#threshold frequency

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

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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

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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

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Photon Energy, Momentum, Power and X-rays

The work function of a metal is \( 2.0 \, \text{eV} \). What is the threshold frequency for photoelectric emission from

**Photon power** light source emits 1.0×10¹⁶ photons/s power 4.0 mW, energy per photon E= P/N =4×10⁻³/10¹⁶=4×10⁻¹⁹ J, wavelength λ= h c/E=6.63×10⁻³⁴×3×10⁸/4×10⁻¹⁹=497 nm, for 6×10¹⁵ photons/s 3.0 mW E=5×10⁻¹⁹ J, for beam 2.5×10¹⁵ photons/s 4×10⁻¹⁹ J each power=10⁻³ W=1 mW, illustrating P = N h f. Work function Φ₀ = 2.0 eV = 2.0 × 1.6 × 10⁻¹⁹ = 3.2 × 10⁻¹⁹ J . Threshold frequency v₀ = (Φ₀/h) = (3.2 × 10⁻¹⁹/6.63 × 10⁻³⁴) ≈ 4.83 × 10¹⁴ Hz . 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 > Photon Energy, Momentum, Power and X-rays

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Photon Energy, Momentum, Power and X-rays

The threshold frequency of a metal is \( 6.0 \times 10^{14} \, \text{Hz} \). What is its work function in eV? (Take \( h

**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 × 6.0 × 10¹⁴ = 3.978 × 10⁻¹⁹ J . Φ₀ = (3.978 × 10⁻¹⁹/1.6 × 10⁻¹⁹) ≈ 2.486 eV . Applying E = h f = h c/λ, p =

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

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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