#stopping potential

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 =

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De Broglie Wavelength and Matter Waves

What is the significance of the stopping potential in the photoelectric effect?

**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. The stopping potential is the minimum negative voltage that stops the most energetic photoelectrons, directly related to their maximum kinetic energy ( e V₀ = Kₘₐₓ ). 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

Light of wavelength \( 500 \, \text{nm} \) is incident on a metal with stopping potential \( 0.6 \, \text{V} \). What is

**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) = (1240/500) = 2.48 eV . Kₘₐₓ = e V₀ = 0.6 eV . Φ₀ = E - Kₘₐₓ = 2.48 - 0.6 = 1.88 eV . λ₀ = (h c/Φ₀) = (1240/1.88) ≈ 659.57 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

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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 frequency \( 8.5 \times 10^{14} \, \text{Hz} \) is incident on a metal with threshold frequency \( 4.0 \times 1

**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 v = 6.63 × 10⁻³⁴ × 8.5 × 10¹⁴ = 5.6355 × 10⁻¹⁹ J . Φ₀ = h v₀ = 6.63 × 10⁻³⁴ × 4.0 × 10¹⁴ = 2.652 × 10⁻¹⁹ J . Kₘₐₓ = E - Φ₀ = 5.6355 × 10⁻¹⁹ - 2.652

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De Broglie Wavelength and Matter Waves

The maximum speed of photoelectrons emitted from a surface is \( 5.0 \times 10^5 \, \text{m/s} \). What is the stopping

**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. Kₘₐₓ = (1/2) m vₘₐₓ² = (1/2) × 9.11 × 10⁻³¹ × (5.0 × 10⁵)² = 1.13875 × 10⁻¹⁹ J . V₀ = (Kₘₐₓ/e) = (1.13875 × 10⁻¹⁹/1.6 × 10⁻¹⁹) ≈ 0.71 V . Applying E = h f = h c/λ, p = h/λ, K_max = h f

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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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Photoelectric Effect - Work Function and Threshold

The stopping potential for photoelectrons emitted from a metal surface is \( 1.2 \, \text{V} \). What is the maximum kin

**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. Maximum kinetic energy Kₘₐₓ = e V₀ . Kₘₐₓ = 1.6 × 10⁻¹⁹ × 1.2 = 1.92 × 10⁻¹⁹ J . 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

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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 wavelength \( 400 \, \text{nm} \) is incident on a metal with a stopping potential of \( 1.2 \, \text{V} \). Wh

**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. E = (h c/λ) = (1240/400) = 3.1 eV . Kₘₐₓ = e V₀ = 1.2 eV . Φ₀ = E - Kₘₐₓ = 3.1 - 1.2 = 1.9 eV . λ₀ = (h c/Φ₀) = (1240/1.9) ≈ 652.63 nm . Applying E = h f = h c/λ, p = h/λ,

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Photoelectric Effect - Work Function and Threshold

What is the nature of the relationship between the stopping potential and the frequency of incident light in the photoel

**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. The stopping potential V₀ varies linearly with frequency ( e V₀ = h v - Φ₀ ), as shown by experimental graphs and Einstein’s equation. 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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Photoelectric Effect - Work Function and Threshold

A photon of energy \( 2.5 \, \text{eV} \) strikes a metal surface. The stopping potential is \( 0.8 \, \text{V} \). What

**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. Kₘₐₓ = e V₀ = 0.8 eV = 0.8 × 1.6 × 10⁻¹⁹ = 1.28 × 10⁻¹⁹ J . Kₘₐₓ = (1/2) m vₘₐₓ² ⇒ vₘₐₓ = √((2 Kₘₐₓ/m)) = √((2 × 1.28 × 10⁻¹⁹/9.11 × 10⁻³¹)) ≈ 5.3 × 10⁵ m/s . Applying E = h f = h c/λ, p = h/λ, K_max

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