Photoelectric effect is possible
with both visible and uv depending on the metal is correct per Photoelectric effect is possible principle; validated by standard physics law.
Ref: NCERT Physics Class 12 Chapter 11 Dual Nature of Radiation and Matter
#light frequency
16 public questions tagged with this topic.
Filter questions
Two identical photo-electrode 1 and photo-electrode 2 receive the light of frequency f 1 and f 2 respectively of the sam
Same intensity different frequency, higher frequency gives higher KE but same photon number? Current same.
Ref: NCERT Physics Class 12 Chapter 3 Current Electricity
The energy of the photon of the monochromatic light is 1.326 × 10 -19 J. Find the frequency of the light.
Frequency f=E/h=1.326e-19/6.6e-34≈2×10^14 Hz.
Ref: NCERT Physics Class 12 Chapter 13 Nuclei
What is the frequency of light with a wavelength of \( 680 \, \text{nm} \) in air, given the speed of light in air is \(
**Wave model predicts** light bends away from normal when entering rarer medium because speed increases, Snell's law n₁ sinθ₁ = n₂ sinθ₂, n₁>n₂ so sinθ₂>sinθ₁ θ₂>θ₁ away from normal, towards normal when denser, wavefront slows in denser, Huygens construction shows bending. Frequency nu = (c/λ) . λ = 6.8 × 10⁻⁷ m , c = 3.0 × 10⁸ m/s . nu = (3.0 × 10⁸/6.8 × 10⁻⁷) ≈ 4.41 × 10¹⁴ Hz . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' = λ/n and A = 2a
Ref: NCERT > Physics Book > Wave Optics > Wave Properties, Frequency and Energy Conservation
What is the frequency of light with a wavelength of \( 595 \, \text{nm} \) in air, given the speed of light in air is \(
**Intensity not depend on speed** when enters denser medium because intensity I ∝ n E₀²? Actually Poynting vector S = E×H, energy density u =½ ε E², for same amplitude E₀ intensity proportional to n, but amplitude changes at interface due to reflection, total energy conserved incident = reflected + transmitted, interference does not destroy energy, it redistributes. Frequency nu = (c/λ) . λ = 5.95 × 10⁻⁷ m , c = 3.0 × 10⁸ m/s . nu = (3.0 × 10⁸/5.95 × 10⁻⁷) ≈ 5.04 × 10¹⁴ Hz . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ,
Ref: NCERT > Physics Book > Wave Optics > Wave Properties, Frequency and Energy Conservation
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
Ref: NCERT > Physics Book > Dual Nature of Matter > 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
Ref: NCERT > Physics Book > Dual Nature of Matter > De Broglie Wavelength and Matter Waves
What happens to photoelectric emission if the frequency of incident light is below the threshold frequency?
**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. No photoelectric emission occurs if the frequency is below the threshold frequency, as the photon energy ( h v ) is less than the work function ( Φ₀ ). Applying E = h f = h c/λ, p = h/λ, K_max = h f - Φ, Φ = h f₀ = h c/λ₀,
Ref: NCERT > Physics Book > Dual Nature of Matter > Photon Energy, Momentum, Power and X-rays
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
Ref: NCERT > Physics Book > Dual Nature of Matter > 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 =
Ref: NCERT > Physics Book > Dual Nature of Matter > 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)
Ref: NCERT > Physics Book > Dual Nature of Matter > Photoelectric Effect - Work Function and Threshold
Light of frequency \( 6.5 \times 10^{14} \, \text{Hz} \) is incident on a metal with work function \( 2.3 \, \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 × 6.5 × 10¹⁴ = 4.3095 × 10⁻¹⁹ J . E = (4.3095 × 10⁻¹⁹/1.6 × 10⁻¹⁹) ≈ 2.693 eV . Kₘₐₓ = E - Φ₀ =
Ref: NCERT > Physics Book > Dual Nature of Matter > Photoelectric Effect - Work Function and Threshold
