#photon momentum

De Broglie Wavelength and Matter Waves

The momentum of a photon is \( 2.21 \times 10^{-27} \, \text{kg m/s} \). What is its wavelength? (Take \( h = 6.63 \time

**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. For a photon, p = (h/λ) . λ = (h/p) = (6.63 × 10⁻³⁴/2.21 × 10⁻²⁷) ≈ 3.0 × 10⁻⁷ m = 300 nm . Applying E = h f = h c/λ, p = h/λ, K_max = h f - Φ, Φ = h f₀ =

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 particle of mass \( 1.5 \times 10^{-30} \, \text{kg} \) has the same momentum as a photon of wavelength \( 500 \, \tex

**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. Photon momentum p = (h/λ) = (6.63 × 10⁻³⁴/500 × 10⁻⁹) = 1.326 × 10⁻²⁷ kg m/s . v = (p/m) = (1.326 × 10⁻²⁷/1.5 × 10⁻³⁰) = 8.84 × 10² = 884 m/s . Applying E = h f = h c/λ, p = h/λ, K_max = h f - Φ, Φ = h f₀ = h c/λ₀, λ

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 a momentum of \( 2.0 \times 10^{-27} \, \text{kg m/s} \). What is its energy in joules? (Take \( c = 3 \tim

**Photon energy** E = h f = h c/λ, momentum p = h/λ = E/c, power P = N E/t, N number of photons per second, wavelength λ = h c/E, for photon momentum 1.5×10⁻²⁷ kg·m/s frequency f = p c/h? Actually E = p c, f = E/h = p c/h =1.5×10⁻²⁷×3×10⁸/6.63×10⁻³⁴=6.79×10¹⁴ Hz, energy E= h f =4.5×10⁻¹⁹ J, for 5.0×10⁻¹⁹ J frequency f=E/h=7.54×10¹⁴ Hz. For a photon, p = (E/c) . E = p c = 2.0 × 10⁻²⁷ × 3 × 10⁸ = 6.0 × 10⁻¹⁹ J . 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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Photon Energy, Momentum, Power and X-rays

A photon has a momentum of \( 1.8 \times 10^{-27} \, \text{kg m/s} \). What is its wavelength? (Take \( h = 6.63 \times

**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. λ = (h/p) = (6.63 × 10⁻³⁴/1.8 × 10⁻²⁷) ≈ 3.683 × 10⁻⁷ m = 368.3 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,

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 of frequency \( 7.0 \times 10^{14} \, \text{Hz} \) has what momentum? (Take \( h = 6.63 \times 10^{-34} \, \tex

**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 = 6.63 × 10⁻³⁴ × 7.0 × 10¹⁴ = 4.641 × 10⁻¹⁹ J . p = (E/c) = (4.641 × 10⁻¹⁹/3 × 10⁸) = 1.547 × 10⁻²⁷ kg m/s . Applying E = h f = h c/λ, p = h/λ, K_max = h f - Φ, Φ = h f₀ = h c/λ₀, λ =

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’s energy is \( 3.0 \, \text{eV} \). What is its momentum? (Take \( c = 3 \times 10^8 \, \text{m/s} \), \( 1 \,

**Photon energy** E = h f = h c/λ, momentum p = h/λ = E/c, power P = N E/t, N number of photons per second, wavelength λ = h c/E, for photon momentum 1.5×10⁻²⁷ kg·m/s frequency f = p c/h? Actually E = p c, f = E/h = p c/h =1.5×10⁻²⁷×3×10⁸/6.63×10⁻³⁴=6.79×10¹⁴ Hz, energy E= h f =4.5×10⁻¹⁹ J, for 5.0×10⁻¹⁹ J frequency f=E/h=7.54×10¹⁴ Hz. E = 3.0 × 1.6 × 10⁻¹⁹ = 4.8 × 10⁻¹⁹ J . For a photon, p = (E/c) = (4.8 × 10⁻¹⁹/3 × 10⁸) = 1.6 × 10⁻²⁷ kg m/s . Applying E = h f = h

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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