#electron

De Broglie Wavelength and Matter Waves

An electron moves with a speed of \( 3.0 \times 10^6 \, \text{m/s} \). What is its de Broglie wavelength? (Take \( h = 6

**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. Momentum p = m v = 9.11 × 10⁻³¹ × 3.0 × 10⁶ = 2.733 × 10⁻²⁴ kg m/s . λ = (h/p) = (6.63 × 10⁻³⁴/2.733 × 10⁻²⁴) ≈ 2.425 × 10⁻¹⁰ m = 0.2425 nm . Applying E = h f = h c/λ, p

Ref: NCERT > Physics Book > Dual Nature of Matter > De Broglie Wavelength and Matter Waves

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Photoelectric Effect - Work Function and Threshold

Which of the following particles was historically identified as a universal constituent of matter through cathode ray ex

**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. J.J. Thomson’s experiments with cathode rays led to the discovery of electrons as fundamental negatively charged particles in matter. 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, resu

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

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De Broglie Wavelength and Matter Waves

An electron is accelerated through \( 50 \, \text{V} \). What is its de Broglie wavelength? (Take \( h = 6.63 \times 10^

**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. K = e V = 1.6 × 10⁻¹⁹ × 50 = 8.0 × 10⁻¹⁸ J . p = √(2 m K) = √(2 × 9.11 × 10⁻³¹ × 8.0 × 10⁻¹⁸) ≈ 3.816 × 10⁻²⁴ kg m/s . λ = (h/p) = (6.63 × 10⁻³⁴/3.816 × 10⁻²⁴)

Ref: NCERT > Physics Book > Dual Nature of Matter > De Broglie Wavelength and Matter Waves

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De Broglie Wavelength and Matter Waves

An electron is accelerated through a potential difference of \( 100 \, \text{V} \). What is its de Broglie wavelength? (

**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. Kinetic energy K = e V = 1.6 × 10⁻¹⁹ × 100 = 1.6 × 10⁻¹⁷ J . Momentum p = √(2 m K) = √(2 × 9.11 × 10⁻³¹ × 1.6 × 10⁻¹⁷) ≈ 5.39 × 10⁻²⁴ kg m/s . λ = (h/p) = (6.63 × 10⁻³⁴/5.39 ×

Ref: NCERT > Physics Book > Dual Nature of Matter > De Broglie Wavelength and Matter Waves

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Hydrogen Atom Properties - Radius, Speed and Energy

An electron in a hydrogen atom has a total energy of -3.4 eV. What is its kinetic energy?

**Energy in hydrogen** total E_n = -13.6/n² eV, kinetic K = +13.6/n² eV, potential U = -27.2/n² eV, ratio K/U = -1/2, total negative indicates bound, zero at ionization, negative total means electron bound, requires energy to free. For n=4 E=-0.85 eV, U=-1.7 eV, K=0.85 eV, potential energy twice total negative. E = -K , K = -E = -(-3.4) = 3.4 eV . 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 3.4 eV, consistent with Bohr model and nuclear binding energy systematics.

Ref: NCERT > Physics Book > Atoms and Nuclei > Hydrogen Atom Properties - Radius, Speed and Energy

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Electric Current, Drift Velocity and Mobility

A copper wire of cross-sectional area \( 8 \times 10^{-7} \, \text{m}^2 \) carries a current of \( 2 \, \text{A} \). If

**Drift velocity** v_d = I/(n e A), I current (A), n number density of conduction electrons (m⁻³) ≈8.5×10²⁸ m⁻³ for copper, e =1.6×10⁻¹⁹ C, A cross-sectional area (m²). Typical v_d ≈10⁻⁴ m/s for 1 A in mm² wire, slow despite fast signal propagation due to electric field establishment. Drift speed: v_d = (I/n e A) . Substitute: v_d = (2/8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 8 × 10⁻⁷) . Calculate: v_d = (2/1.088 × 10⁴) ≈ 1.84 × 10⁻⁴ m/s . 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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Torque on Current Loop, Magnetic Moment and Galvanometer

An electron moves at \( 7 \times 10^6 \, \text{m/s} \) perpendicular to a field of \( 0.2 \, \text{T} \). What is the ma

**SI unit of magnetic field** is tesla (T), defined as force 1 N on 1 A·m wire perpendicular to field. Moving coil galvanometer uses torque τ = N I A B balanced by spring torque k φ, so deflection φ ∝ I, enabling current measurement, with radial field ensuring τ = N I A B always maximum. Force F = q v B sin θ , θ = 90° , so sin θ = 1 . F = 1.6 × 10⁻¹⁹ × 7 × 10⁶ × 0.2 = 2.24 × 10⁻¹³ N . Using F = q v B sinθ, F = I l B

