#frequency calculation

Photon Energy, Momentum, Power and X-rays

A photon has an energy of \( 5.0 \times 10^{-19} \, \text{J} \). What is its frequency? (Take \( h = 6.63 \times 10^{-34

**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. E = h v . v = (E/h) = (5.0 × 10⁻¹⁹/6.63 × 10⁻³⁴) ≈ 7.54 × 10¹⁴ Hz . 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 7.54 × 10¹⁴ Hz

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

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Photon Energy, Momentum, Power and X-rays

A photon has momentum \( 1.5 \times 10^{-27} \, \text{kg m/s} \). What is its frequency? (Take \( h = 6.63 \times 10^{-3

**X-rays from tube** maximum frequency f_max = e V/h, minimum wavelength λ_min = h c/(e V), Duane-Hunt law, for 10 kV f_max=1.6×10⁻¹⁹×10⁴/6.63×10⁻³⁴=2.41×10¹⁸ Hz, λ_min= c/f_max=0.124 nm, for 35 kV λ_min=1240/(35000) eV·nm? Actually 35 keV photon λ=1240/35000=0.0354 nm, for 15 kV 0.0827 nm, for 25 kV 0.0496 nm, for 30 kV 0.0413 nm. For a photon, p = (h v/c) , but also p = (h/λ) , and v = (c/λ) . So, v = (p c/h) , but directly, λ = (h/p) = (6.63 × 10⁻³⁴/1.5 × 10⁻²⁷) = 4.42 × 10⁻⁷ m . v = (c/λ) = (3 × 10⁸/4.42 × 10⁻⁷) ≈ 6.79

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

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Ultraviolet, X-rays, Gamma Rays and Their Properties

An electromagnetic wave in vacuum has a wavelength of \( 8 \, \text{m} \). What is its frequency? (Given \( c = 3 \times

**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 v = (c/λ) = (3 × 10⁸/8) = 3.75 × 10⁷ Hz . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields 3.75 × 10⁷ Hz, 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 wavelength of \( 6 \, \text{m} \). What is its frequency? (Given \( c = 3 \times

**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 v = (c/λ) = (3 × 10⁸/6) = 5 × 10⁷ Hz . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields 5 × 10⁷ Hz, 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 wavelength of \( 5 \, \text{m} \). What is its frequency? (Given \( c = 3 \times

**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 v = (c/λ) = (3 × 10⁸/5) = 6 × 10⁷ Hz . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields 6 × 10⁷ Hz, 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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Equations of SHM, Phase and Angular Frequency

A simple pendulum has a frequency of \( 0.25 \, \text{Hz} \) on Earth (\( g = 9.8 \, \text{m/s}^2 \)). What is its lengt

**Angular frequency** ω = 2πf = 2π/T characterizes rapidity, independent of amplitude for SHM. Phase constant φ shifts sine/cosine, allowing any initial condition, e.g., x(0)=0 requires φ=0 for sine form. Displacement, velocity, acceleration share ω but differ in phase by 90° and 180°. Period: T = (1/v) = (1/0.25) = 4 s . T = 2π √((L/g)) ⇒ 4 = 2π √((L/9.8)) . √((L/9.8)) = (4/2π) ≈ 0.637 ⇒ (L/9.8) = (0.637)² ⇒ L ≈ 3.98 m . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 3.98 m follows, reflecting

Ref: NCERT > Physics Book > Oscillations > Equations of SHM, Phase and Angular Frequency

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Beats Phenomenon

Two strings produce beats of 8 Hz. One has a frequency of 440 Hz. When the tension in the second string is increased, th

**Beats arise** when two waves of slightly different frequencies f₁ and f₂ superpose, producing amplitude modulation at beat frequency f_beat = |f₁ - f₂| (Hz). Intensity waxes and wanes periodically, number of beats in interval Δt equals f_beat·Δt, loudness variation audible when f_beat < 10 Hz. Let v₂ be the original frequency. |440 - v₂| = 8 ⇒ v₂ = 432 Hz or 448 Hz . Increasing tension increases frequency. If v₂ = 432 , new v₂’ > 432 , beat = 440 - v₂’ < 8 , becomes 6 Hz ( v₂’ = 434 ), consistent. If v₂ = 448 , beat increases, contradicts.

