#photon energy

Superposition, Resultant Amplitude and Intensity

What is the intensity of light in the photon picture, according to the wave optics concept?

**Intensity at point** in double-slit I = I_max cos²(φ/2), φ = (2π/λ)Δ, for Δ=λ/4 φ=π/2 I= I_max cos²(π/4)= I_max/2 =2I₀, for Δ=λ/3 φ=2π/3 I= I_max cos²(π/3)= I_max×0.25= I₀, for Δ=5λ/8 φ=5π/4? Actually φ=2π×5/8=5π/4, cos²(5π/8)=?, path difference for destructive φ=(2n+1)π, constructive 2nπ. In the photon picture, intensity is determined by the number of photons crossing a unit area per unit time. 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 cos(φ/2), calculation gives Number of photons per unit area per unit time, illustrating interferenc

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Wave Properties, Frequency and Energy Conservation

What is the effect on the energy of a light wave when its speed decreases in a denser medium?

**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. Energy of a light wave depends on its amplitude, not its speed, so it remains unchanged when speed decreases. 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

Ref: NCERT > Physics Book > Wave Optics > Wave Properties, Frequency and Energy Conservation

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Photon Energy, Momentum, Power and X-rays

The maximum frequency of X-rays produced by a \( 20 \, \text{kV} \) tube is: (Take \( h = 6.63 \times 10^{-34} \, \text{

**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. Energy E = e V = 1.6 × 10⁻¹⁹ × 20 × 10³ = 3.2 × 10⁻¹⁵ J . vₘₐₓ = (E/h) = (3.2 × 10⁻¹⁵/6.63 × 10⁻³⁴) ≈ 4.83 × 10¹⁸ Hz . Applying E = h f = h c/λ, p = h/λ, K_max = h f - Φ, Φ =

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 minimum wavelength of X-rays from a \( 15 \, \text{kV} \) tube is: (Take \( h = 6.63 \times 10^{-34} \, \text{J s} \

**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. E = e V = 1.6 × 10⁻¹⁹ × 15 × 10³ = 2.4 × 10⁻¹⁵ J . λₘiₙ = (h c/E) = (6.63 × 10⁻³⁴ × 3 × 10⁸/2.4 × 10⁻¹⁵) ≈ 8.2875 × 10⁻¹¹ m = 0.082875 nm . Applying E = h f = h c/λ, p = h/λ, K_max

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

A light source emits \( 5.0 \times 10^{15} \) photons per second with a power of \( 2.0 \, \text{mW} \). What is the ene

**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. Power P = N × E . E = (P/N) = (2.0 × 10⁻³/5.0 × 10¹⁵) = 4.0 × 10⁻¹⁹ J . Applying E = h f = h c/λ, p = h/λ, K_max = h f - Φ,

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

A light source emits photons of energy \( 3.0 \times 10^{-19} \, \text{J} \) at a rate of \( 4.0 \times 10^{15} \) photo

**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. Power P = N × E . P = 4.0 × 10¹⁵ × 3.0 × 10⁻¹⁹ = 1.2 × 10⁻³ W = 1.2 mW . 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,

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

A photon of wavelength \( 300 \, \text{nm} \) is incident on a metal. What is its energy in eV? (Take \( h c = 1240 \, \

**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. Energy E = (h c/λ) . E = (1240/300) ≈ 4.13 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 4.13 eV 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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Photon Energy, Momentum, Power and X-rays

A light source emits photons of energy \( 4.5 \times 10^{-19} \, \text{J} \) at a rate of \( 2.0 \times 10^{16} \) photo

**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. λ = (h c/E) = (6.63 × 10⁻³⁴ × 3 × 10⁸/4.5 × 10⁻¹⁹) = 4.42 × 10⁻⁷ m = 442 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)

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

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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Photon Energy, Momentum, Power and X-rays

What characteristic of light quanta (photons) is directly proportional to their frequency?

**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. The energy of a photon is given by E = h v , making it directly proportional to frequency. Applying E = h f = h c/λ, p = h/λ, K_max = h f - Φ, Φ = h

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

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

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

Ref: NCERT > Physics Book > Dual Nature of Matter > Photoelectric Effect - Work Function and Threshold

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