#photoelectric effect

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

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

View Details

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

View Details

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 =

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

View Details

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

View Details

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

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

View Details

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.

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

View Details

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

View Details

De Broglie Wavelength and Matter Waves

Light of wavelength \( 450 \, \text{nm} \) is incident on a metal with work function \( 1.9 \, \text{eV} \). What is the

**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) = (1240/450) ≈ 2.756 eV . Kₘₐₓ = E - Φ₀ = 2.756 - 1.9 ≈ 0.856 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 0.856 eV follows, reflecting photoelectric and de Broglie relations.

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

View Details

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

View Details

Photon Energy, Momentum, Power and X-rays

The threshold frequency of a metal is \( 4.8 \times 10^{14} \, \text{Hz} \). What is the maximum kinetic energy for ligh

**Photon energy** E = h f = h c/λ, momentum p = h/λ = E/c, power P = N E/t, N number of photons per second, wavelength λ = h c/E, for photon momentum 1.5×10⁻²⁷ kg·m/s frequency f = p c/h? Actually E = p c, f = E/h = p c/h =1.5×10⁻²⁷×3×10⁸/6.63×10⁻³⁴=6.79×10¹⁴ Hz, energy E= h f =4.5×10⁻¹⁹ J, for 5.0×10⁻¹⁹ J frequency f=E/h=7.54×10¹⁴ Hz. Φ₀ = h v₀ = 6.63 × 10⁻³⁴ × 4.8 × 10¹⁴ = 3.1824 × 10⁻¹⁹ J . E = h v = 6.63 × 10⁻³⁴ × 6.8 × 10¹⁴ = 4.5084 × 10⁻¹⁹ J . Kₘₐₓ = E

Ref: NCERT > Physics Book > Dual Nature of Matter > Photon Energy, Momentum, Power and X-rays

View Details

Photon Energy, Momentum, Power and X-rays

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

View Details

De Broglie Wavelength and Matter Waves

The maximum speed of photoelectrons emitted from a surface is \( 5.0 \times 10^5 \, \text{m/s} \). What is the stopping

**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. Kₘₐₓ = (1/2) m vₘₐₓ² = (1/2) × 9.11 × 10⁻³¹ × (5.0 × 10⁵)² = 1.13875 × 10⁻¹⁹ J . V₀ = (Kₘₐₓ/e) = (1.13875 × 10⁻¹⁹/1.6 × 10⁻¹⁹) ≈ 0.71 V . Applying E = h f = h c/λ, p = h/λ, K_max = h f

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

View Details