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53 public questions tagged with this topic.

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

Ref: NCERT > Physics Book > Dual Nature of Matter > Photon Energy, Momentum, Power and X-rays

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Mass Defect, Binding Energy and Binding Energy per Nucleon

What is the primary reason nuclear reactions release more energy than chemical reactions?

**Binding energy calculation** from mass defect, for nucleus mass number 28 BE 224 MeV BE/A=8 MeV, A=18 BE 144 MeV BE/A=8 MeV, BE/A indicates stability, fusion of light nuclei and fission of heavy release energy because product has higher BE/A, difference released. Nuclear reactions involve changes in nuclear binding energy, which operates on a scale of MeV, vastly exceeding the eV scale of chemical bond energy, due to the strong nuclear force. 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 Higher nuclear density, consistent

Ref: NCERT > Physics Book > Atoms and Nuclei > Mass Defect, Binding Energy and Binding Energy per Nucleon

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Energy Stored in Capacitor and Energy Density

A \( 1 \, \mu\text{F} \) capacitor is charged to \( 400 \, \text{V} \). What is the energy stored in it?

**Energy stored in capacitor** U = ½ C V² = ½ Q V = Q²/(2C) (J), C capacitance (F), V voltage (V), Q charge (C). For 4 μF charged to 250 V, U=0.5×4×10⁻⁶×62500=0.125 J. Energy resides in electric field, energy density u = ½ ε₀ E² (J/m³), E field between plates. U = (1/2) C V² = (1/2) × 1 × 10⁻⁶ × (400)² . U = (1/2) × 1 × 10⁻⁶ × 16 × 10⁴ = 0.08 J . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC parallel, U = ½ C V²

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Energy Stored in Capacitor and Energy Density

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Energy Stored in Capacitor and Energy Density

A \( 3 \, \mu\text{F} \) capacitor is charged to \( 300 \, \text{V} \). What is the energy stored in it?

**Work done charging capacitor** is integral ∫ V dQ = ∫ Q/C dQ = Q²/2C, stored as electrostatic energy. When capacitor discharges, energy released as heat or work, explaining spark when shorted, energy proportional to V². U = (1/2) C V² = (1/2) × 3 × 10⁻⁶ × (300)² = (1/2) × 3 × 10⁻⁶ × 9 × 10⁴ = 0.135 J . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC parallel, U = ½ C V² and common potential V = Q_total/C_total, result 0.135 J follows, reflecting potential-capacitance relations.

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Energy Stored in Capacitor and Energy Density

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Combination of Capacitors - Series and Parallel

Why does the total energy stored in a system of capacitors connected in series decrease if one of the capacitors is remo

**Charge conservation in series** explains same Q: when battery charges first plate, it induces -Q on second plate of same capacitor, which comes from next capacitor's plate, etc., so all have same magnitude Q, potential divides as per 1/C. In series, the equivalent capacitance Cₑq decreases when a capacitor is removed ( (1/Cₑq) = sum (1/C_i) ). With a battery maintaining constant voltage V , the energy stored is U = (1/2) Cₑq V² . A smaller Cₑq reduces U , as energy is directly proportional to capacitance under constant V . Additionally, removing a capacitor redistributes charges, potentially

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Combination of Capacitors - Series and Parallel

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Forced Oscillations and Resonance

A spring of \( k = 450 \, \text{N/m} \) has a \( 1.5 \, \text{kg} \) mass. If \( E = 2.25 \, \text{J} \), what is the am

**Resonance phenomenon** amplifies response when driving frequency matches natural frequency ω_d ≈ ω₀, large amplitude even with small F₀, as damping limits growth. Natural frequency determined by system parameters, resonance condition crucial for understanding vibrations and energy absorption, e.g., bridge collapse, tuning. Total energy: E = (1/2) k A² . 2.25 = 0.5 × 450 × A² ⇒ 2.25 = 225 A² ⇒ A² = 0.01 ⇒ A = 0.1 m . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.1 m follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Forced Oscillations and Resonance

