#work function

De Broglie Wavelength and Matter Waves

Light of frequency \( 5.5 \times 10^{14} \, \text{Hz} \) produces photoelectrons with a maximum speed of \( 4.0 \times 1

**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. E = h v = 6.63 × 10⁻³⁴ × 5.5 × 10¹⁴ = 3.6465 × 10⁻¹⁹ J . E = (3.6465 × 10⁻¹⁹/1.6 × 10⁻¹⁹) ≈ 2.28 eV . Kₘₐₓ = (1/2) m vₘₐₓ² = (1/2) × 9.11 × 10⁻³¹ × (4.0 × 10⁵)² = 7.288 × 10⁻²⁰ J . Kₘₐₓ = (7.288 × 10⁻²⁰/1.6 × 10⁻¹⁹) ≈ 0.455 eV . Φ₀ = E - Kₘₐₓ = 2.28 - 0.455 ≈ 1.825 eV . Applying

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De Broglie Wavelength and Matter Waves

The stopping potential for photoelectrons from a metal is \( 2.0 \, \text{V} \) when illuminated with light of frequency

**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 = h v = 6.63 × 10⁻³⁴ × 6.0 × 10¹⁴ = 3.978 × 10⁻¹⁹ J . E = (3.978 × 10⁻¹⁹/1.6 × 10⁻¹⁹) ≈ 2.486 eV . Kₘₐₓ = e V₀ = 2.0 eV . Φ₀ = E - Kₘₐₓ = 2.486 - 2.0 = 0.486 eV

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De Broglie Wavelength and Matter Waves

Light of wavelength \( 450 \, \text{nm} \) is incident on a metal with work function \( 1.9 \, \text{eV} \). What is the

**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. E = (h c/λ) = (1240/450) ≈ 2.756 eV . Kₘₐₓ = E - Φ₀ = 2.756 - 1.9 ≈ 0.856 eV . 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 0.856 eV follows, reflecting photoelectric and de Broglie relations.

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De Broglie Wavelength and Matter Waves

The threshold wavelength for a metal is \( 600 \, \text{nm} \). What is its work function in eV? (Take \( h c = 1240 \,

**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. Work function Φ₀ = (h c/λ₀) . Φ₀ = (1240/600) ≈ 2.07 eV . 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 2.07 eV follows, reflecting photoelectric and de Broglie relations.

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De Broglie Wavelength and Matter Waves

The work function of a metal is \( 1.5 \, \text{eV} \). Light of wavelength \( 400 \, \text{nm} \) is incident on it. Wh

**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 = (h c/λ) = (1240/400) = 3.1 eV . Kₘₐₓ = E - Φ₀ = 3.1 - 1.5 = 1.6 eV . 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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De Broglie Wavelength and Matter Waves

The threshold wavelength of a metal is \( 620 \, \text{nm} \). What is the work function in eV? (Take \( h c = 1240 \, \

**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. Φ₀ = (h c/λ₀) = (1240/620) = 2.0 eV . 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 2.0

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Photon Energy, Momentum, Power and X-rays

Light of wavelength \( 400 \, \text{nm} \) produces photoelectrons with a stopping potential of \( 0.8 \, \text{V} \) fr

**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. Photon energy E = (h c/λ) = (1240/400) = 3.1 eV . Kₘₐₓ = e V₀ = 0.8 eV . Φ₀ = E - Kₘₐₓ = 3.1 - 0.8 = 2.3 eV . Applying E = h f = h c/λ, p = h/λ, K_max = h f - Φ, Φ = h

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Photon Energy, Momentum, Power and X-rays

The work function of a metal is \( 2.0 \, \text{eV} \). What is the threshold frequency for photoelectric emission from

**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. Work function Φ₀ = 2.0 eV = 2.0 × 1.6 × 10⁻¹⁹ = 3.2 × 10⁻¹⁹ J . Threshold frequency v₀ = (Φ₀/h) = (3.2 × 10⁻¹⁹/6.63 × 10⁻³⁴) ≈ 4.83 × 10¹⁴ Hz . 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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Photon Energy, Momentum, Power and X-rays

The threshold frequency of a metal is \( 6.0 \times 10^{14} \, \text{Hz} \). What is its work function in eV? (Take \( h

**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. Φ₀ = h v₀ = 6.63 × 10⁻³⁴ × 6.0 × 10¹⁴ = 3.978 × 10⁻¹⁹ J . Φ₀ = (3.978 × 10⁻¹⁹/1.6 × 10⁻¹⁹) ≈ 2.486 eV . Applying E = h f = h c/λ, p =

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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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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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Photoelectric Effect - Work Function and Threshold

The work function of a metal is \( 2.4 \, \text{eV} \). Light of wavelength \( 350 \, \text{nm} \) is incident on it. Wh

**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 c/λ) = (1240/350) ≈ 3.543 eV . Kₘₐₓ = E - Φ₀ = 3.543 - 2.4 = 1.143 eV = 1.143 × 1.6 × 10⁻¹⁹ = 1.8288 × 10⁻¹⁹ J . vₘₐₓ = √((2 Kₘₐₓ/m)) = √((2 × 1.8288 × 10⁻¹⁹/9.11 × 10⁻³¹)) ≈ 6.33 × 10⁵ m/s . Applying E =

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