#electron motion

Hydrogen Atom Properties - Radius, Speed and Energy

What is the orbital period of an electron in the \( n = 3 \) orbit if \( v_1 = 2.2 \times 10^6 \, \text{m/s} \) and \( r

**Hydrogen atom radius** r_n = n² a₀, a₀=5.3×10⁻¹¹ m first Bohr radius, r₂=4a₀=2.12×10⁻¹⁰ m, ratio r₄/r₂ =16/4=4, r₃=9a₀, circumference 2πr_n =2π n² a₀, for n=3 circumference=2π×9×5.3×10⁻¹¹=3×10⁻⁹ m. Orbital period T =2πr/v, v_n = v₁/n, v₁=2.2×10⁶ m/s, T₂=2πr₂/v₂, v₂=1.1×10⁶ m/s, T₂≈1.21×10⁻¹⁵ s. v₃ = (2.2 × 10⁶/3) ≈ 7.33 × 10⁵ m/s . r₃ = 9 × 5.3 × 10⁻¹¹ = 4.77 × 10⁻¹⁰ m . T = (2π r₃/v₃) = (2 × 3.14 × 4.77 × 10⁻¹⁰/7.33 × 10⁵) ≈ 4.09 × 10⁻¹⁵ s . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm

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

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Potentiometer, Conductivity and Special Cases

A copper wire of cross-sectional area \( 5 \times 10^{-7} \, \text{m}^2 \) carries a current of \( 1 \, \text{A} \). If

**Conductivity** σ = 1/ρ = n e² τ/m (S/m), τ relaxation time, m electron mass. Resistivity deviation at high fields occurs when τ depends on E or n changes due to impact ionization, breaking Ohm's law, seen in varistors, gas discharge. Drift speed: v_d = (I/n e A) . Substitute: v_d = (1/8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 5 × 10⁻⁷) . Calculate: v_d = (1/6.8 × 10³) ≈ 1.47 × 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 - I r and P = I²R,

Ref: NCERT > Physics Book > Current Electricity > Potentiometer, Conductivity and Special Cases

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Potentiometer, Conductivity and Special Cases

What happens to the drift velocity of electrons in a conductor if the conductor’s temperature increases while the applie

**Potentiometer** measures potential difference without drawing current, using null deflection, principle V ∝ l, l balance length, accurate because no I r drop. Potential drop across resistor V = I R arises because electric field does work on charges, energy converted to heat, maintaining E = -dV/dx along wire. Drift velocity v_d = e E tau / m . As temperature increases, tau (relaxation time) decreases due to more collisions, reducing v_d if E is constant. 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 - I r and P = I²R,

Ref: NCERT > Physics Book > Current Electricity > Potentiometer, Conductivity and Special Cases

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Temperature Dependence of Resistance and Resistivity

In a conductor, if the electric field is suddenly doubled while keeping the conductor's properties unchanged, what happe

**Temperature dependence** of resistance R_t = R₀[1+α(T-T₀)], α temperature coefficient (per °C), R₀ resistance at T₀ (Ω). For metals α positive ≈10⁻³ /°C, resistance increases with temperature because τ decreases due to increased phonon scattering, n nearly constant. Drift velocity ( v_d ) is given by v_d = (e E tau/m) , where E is the electric field, e is the electron charge, tau is the relaxation time, and m is the electron mass. If E is doubled, v_d becomes 2v_d , assuming tau and other properties remain constant. Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0,

Ref: NCERT > Physics Book > Current Electricity > Temperature Dependence of Resistance and Resistivity

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Electric Current, Drift Velocity and Mobility

Why does the drift velocity of electrons in a conductor remain constant despite continuous acceleration by an electric f

**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. Electrons accelerate due to the electric field but collide with lattice ions, losing momentum. These collisions occur at random intervals, and the average time between collisions ( tau ) stabilizes the drift velocity ( v_d = e E tau / m ), balancing acceleration with energy loss. Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0,

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 with a speed of \( 2 \times 10^6 \, \text{m/s} \) perpendicular to a magnetic field of \( 0.7 \, \text

**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. Radius r = (mv/qB) . r = (9.1 × 10⁻³¹ × 2 × 10⁶/1.6 × 10⁻¹⁹ × 0.7) = (1.82 × 10⁻²⁴/1.12 × 10⁻¹⁹) = 1.625 × 10⁻⁵ m = 1.625 × 10⁻³ cm . Using F = q v B sinθ, F = I l B sinθ,

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

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Magnetic Force on Moving Charge - Lorentz Force and Motion

An electron moves with a speed of \( 3.5 \times 10^6 \, \text{m/s} \) perpendicular to a magnetic field of \( 0.25 \, \t

**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. Radius r = (mv/qB) . r = (9.1 × 10⁻³¹ × 3.5 × 10⁶/1.6 × 10⁻¹⁹ × 0.25) = (3.185 × 10⁻²⁴/4 × 10⁻²⁰) = 7.9625 × 10⁻⁵ m = 7.96 × 10⁻³ cm . Using F = q v

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

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PHYSICS

An electron moves with a speed of 2 × 10⁶ m/s perpendicular to a magnetic field of 0.7 T . What is the radius of its

Given: An electron moves with a speed of 2 × 10⁶ m/s perpendicular to a magnetic field of 0.7 T . What is the radius of its path? (Mass = 9.1 × 10⁻³¹ kg, charge = 1.6 × 10⁻¹⁹ C ) These values define the system as per NCERT data. Formula: Radius r = mv/qB. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: r = frac9.1 × 10⁻³¹ × 2 × 10⁶¹.6 × 10⁻¹⁹ × 0.7 = frac1.82 × 10⁻²⁴¹.12 × 10⁻¹⁹= 1.625 × 10⁻⁵ m = 1.625 × 10⁻³ cm . Result: The computed value matches the expected outcome and confirms the corr

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Moving Charges and Magnetism and Magnetism and Matter, Topic: Magnetic field due to current loop, solenoid and magnetic dipole moment. Page number should be added only after verification from the.

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