Ref: NCERT > Physics Book > Waves > Beats Phenomenon

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Sound Waves, Reflection and Characteristics

A pipe closed at one end has a length of 0.6 m and resonates at its second harmonic with a speed of sound of 360 m/s. Wh

**Sound wave reflection** at rigid wall behaves like string fixed end, displacement inverted. Equation of reflected wave includes sign change and direction reversal, amplitude unchanged but sign may flip, explaining standing wave formation with incident wave. For closed pipe: v_n = (n + (1/2)) (v/2L) , n = 1 for second harmonic. v₁ = (1 + (1/2)) (360/2 × 0.6) = 1.5 × (360/1.2) = 1.5 × 300 = 450 Hz . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 450 Hz, illustrating frequency-length-speed interdependence and quantization by boundaries.

Ref: NCERT > Physics Book > Waves > Sound Waves, Reflection and Characteristics

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Beats Phenomenon

A stationary wave on a string fixed at both ends has a frequency of 90 Hz and a wave speed of 36 m/s. What is the length

**Superposition of nearly equal frequencies** creates resultant y = 2A cos(Δω·t/2) sin(ω_avg·t), envelope frequency Δf/2, beat frequency Δf. Total beats heard in Δt is f_beat·Δt, explaining counting over seconds. Fundamental: v₁ = (v/2L) . 90 = (36/2L) ⇒ 2L = (36/90) ⇒ 2L = 0.4 ⇒ L = 0.2 m . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 0.2 m, illustrating frequency-length-speed interdependence and quantization by boundaries. This aligns with NCERT Class 11 treatment, emphasizing conservation, symmetry and dimensional consistency useful for

Ref: NCERT > Physics Book > Waves > Beats Phenomenon

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Sound Waves, Reflection and Characteristics

A pipe open at both ends has a length of 0.25 m and a speed of sound of 340 m/s. What is the frequency of its second har

**Sound wave reflection** at rigid wall behaves like string fixed end, displacement inverted. Equation of reflected wave includes sign change and direction reversal, amplitude unchanged but sign may flip, explaining standing wave formation with incident wave. For open pipe: v_n = (n v/2L) . Second harmonic ( n = 2 ): v₂ = (2 × 340/2 × 0.25) = (680/0.5) = 1360 Hz . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 1360 Hz, illustrating frequency-length-speed interdependence and quantization by boundaries.

Ref: NCERT > Physics Book > Waves > Sound Waves, Reflection and Characteristics

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Superposition and Interference of Waves

A string of length 1.5 m fixed at both ends has a wave speed of 45 m/s. What is the frequency of its fifth harmonic?

**Interference of waves** produces enhancement or cancellation based on phase. Two equal amplitude waves out of phase by π cancel completely, A = 0, while in-phase superposition doubles amplitude to 2a, demonstrating energy redistribution without violation of conservation. v_n = (n v/2L) . Fifth harmonic ( n = 5 ): v₅ = (5 × 45/2 × 1.5) = (225/3) = 75 Hz . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 75 Hz, illustrating frequency-length-speed interdependence and quantization by boundaries.

Ref: NCERT > Physics Book > Waves > Superposition and Interference of Waves

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Beats Phenomenon

Two strings produce beats of 7 Hz. One has a frequency of 392 Hz. When the tension in the second string is increased, th

**Superposition of nearly equal frequencies** creates resultant y = 2A cos(Δω·t/2) sin(ω_avg·t), envelope frequency Δf/2, beat frequency Δf. Total beats heard in Δt is f_beat·Δt, explaining counting over seconds. Let v₂ be the original frequency. |392 - v₂| = 7 ⇒ v₂ = 385 Hz or 399 Hz . Increasing tension increases frequency. If v₂ = 385 , new v₂’ > 385 , beat = 392 - v₂’ < 7 , becomes 5 Hz ( v₂’ = 387 ), consistent. If v₂ = 399 , beat increases, contradicts. So, v₂ = 385 Hz . Using v = fλ and standing-wave condition fₙ = n v/(2L)

Ref: NCERT > Physics Book > Waves > Beats Phenomenon

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