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#photon flux

2 public questions tagged with this topic.

De Broglie Wavelength and Matter Waves

A light beam emits \( 2.5 \times 10^{15} \) photons per second, each of energy \( 4.0 \times 10^{-19} \, \text{J} \). Wh

**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. Power P = N × E . P = 2.5 × 10¹⁵ × 4.0 × 10⁻¹⁹ = 1.0 × 10⁻³ W = 1.0 mW . 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 > De Broglie Wavelength and Matter Waves

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Photon Energy, Momentum, Power and X-rays

A light source emits \( 6.0 \times 10^{15} \) photons per second with a power of \( 3.0 \, \text{mW} \). What is the ene

**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. E = (P/N) = (3.0 × 10⁻³/6.0 × 10¹⁵) = 5.0 × 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 V/h, result 5.0 × 10⁻¹⁹ J follows, reflecting photoelectric and de Broglie relations.

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

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