De Broglie Wavelength and Matter Waves

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

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 de Broglie wavelength of a particle of mass \( 2.0 \times 10^{-30} \, \text{kg} \) moving at \( 1.5 \times 10^6 \, \

**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. p = m v = 2.0 × 10⁻³⁰ × 1.5 × 10⁶ = 3.0 × 10⁻²⁴ kg m/s . λ = (h/p) = (6.63 × 10⁻³⁴/3.0 × 10⁻²⁴) = 2.21 × 10⁻¹⁰ m = 0.221 nm . Applying E = h f = h c/λ, p =

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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

Which of the following properties of photoelectrons is independent of the intensity of incident light?

**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. The maximum kinetic energy depends on frequency ( Kₘₐₓ = h v - Φ₀ ), not intensity, which only affects the number of electrons. 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 Maximum kinetic energy follows, reflecting photoelectric and de Broglie relations.

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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

A light beam emits \( 2.5 \times 10^{15} \) photons per second, each of energy \( 4.0 \times 10^{-19} \, \text{J} \). 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. Power P = N × E . P = 2.5 × 10¹⁵ × 4.0 × 10⁻¹⁹ = 1.0 × 10⁻³ W = 1.0 mW . Applying E = h f = h c/λ, p = h/λ, K_max = h f - Φ, Φ = h f₀ = h c/λ₀,

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De Broglie Wavelength and Matter Waves

Which of the following is a key feature of photons that distinguishes them from charged particles?

**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. Photons are electrically neutral and thus not deflected by electric or magnetic fields, unlike charged particles such as electrons. 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 Are neutral follows, reflecting photoelectric and de Broglie relations.

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De Broglie Wavelength and Matter Waves

The de Broglie wavelength of an electron is \( 0.2 \, \text{nm} \). What is its speed? (Take \( h = 6.63 \times 10^{-34}

**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. p = (h/λ) = (6.63 × 10⁻³⁴/0.2 × 10⁻⁹) = 3.315 × 10⁻²⁴ kg m/s . v = (p/m) = (3.315 × 10⁻²⁴/9.11 × 10⁻³¹) ≈ 3.64 × 10⁶ m/s . Applying E = h f = h c/λ, p = h/λ, K_max = h f -

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De Broglie Wavelength and Matter Waves

Which scientist proposed that moving particles of matter exhibit wave-like properties?

**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. Louis de Broglie proposed the wave-particle duality of matter, suggesting that particles like electrons have associated wavelengths ( λ = (h/p) ). Applying E = h f = h c/λ, p = h/λ, K_max = h f - Φ, Φ = h f₀ = h c/λ₀, λ = h/√(2

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