Photon Energy, Momentum, Power and X-rays

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 =

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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

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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,

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Photon Energy, Momentum, Power and X-rays

A photon of frequency \( 8.0 \times 10^{14} \, \text{Hz} \) has what energy in joules? (Take \( h = 6.63 \times 10^{-34}

**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 = h v = 6.63 × 10⁻³⁴ × 8.0 × 10¹⁴ = 5.304 × 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 5.304 × 10⁻¹⁹ J follows, reflecting photoelectric and de Broglie relations.

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Photon Energy, Momentum, Power and X-rays

A photon has a wavelength of \( 250 \, \text{nm} \). What is its energy in joules? (Take \( h = 6.63 \times 10^{-34} \,

**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 = (h c/λ) = (6.63 × 10⁻³⁴ × 3 × 10⁸/250 × 10⁻⁹) = 7.974 × 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 7.974 × 10⁻¹⁹ J

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Photon Energy, Momentum, Power and X-rays

A photon has a wavelength of \( 200 \, \text{nm} \). What is its momentum? (Take \( h = 6.63 \times 10^{-34} \, \text{J

**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. p = (h/λ) = (6.63 × 10⁻³⁴/200 × 10⁻⁹) = 3.315 × 10⁻²⁷ kg m/s . Applying E = h f = h c/λ, p = h/λ, K_max = h f - Φ, Φ = h f₀ = h

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Photon Energy, Momentum, Power and X-rays

A particle of mass \( 1.5 \times 10^{-30} \, \text{kg} \) has the same momentum as a photon of wavelength \( 500 \, \tex

**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. Photon momentum p = (h/λ) = (6.63 × 10⁻³⁴/500 × 10⁻⁹) = 1.326 × 10⁻²⁷ kg m/s . v = (p/m) = (1.326 × 10⁻²⁷/1.5 × 10⁻³⁰) = 8.84 × 10² = 884 m/s . Applying E = h f = h c/λ, p = h/λ, K_max = h f - Φ, Φ = h f₀ = h c/λ₀, λ

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Photon Energy, Momentum, Power and X-rays

Light of wavelength \( 400 \, \text{nm} \) produces photoelectrons with a stopping potential of \( 0.8 \, \text{V} \) fr

**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. Photon energy E = (h c/λ) = (1240/400) = 3.1 eV . Kₘₐₓ = e V₀ = 0.8 eV . Φ₀ = E - Kₘₐₓ = 3.1 - 0.8 = 2.3 eV . Applying E = h f = h c/λ, p = h/λ, K_max = h f - Φ, Φ = h

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Photon Energy, Momentum, Power and X-rays

A photon has a momentum of \( 2.0 \times 10^{-27} \, \text{kg m/s} \). What is its energy in joules? (Take \( c = 3 \tim

**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. For a photon, p = (E/c) . E = p c = 2.0 × 10⁻²⁷ × 3 × 10⁸ = 6.0 × 10⁻¹⁹ J . Applying E = h f = h c/λ, p = h/λ, K_max = h

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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

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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

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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 =

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