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Damped Oscillations

A mass of \( 0.5 \, \text{kg} \) on a spring has \( E = 2 \, \text{J} \) at \( A = 20 \, \text{cm} \). What is the sprin

**Damped oscillations** occur when resistive forces dissipate energy, amplitude decays exponentially as A(t) = A₀ e^(-b t/2m), b damping coefficient (kg/s), frequency slightly reduced ω' = √(ω₀² - (b/2m)²). Damping arises from friction or viscosity, energy loss per cycle proportional to velocity squared, motion eventually stops. Total energy: E = (1/2) k A² . 2 = (1/2) k (0.2)² ⇒ 2 = 0.02 k ⇒ k = (2/0.02) = 100 N/m . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 100 N/m follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Damped Oscillations

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Damped Oscillations

A spring of \( k = 200 \, \text{N/m} \) has a \( 0.5 \, \text{kg} \) mass. If \( E = 1 \, \text{J} \), what is the ampli

**Real oscillators** experience damping, amplitude decreasing with time. Critical damping returns to equilibrium fastest without oscillation, overdamping slows return, underdamping shows decaying oscillations, classification based on b relative to 2mω₀, important for practical systems. Total energy: E = (1/2) k A² . 1 = (1/2) × 200 × A² ⇒ 1 = 100 A² ⇒ A² = 0.01 ⇒ A = 0.1 m . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.1 m follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Damped Oscillations

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Damped Oscillations

A mass of \( 4 \, \text{kg} \) is attached to a spring with \( k = 1600 \, \text{N/m} \) and displaced by \( 5 \, \text{

**Damped oscillations** occur when resistive forces dissipate energy, amplitude decays exponentially as A(t) = A₀ e^(-b t/2m), b damping coefficient (kg/s), frequency slightly reduced ω' = √(ω₀² - (b/2m)²). Damping arises from friction or viscosity, energy loss per cycle proportional to velocity squared, motion eventually stops. Total energy: E = (1/2) k A² . A = 0.05 m, k = 1600 N/m . E = 0.5 × 1600 × (0.05)² = 0.5 × 1600 × 0.0025 = 2 J . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 2.0 J

Ref: NCERT > Physics Book > Oscillations > Damped Oscillations

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Spring-Mass System and Combination of Springs

A spring-mass system has \( m = 0.6 \, \text{kg}, k = 240 \, \text{N/m} \). If displaced by \( 7 \, \text{cm} \), what i

**Spring-mass system** has period T = 2π√(m/k), frequency f = (1/2π)√(k/m), ω = √(k/m), where k spring constant (N/m) and m mass (kg). For parallel combination, k_eff = k₁ + k₂, series gives 1/k_eff = 1/k₁ + 1/k₂, affecting ω = √(k_eff/m) and T = 2π√(m/k_eff). Total energy: E = (1/2) k A² . A = 0.07 m, k = 240 N/m . E = 0.5 × 240 × (0.07)² = 0.5 × 240 × 0.0049 = 0.588 J . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.588 J

Ref: NCERT > Physics Book > Oscillations > Spring-Mass System and Combination of Springs

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Spring-Mass System and Combination of Springs

A spring-mass system has \( m = 0.25 \, \text{kg}, k = 100 \, \text{N/m} \). If displaced by \( 6 \, \text{cm} \), what

**Mass-spring dynamics** show T depends on mass and stiffness, independent of amplitude for ideal spring. Given T and m, k = 4π² m/T² extracted, and energy E = ½ k A² connects amplitude to total mechanical energy, illustrating isochronism. Total energy: E = (1/2) k A² . A = 0.06 m, k = 100 N/m . E = 0.5 × 100 × (0.06)² = 0.5 × 100 × 0.0036 = 0.18 J . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.18 J follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Spring-Mass System and Combination of Springs

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