#photoelectric effect

Photoelectric Effect - Work Function and Threshold

The maximum speed of photoelectrons is \( 6.0 \times 10^5 \, \text{m/s} \) when light of wavelength \( 400 \, \text{nm}

**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. Kₘₐₓ = (1/2) m vₘₐₓ² = (1/2) × 9.11 × 10⁻³¹ × (6.0 × 10⁵)² = 1.6398 × 10⁻¹⁹ J . Kₘₐₓ = (1.6398 × 10⁻¹⁹/1.6 × 10⁻¹⁹) ≈ 1.025 eV . E = (h c/λ) = (1240/400) = 3.1 eV . Φ₀ = E - Kₘₐₓ = 3.1 - 1.025 ≈

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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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Photoelectric Effect - Work Function and Threshold

The maximum kinetic energy of photoelectrons is \( 2.5 \times 10^{-19} \, \text{J} \). What is the stopping potential? (

**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. Kₘₐₓ = e V₀ . V₀ = (Kₘₐₓ/e) = (2.5 × 10⁻¹⁹/1.6 × 10⁻¹⁹) ≈ 1.5625 V . 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 wavelength \( 300 \, \text{nm} \) produces a stopping potential of \( 1.5 \, \text{V} \). What is the work func

**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) = (1240/300) ≈ 4.13 eV . Kₘₐₓ = e V₀ = 1.5 eV . Φ₀ = E - Kₘₐₓ = 4.13 - 1.5 = 2.63 eV . 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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Photoelectric Effect - Work Function and Threshold

The work function of a metal is \( 3.0 \, \text{eV} \). Light of frequency \( 8.0 \times 10^{14} \, \text{Hz} \) is inci

**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 v = 6.63 × 10⁻³⁴ × 8.0 × 10¹⁴ = 5.304 × 10⁻¹⁹ J . E = (5.304 × 10⁻¹⁹/1.6 × 10⁻¹⁹) ≈ 3.315 eV . Kₘₐₓ = E - Φ₀ = 3.315 - 3.0 = 0.315 eV . V₀ = (Kₘₐₓ/e) = 0.315 V . Applying E =

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Photoelectric Effect - Work Function and Threshold

The work function of a metal is \( 1.8 \, \text{eV} \). What is the threshold frequency for this metal? (Take \( h = 6.6

**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 = 1.8 eV = 1.8 × 1.6 × 10⁻¹⁹ = 2.88 × 10⁻¹⁹ J . v₀ = (Φ₀/h) = (2.88 × 10⁻¹⁹/6.63 × 10⁻³⁴) ≈ 4.34 × 10¹⁴ Hz . Applying E = h f = h c/λ, p = h/λ, K_max = h f - Φ, Φ = h f₀

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Photoelectric Effect - Work Function and Threshold

Which of the following is a characteristic of the photoelectric effect that the classical wave theory fails to explain?

**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 existence of a threshold frequency, below which no emission occurs regardless of intensity, contradicts the wave theory’s continuous energy absorption model. 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

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Photoelectric Effect - Work Function and Threshold

The maximum kinetic energy of photoelectrons is \( 0.5 \, \text{eV} \) when light of frequency \( 6.0 \times 10^{14} \,

**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 × 6.0 × 10¹⁴ = 3.978 × 10⁻¹⁹ J . E = (3.978 × 10⁻¹⁹/1.6 × 10⁻¹⁹) ≈ 2.486 eV . Φ₀ = E - Kₘₐₓ = 2.486 - 0.5 = 1.986 eV . v₀ = (Φ₀/h) = (1.986 × 1.6 × 10⁻¹⁹/6.63 × 10⁻³⁴) ≈ 4.79

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Photoelectric Effect - Work Function and Threshold

The maximum kinetic energy of photoelectrons is \( 1.0 \, \text{eV} \) when light of wavelength \( 500 \, \text{nm} \) i

**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) = (1240/500) = 2.48 eV . Φ₀ = E - Kₘₐₓ = 2.48 - 1.0 = 1.48 eV . v₀ = (Φ₀/h) = (1.48 × 1.6 × 10⁻¹⁹/6.63 × 10⁻³⁴) ≈ 3.57 × 10¹⁴ Hz . Applying E = h f = h c/λ, p = h/λ, K_max

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Photoelectric Effect - Work Function and Threshold

The stopping potential for photoelectrons is \( 0.7 \, \text{V} \). What is the maximum kinetic energy in eV?

**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 = 0.7 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 0.7 eV follows, reflecting photoelectric and de Broglie relations.

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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

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Photoelectric Effect - Work Function and Threshold

Light of wavelength \( 450 \, \text{nm} \) produces a photocurrent that stops at \( 0.9 \, \text{V} \). What is the thre

**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 c/λ) = (1240/450) ≈ 2.76 eV . Kₘₐₓ = e V₀ = 0.9 eV . Φ₀ = E - Kₘₐₓ = 2.76 - 0.9 = 1.86 eV . v₀ = (Φ₀/h) =

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