Photoelectric Effect - Work Function and Threshold

Photoelectric Effect - Work Function and Threshold

In the photoelectric effect, what does the saturation current depend on, assuming a fixed frequency above the threshold?

**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. Saturation current depends on the intensity of light, as it determines the number of photons and thus the number of photoelectrons emitted per second. 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 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

Which of the following particles was historically identified as a universal constituent of matter through cathode ray ex

**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. J.J. Thomson’s experiments with cathode rays led to the discovery of electrons as fundamental negatively charged particles in matter. 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

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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

Which property of incident light determines the maximum kinetic energy of photoelectrons, assuming the metal remains the

**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. According to Einstein’s photoelectric equation, Kₘₐₓ = h v - Φ₀ , the maximum kinetic energy depends on the frequency of the incident light. Applying E = h f = h c/λ, p = h/λ,

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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

The threshold frequency of a metal is \( 3.0 \times 10^{14} \, \text{Hz} \). What is the maximum kinetic energy of elect

**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 = h v₀ = 6.63 × 10⁻³⁴ × 3.0 × 10¹⁴ = 1.989 × 10⁻¹⁹ J . E = (h c/λ) = (6.63 × 10⁻³⁴ × 3 × 10⁸/500 × 10⁻⁹) = 3.978 × 10⁻¹⁹ J . Kₘₐₓ = E - Φ₀ = 3.978 × 10⁻¹⁹ - 1.989 × 10⁻¹⁹ =

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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

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

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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Photoelectric Effect - Work Function and Threshold

The threshold frequency of a metal is \( 5.0 \times 10^{14} \, \text{Hz} \). What is the stopping potential for light of

**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. Φ₀ = h v₀ = 6.63 × 10⁻³⁴ × 5.0 × 10¹⁴ = 3.315 × 10⁻¹⁹ J . E = h v = 6.63 × 10⁻³⁴ × 7.0 × 10¹⁴ = 4.641 × 10⁻¹⁹ J . Kₘₐₓ = E - Φ₀ = 4.641 × 10⁻¹⁹ - 3.315 × 10⁻¹⁹ = 1.326 × 10⁻¹⁹ J .

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Photoelectric Effect - Work Function and Threshold

Which phenomenon provided evidence that light behaves as if it consists of quanta of energy?

**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. The photoelectric effect demonstrated that light interacts with matter in discrete packets (photons), explained by Einstein’s quantum theory. Applying E = h f = h c/λ, p = h/λ, K_max = h f - Φ,

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