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Torque on Current Loop, Magnetic Moment and Galvanometer

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Torque on Current Loop, Magnetic Moment and Galvanometer

An electron moves at \( 6.5 \times 10^6 \, \text{m/s} \) perpendicular to a field of \( 0.2 \, \text{T} \). What is the

**Torque on current loop** in magnetic field B is τ = N I A × B, magnitude τ = N I A B sinθ, N turns, I current (A), A area (m²) = l×b for rectangular, θ angle between normal to plane and B. Maximum when plane parallel to B (θ=90°), zero when perpendicular (θ=0°), magnetic moment m = N I A direction along normal via right-hand rule. r = (mv/qB) . r = (9.1 × 10⁻³¹ × 6.5 × 10⁶/1.6 × 10⁻¹⁹ × 0.2) = (5.915 × 10⁻²⁴/3.2 × 10⁻²⁰) = 1.848 × 10⁻⁴ m ≈ 0.0185 cm . Using F = q v

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Torque on Current Loop, Magnetic Moment and Galvanometer

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Torque on Current Loop, Magnetic Moment and Galvanometer

An electron moves at \( 2.5 \times 10^6 \, \text{m/s} \) perpendicular to a field of \( 0.5 \, \text{T} \). What is the

**SI unit of magnetic field** is tesla (T), defined as force 1 N on 1 A·m wire perpendicular to field. Moving coil galvanometer uses torque τ = N I A B balanced by spring torque k φ, so deflection φ ∝ I, enabling current measurement, with radial field ensuring τ = N I A B always maximum. r = (mv/qB) . r = (9.1 × 10⁻³¹ × 2.5 × 10⁶/1.6 × 10⁻¹⁹ × 0.5) = (2.275 × 10⁻²⁴/8 × 10⁻²⁰) = 2.84375 × 10⁻⁵ m ≈ 2.84 × 10⁻³ cm . Using F = q v B sinθ, F = I l B sinθ, B

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Torque on Current Loop, Magnetic Moment and Galvanometer

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Solenoid, Toroid and Ampere's Law

An electron moves at \( 5 \times 10^6 \, \text{m/s} \) perpendicular to a magnetic field of \( 0.2 \, \text{T} \). What

**Solenoid field** inside long solenoid is B = μ₀ n I, n = N/L turns per meter (m⁻¹), uniform and parallel to axis, outside negligible for long solenoid because fields from opposite sides cancel. For n = 1200 m⁻¹, I = 1 A, B = 4π×10⁻⁷×1200 = 1.51×10⁻³ T = 1.51 mT. Force F = q v B sin θ , θ = 90° , so sin θ = 1 . F = 1.6 × 10⁻¹⁹ × 5 × 10⁶ × 0.2 = 1.6 × 10⁻¹³ N . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Solenoid, Toroid and Ampere's Law

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Magnetic Force on Moving Charge - Lorentz Force and Motion

An electron moves at \( 6 \times 10^6 \, \text{m/s} \) perpendicular to a field of \( 0.4 \, \text{T} \). What is the ma

**Effect of doubling velocity** on magnetic force F = q v B sinθ is linear increase, F doubles for same θ and B. Electron with charge 1.6×10⁻¹⁹ C, v = 4.5×10⁶ m/s, B = 0.35 T, θ = 90°, F = 1.6×10⁻¹⁹×4.5×10⁶×0.35 = 2.52×10⁻¹³ N, illustrating magnitude for typical lab values. Force F = q v B sin θ , θ = 90° , so sin θ = 1 . F = 1.6 × 10⁻¹⁹ × 6 × 10⁶ × 0.4 = 3.84 × 10⁻¹³ N . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r),

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Magnetic Force on Moving Charge - Lorentz Force and Motion

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Magnetic Force on Moving Charge - Lorentz Force and Motion

An electron moves at \( 4.5 \times 10^6 \, \text{m/s} \) perpendicular to a field of \( 0.35 \, \text{T} \). What is the

**Lorentz force** on charge q moving with velocity v in magnetic field B is F = q v × B, magnitude F = q v B sinθ, θ angle between v and B (degrees), unit N. Direction perpendicular to both v and B via right-hand rule. When v ⊥ B, motion circular with radius r = m v/(q B), centripetal force provided by magnetic force. Force F = q v B sin θ , θ = 90° , so sin θ = 1 . F = 1.6 × 10⁻¹⁹ × 4.5 × 10⁶ × 0.35 = 2.52 × 10⁻¹³ N . Using F = q

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Magnetic Force on Moving Charge - Lorentz Force and Motion

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