#Planck constant

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 =

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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,

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De Broglie Wavelength and Matter Waves

An electron moves with a speed of \( 3.0 \times 10^6 \, \text{m/s} \). What is its de Broglie wavelength? (Take \( h = 6

**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. Momentum p = m v = 9.11 × 10⁻³¹ × 3.0 × 10⁶ = 2.733 × 10⁻²⁴ kg m/s . λ = (h/p) = (6.63 × 10⁻³⁴/2.733 × 10⁻²⁴) ≈ 2.425 × 10⁻¹⁰ m = 0.2425 nm . Applying E = h f = h c/λ, p

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Photon Energy, Momentum, Power and X-rays

A photon has a wavelength of \( 200 \, \text{nm} \). What is its momentum? (Take \( h = 6.63 \times 10^{-34} \, \text{J

**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. p = (h/λ) = (6.63 × 10⁻³⁴/200 × 10⁻⁹) = 3.315 × 10⁻²⁷ kg m/s . Applying E = h f = h c/λ, p = h/λ, K_max = h f - Φ, Φ = h f₀ = h

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Photoelectric Effect - Work Function and Threshold

Light of frequency \( 6.2 \times 10^{14} \, \text{Hz} \) is incident on a metal with work function \( 2.1 \, \text{eV} \

**Photoelectric effect** demonstrates particle nature of light, photon energy E = h f, work function Φ = h f₀ minimum energy to eject electron, threshold frequency f₀ = Φ/h, threshold wavelength λ₀ = hc/Φ, hc=1240 eV·nm, maximum kinetic energy K_max = h f - Φ = e V₀, V₀ stopping potential, photocurrent proportional to intensity (number of photons) when f>f₀, saturation current depends on intensity, frequency determines K_max not current. E = h v = 6.63 × 10⁻³⁴ × 6.2 × 10¹⁴ = 4.1106 × 10⁻¹⁹ J . E = (4.1106 × 10⁻¹⁹/1.6 × 10⁻¹⁹) ≈ 2.569 eV . Kₘₐₓ = E - Φ₀ =

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Photon Energy, Momentum, Power and X-rays

A light source emits photons of energy \( 4.5 \times 10^{-19} \, \text{J} \) at a rate of \( 2.0 \times 10^{16} \) photo

**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 c/E) = (6.63 × 10⁻³⁴ × 3 × 10⁸/4.5 × 10⁻¹⁹) = 4.42 × 10⁻⁷ m = 442 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)

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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,

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Photoelectric Effect - Work Function and Threshold

The threshold frequency of a metal is \( 3.0 \times 10^{14} \, \text{Hz} \). What is the maximum kinetic energy of elect

**Classical wave theory fails** to explain instant emission, threshold existence, K_max dependence on frequency not intensity, saturation current dependence on intensity. Observations: K_max independent of intensity, exists threshold frequency, emission instantaneous, all explained by photon model E = h f, one photon ejects one electron, energy conservation h f = Φ + K_max. Φ₀ = h v₀ = 6.63 × 10⁻³⁴ × 3.0 × 10¹⁴ = 1.989 × 10⁻¹⁹ J . E = (h c/λ) = (6.63 × 10⁻³⁴ × 3 × 10⁸/500 × 10⁻⁹) = 3.978 × 10⁻¹⁹ J . Kₘₐₓ = E - Φ₀ = 3.978 × 10⁻¹⁹ - 1.989 × 10⁻¹⁹ =

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Photoelectric Effect - Work Function and Threshold

Light of frequency \( 7.5 \times 10^{14} \, \text{Hz} \) is incident on a metal with work function \( 2.5 \, \text{eV} \

**Photoelectric effect** demonstrates particle nature of light, photon energy E = h f, work function Φ = h f₀ minimum energy to eject electron, threshold frequency f₀ = Φ/h, threshold wavelength λ₀ = hc/Φ, hc=1240 eV·nm, maximum kinetic energy K_max = h f - Φ = e V₀, V₀ stopping potential, photocurrent proportional to intensity (number of photons) when f>f₀, saturation current depends on intensity, frequency determines K_max not current. Photon energy E = h v = 6.63 × 10⁻³⁴ × 7.5 × 10¹⁴ = 4.9725 × 10⁻¹⁹ J . E = (4.9725 × 10⁻¹⁹/1.6 × 10⁻¹⁹) ≈ 3.11 eV . Kₘₐₓ = E -

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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 =

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Photoelectric Effect - Work Function and Threshold

The threshold frequency of a metal is \( 5.0 \times 10^{14} \, \text{Hz} \). What is the stopping potential for light of

**Work function** Φ minimum energy to escape, for Φ=2.2 eV λ₀=1240/2.2=563.6 nm, Φ=3.0 eV threshold frequency f₀=Φ/h=3×1.6×10⁻¹⁹/6.63×10⁻³⁴=7.24×10¹⁴ Hz, stopping potential V₀ = (h f - Φ)/e, for 0.7 V K_max=0.7 eV, for 1.5 V K_max=1.5 eV, intensity increase increases photocurrent not K_max, frequency below threshold no emission regardless of intensity. Φ₀ = h v₀ = 6.63 × 10⁻³⁴ × 5.0 × 10¹⁴ = 3.315 × 10⁻¹⁹ J . E = h v = 6.63 × 10⁻³⁴ × 7.0 × 10¹⁴ = 4.641 × 10⁻¹⁹ J . Kₘₐₓ = E - Φ₀ = 4.641 × 10⁻¹⁹ - 3.315 × 10⁻¹⁹ = 1.326 × 10⁻¹⁹ J .

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Photoelectric Effect - Work Function and Threshold

Light of frequency \( 7.5 \times 10^{14} \, \text{Hz} \) produces a stopping potential of \( 0.9 \, \text{V} \). What is

**Work function** Φ minimum energy to escape, for Φ=2.2 eV λ₀=1240/2.2=563.6 nm, Φ=3.0 eV threshold frequency f₀=Φ/h=3×1.6×10⁻¹⁹/6.63×10⁻³⁴=7.24×10¹⁴ Hz, stopping potential V₀ = (h f - Φ)/e, for 0.7 V K_max=0.7 eV, for 1.5 V K_max=1.5 eV, intensity increase increases photocurrent not K_max, frequency below threshold no emission regardless of intensity. E = h v = 6.63 × 10⁻³⁴ × 7.5 × 10¹⁴ = 4.9725 × 10⁻¹⁹ J . E = (4.9725 × 10⁻¹⁹/1.6 × 10⁻¹⁹) ≈ 3.108 eV . Kₘₐₓ = e V₀ = 0.9 eV . Φ₀ = E - Kₘₐₓ = 3.108 - 0.9 ≈ 2.208 eV . Applying E = h f

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