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#quantum physics

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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 =

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De Broglie Wavelength and Matter Waves

The momentum of a photon is \( 2.21 \times 10^{-27} \, \text{kg m/s} \). What is its wavelength? (Take \( h = 6.63 \time

**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. For a photon, p = (h/λ) . λ = (h/p) = (6.63 × 10⁻³⁴/2.21 × 10⁻²⁷) ≈ 3.0 × 10⁻⁷ m = 300 nm . Applying E = h f = h c/λ, p = h/λ, K_max = h f - Φ, Φ = h f₀ =

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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.

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 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

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

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 =

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De Broglie Wavelength and Matter Waves

A particle of mass \( 2.5 \times 10^{-29} \, \text{kg} \) has a de Broglie wavelength of \( 0.5 \, \text{nm} \). 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. p = (h/λ) = (6.63 × 10⁻³⁴/0.5 × 10⁻⁹) = 1.326 × 10⁻²⁴ kg m/s . v = (p/m) = (1.326 × 10⁻²⁴/2.5 × 10⁻²⁹) = 5.304 × 10⁴ m/s . 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 × 10⁴ m/s follows, reflecting photoelectric and de Broglie relations.

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De Broglie Wavelength and Matter Waves

The de Broglie wavelength of an electron is \( 0.15 \, \text{nm} \). What is its speed? (Take \( h = 6.63 \times 10^{-34

**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. p = (h/λ) = (6.63 × 10⁻³⁴/0.15 × 10⁻⁹) = 4.42 × 10⁻²⁴ kg m/s . v = (p/m) = (4.42 × 10⁻²⁴/9.11 × 10⁻³¹) ≈ 4.852 × 10⁶ m/s . 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

A photon of frequency \( 7.0 \times 10^{14} \, \text{Hz} \) has what momentum? (Take \( h = 6.63 \times 10^{-34} \, \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. E = h v = 6.63 × 10⁻³⁴ × 7.0 × 10¹⁴ = 4.641 × 10⁻¹⁹ J . p = (E/c) = (4.641 × 10⁻¹⁹/3 × 10⁸) = 1.547 × 10⁻²⁷ kg 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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