating EM wave transverse nature and Maxwell's displacement current concept.

Ref: NCERT > Physics Book > Electromagnetic Waves > Energy, Intensity and Momentum of EM Waves

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Displacement Current and Ampere-Maxwell Law

A radio tunes to stations in the \( 7.5 \, \text{MHz} \) to \( 12 \, \text{MHz} \) band. What is the corresponding wavel

**Ampere-Maxwell law** ∮ B·dl = μ₀(I_c + ε₀ dΦ_E/dt) generalizes Ampere's law, displacement current arises from time-varying electric field, source of magnetic field like conduction current. For rate of change of flux 2×10¹¹ V·m/s, I_d = ε₀×2×10¹¹ =8.85×10⁻¹²×2×10¹¹=1.77 A. Using λ = (c/v) , for 7.5 MHz , λ = (3 × 10⁸/7.5 × 10⁶) = 40 m . For 12 MHz , λ = (3 × 10⁸/12 × 10⁶) = 25 m . So the range is 25 m to 40 m . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields 25 m

Ref: NCERT > Physics Book > Electromagnetic Waves > Displacement Current and Ampere-Maxwell Law

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Ultraviolet, X-rays, Gamma Rays and Their Properties

An electromagnetic wave travels in vacuum with frequency \( 10 \, \text{MHz} \). What is its wavelength? (Given \( c = 3

**X-rays in medicine** treat certain cancers because high-energy photons ionize and damage DNA of malignant cells, also imaging because bones absorb more than tissue, contrast. Gamma rays from Co-60 used for radiotherapy, energy ≈1.17 and 1.33 MeV, highly penetrating. Using v λ = c , we have λ = (c/v) = (3 × 10⁸/10 × 10⁶) = 30 m . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields 30 m, illustrating EM wave transverse nature and Maxwell's displacement current concept.

Ref: NCERT > Physics Book > Electromagnetic Waves > Ultraviolet, X-rays, Gamma Rays and Their Properties

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Ultraviolet, X-rays, Gamma Rays and Their Properties

An electromagnetic wave in vacuum has a frequency of \( 75 \, \text{MHz} \). What is its wavelength? (Given \( c = 3 \ti

**Gamma rays** λ10¹⁹ Hz, produced by nuclear transitions and radioactive decay, associated with nuclear processes because nuclear energy levels MeV vs atomic eV, highest photon energy, used in cancer treatment due to high penetration and cell damage, also sterilization, astronomy. Using v λ = c , we have λ = (c/v) = (3 × 10⁸/75 × 10⁶) = 4 m . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields 4 m, illustrating EM wave transverse nature and Maxwell's displacement current concept.

Ref: NCERT > Physics Book > Electromagnetic Waves > Ultraviolet, X-rays, Gamma Rays and Their Properties

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Ultraviolet, X-rays, Gamma Rays and Their Properties

An electromagnetic wave in vacuum has a frequency of \( 30 \, \text{MHz} \). What is its wavelength? (Given \( c = 3 \ti

**X-rays in medicine** treat certain cancers because high-energy photons ionize and damage DNA of malignant cells, also imaging because bones absorb more than tissue, contrast. Gamma rays from Co-60 used for radiotherapy, energy ≈1.17 and 1.33 MeV, highly penetrating. Using v λ = c , we have λ = (c/v) = (3 × 10⁸/30 × 10⁶) = 10 m . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields 10 m, illustrating EM wave transverse nature and Maxwell's displacement current concept.

Ref: NCERT > Physics Book > Electromagnetic Waves > Ultraviolet, X-rays, Gamma Rays and Their Properties

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