#electron speed

De Broglie Wavelength and Matter Waves

Light of frequency \( 5.5 \times 10^{14} \, \text{Hz} \) produces photoelectrons with a maximum speed of \( 4.0 \times 1

**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. E = h v = 6.63 × 10⁻³⁴ × 5.5 × 10¹⁴ = 3.6465 × 10⁻¹⁹ J . E = (3.6465 × 10⁻¹⁹/1.6 × 10⁻¹⁹) ≈ 2.28 eV . Kₘₐₓ = (1/2) m vₘₐₓ² = (1/2) × 9.11 × 10⁻³¹ × (4.0 × 10⁵)² = 7.288 × 10⁻²⁰ J . Kₘₐₓ = (7.288 × 10⁻²⁰/1.6 × 10⁻¹⁹) ≈ 0.455 eV . Φ₀ = E - Kₘₐₓ = 2.28 - 0.455 ≈ 1.825 eV . Applying

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De Broglie Wavelength and Matter Waves

The maximum speed of photoelectrons emitted from a surface is \( 5.0 \times 10^5 \, \text{m/s} \). What is the stopping

**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. Kₘₐₓ = (1/2) m vₘₐₓ² = (1/2) × 9.11 × 10⁻³¹ × (5.0 × 10⁵)² = 1.13875 × 10⁻¹⁹ J . V₀ = (Kₘₐₓ/e) = (1.13875 × 10⁻¹⁹/1.6 × 10⁻¹⁹) ≈ 0.71 V . Applying E = h f = h c/λ, p = h/λ, K_max = h f

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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

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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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Photoelectric Effect - Work Function and Threshold

The work function of a metal is \( 2.4 \, \text{eV} \). Light of wavelength \( 350 \, \text{nm} \) is incident on it. Wh

**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. E = (h c/λ) = (1240/350) ≈ 3.543 eV . Kₘₐₓ = E - Φ₀ = 3.543 - 2.4 = 1.143 eV = 1.143 × 1.6 × 10⁻¹⁹ = 1.8288 × 10⁻¹⁹ J . vₘₐₓ = √((2 Kₘₐₓ/m)) = √((2 × 1.8288 × 10⁻¹⁹/9.11 × 10⁻³¹)) ≈ 6.33 × 10⁵ m/s . Applying E =

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De Broglie Wavelength and Matter Waves

The maximum speed of photoelectrons is \( 7.0 \times 10^5 \, \text{m/s} \). What is the maximum kinetic energy in joules

**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. Kₘₐₓ = (1/2) m vₘₐₓ² = (1/2) × 9.11 × 10⁻³¹ × (7.0 × 10⁵)² . Kₘₐₓ = (1/2) × 9.11 × 10⁻³¹ × 4.9 × 10¹¹ ≈ 2.231 × 10⁻¹⁹ J . Applying E = h f = h c/λ, p = h/λ, K_max = h

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De Broglie Wavelength and Matter Waves

The de Broglie wavelength of an electron moving at \( 2.0 \times 10^6 \, \text{m/s} \) is calculated. What is its value?

**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. Momentum p = m v = 9.11 × 10⁻³¹ × 2.0 × 10⁶ = 1.822 × 10⁻²⁴ kg m/s . λ = (h/p) = (6.63 × 10⁻³⁴/1.822 × 10⁻²⁴) ≈ 3.64 × 10⁻¹⁰ m = 0.364 nm . Applying E = h f = h c/λ, p = h/λ,

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De Broglie Hypothesis and Quantization in Bohr Model

What is the speed of an electron in the \( n = 5 \) orbit of a hydrogen atom if its speed in \( n = 1 \) is \( 2.2 \time

**Bohr's quantization** angular momentum L = m v r = n h/2π, for n=2 L=2h/2π= h/π=2.11×10⁻³⁴ J·s, for n=5 L=5h/2π, de Broglie λ = h/p, p= m v, for first orbit v=2.2×10⁶ m/s, λ= h/(m v)=6.6×10⁻³⁴/(9.1×10⁻³¹×2.2×10⁶)=3.3×10⁻¹⁰ m, circumference 2πr=3.33×10⁻¹⁰ m, one wavelength fits for n=1. v_n = (v₁/n) . For n = 5 : v₅ = (2.2 × 10⁶/5) = 4.4 × 10⁵ m/s . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.5 MeV, evaluation yields 4.4 × 10⁵ m/s, consistent with Bohr model and nuclear binding energy systematics.

Ref: NCERT > Physics Book > Atoms and Nuclei > De Broglie Hypothesis and Quantization in Bohr Model

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De Broglie Hypothesis and Quantization in Bohr Model

What is the speed of an electron in the \( n = 3 \) orbit of a hydrogen atom if its speed in \( n = 1 \) is \( 2.2 \time

**De Broglie hypothesis** λ = h/p, p=mv momentum, suggests electron as wave, in Bohr model circumference 2πr = n λ, standing wave condition, n wavelengths fit into orbit, for n=6, 6 wavelengths, for n=4, 4 wavelengths, explains quantization of angular momentum L = r p = r h/λ = r h n/(2πr)= n h/2π = n ħ, physical basis for Bohr quantization. v_n = (v₁/n) . For n = 3 : v₃ = (2.2 × 10⁶/3) ≈ 7.33 × 10⁵ m/s . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1

Ref: NCERT > Physics Book > Atoms and Nuclei > De Broglie Hypothesis and Quantization in Bohr Model

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Electric Current, Drift Velocity and Mobility

Why does the drift velocity of electrons in a conductor remain much smaller than their thermal velocity?

**Current and drift relation** I = n e A v_d shows current proportional to drift velocity and area. For A=6×10⁻⁷ m², I=1.8 A, n=8.5×10²⁸ m⁻³, v_d =1.8/(8.5×10²⁸×1.6×10⁻¹⁹×6×10⁻⁷)=2.2×10⁻⁴ m/s, illustrating small drift speed even for ampere currents. Drift velocity ( v_d = e E tau / m ) is small because E is typically weak and tau is short due to frequent collisions, whereas thermal velocity arises from random motion at high speeds (proportional to √(k T / m) ), unaffected by the field. Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq series/parallel, V = ε

Ref: NCERT > Physics Book > Current Electricity > Electric Current, Drift Velocity and Mobility

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