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#threshold wavelength

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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 =

Ref: NCERT > Physics Book > Dual Nature of Matter > De Broglie Wavelength and Matter Waves

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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

Ref: NCERT > Physics Book > Dual Nature of Matter > De Broglie Wavelength and Matter Waves

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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.

Ref: NCERT > Physics Book > Dual Nature of Matter > De Broglie Wavelength and Matter Waves

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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

Ref: NCERT > Physics Book > Dual Nature of Matter > De Broglie Wavelength and Matter Waves

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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,

Ref: NCERT > Physics Book > Dual Nature of Matter > Photoelectric Effect - Work Function and Threshold

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Photoelectric Effect - Work Function and Threshold

The work function of a metal is \( 2.2 \, \text{eV} \). What is the threshold wavelength in nm? (Take \( h c = 1240 \, \

**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. λ₀ = (h c/Φ₀) = (1240/2.2) ≈ 563.64 nm . 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 > Photoelectric Effect - Work Function and Threshold

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