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#physics concept

174 public questions tagged with this topic.

Wave Properties, Frequency and Energy Conservation

What causes the intensity of light to be zero at certain points in a diffraction pattern?

**Wave model predicts** light bends away from normal when entering rarer medium because speed increases, Snell's law n₁ sinθ₁ = n₂ sinθ₂, n₁>n₂ so sinθ₂>sinθ₁ θ₂>θ₁ away from normal, towards normal when denser, wavefront slows in denser, Huygens construction shows bending. Complete destructive interference occurs when secondary wavelets from different parts of the slit cancel each other out, resulting in zero intensity at minima. Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' = λ/n and A = 2a cos(φ/2), calculation gives Destructive interference, illustrating

Ref: NCERT > Physics Book > Wave Optics > Wave Properties, Frequency and Energy Conservation

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De Broglie Wavelength and Matter Waves

Which of the following properties of photoelectrons is independent of the intensity of incident light?

**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. The maximum kinetic energy depends on frequency ( Kₘₐₓ = h v - Φ₀ ), not intensity, which only affects the number of electrons. 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 Maximum kinetic energy follows, reflecting photoelectric and de Broglie relations.

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

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Photon Energy, Momentum, Power and X-rays

What happens to photoelectric emission if the frequency of incident light is below the threshold frequency?

**X-rays from tube** maximum frequency f_max = e V/h, minimum wavelength λ_min = h c/(e V), Duane-Hunt law, for 10 kV f_max=1.6×10⁻¹⁹×10⁴/6.63×10⁻³⁴=2.41×10¹⁸ Hz, λ_min= c/f_max=0.124 nm, for 35 kV λ_min=1240/(35000) eV·nm? Actually 35 keV photon λ=1240/35000=0.0354 nm, for 15 kV 0.0827 nm, for 25 kV 0.0496 nm, for 30 kV 0.0413 nm. No photoelectric emission occurs if the frequency is below the threshold frequency, as the photon energy ( h v ) is less than the work function ( Φ₀ ). Applying E = h f = h c/λ, p = h/λ, K_max = h f - Φ, Φ = h f₀ = h c/λ₀,

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

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Photoelectric Effect - Work Function and Threshold

In the photoelectric effect, what does the saturation current depend on, assuming a fixed frequency above the threshold?

**Classical wave theory fails** to explain instant emission, threshold existence, K_max dependence on frequency not intensity, saturation current dependence on intensity. Observations: K_max independent of intensity, exists threshold frequency, emission instantaneous, all explained by photon model E = h f, one photon ejects one electron, energy conservation h f = Φ + K_max. Saturation current depends on the intensity of light, as it determines the number of photons and thus the number of photoelectrons emitted per second. 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 > Photoelectric Effect - Work Function and Threshold

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Nuclear Reactions - Fission, Fusion and Energy Release

What is the primary outcome of nuclear fission in terms of energy?

**Energy release in nuclear processes** always because final BE/A higher than initial, mass defect difference appears as kinetic energy of fragments and radiation, 1 u =931.5 MeV, high temperature in fusion provides kinetic energy to overcome Coulomb barrier, confinement needed, Sun's core temperature ~1.5×10⁷ K enables fusion. Nuclear fission converts mass defect into energy, initially as kinetic energy of fragments and neutrons, which is eventually transferred as heat, due to the increase in binding energy per nucleon. Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c

Ref: NCERT > Physics Book > Atoms and Nuclei > Nuclear Reactions - Fission, Fusion and Energy Release

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Hydrogen Atom Properties - Radius, Speed and Energy

Which of the following is true about the nuclear force at very short distances (less than 0.8 fm)?

**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. The nuclear force becomes strongly repulsive at distances less than 0.8 fm, preventing nucleons from collapsing into each other, as shown in the potential energy plot. 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 It is strongly repul

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

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Mass Defect, Binding Energy and Binding Energy per Nucleon

Why is the mass of a nucleus less than the sum of its constituent nucleons?

**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. The mass defect occurs because some mass is converted into binding energy, which holds the nucleus together, as per Einstein’s E = m c² . 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 Mass converted to binding energy, consistent with Bohr model and nuclear binding e

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

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Bohr Model Energy Levels and Hydrogen Spectrum

What does the line spectrum of hydrogen indicate about its atomic structure?

**Hydrogen transitions** example n=3→n=1 ΔE=13.6(1-1/9)=12.09 eV, photon 12.09 eV, λ=1240/12.09≈102.6 nm Lyman series, n=3→n=2 ΔE=1.89 eV Balmer visible Hα 656 nm. Absorption photon energy must match difference, if atom in n=2 absorbs 1.89 eV jumps to n=3, if absorbs 12.75 eV from ground 1→4 because -13.6+12.75=-0.85 eV = -13.6/16. The line spectrum suggests that electrons occupy discrete energy levels, emitting photons of specific wavelengths when transitioning between them. 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 Discr

Ref: NCERT > Physics Book > Atoms and Nuclei > Bohr Model Energy Levels and Hydrogen Spectrum

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AC Generator, Back EMF and Time Duration of EMF

In an AC generator, what happens to the induced emf when the rotational speed of the coil doubles?

**Back emf** in motor opposes applied voltage, e_b = N B A ω sin ωt, reduces net current, at start ω=0 e_b=0 current large, as speed increases e_b increases limiting current, power conversion mechanical, principle of motor and generator reciprocity. The emf is proportional to the angular speed ( ε = N B A ω ), so doubling the rotational speed doubles the emf. Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B A ω sinωt, L = μ₀ N²A/l, M = e/(dI/dt) and U = ½ L I², result Doubles follows, reflecting Faraday's law and Lenz's opposition.

Ref: NCERT > Physics Book > Electromagnetic Induction > AC Generator, Back EMF and Time Duration of EMF

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Lenz's Law, Eddy Currents and Applications

In a system of two coils, the mutual inductance depends on which property of the setup?

**Lenz's law** induced current direction opposes change in flux causing it, e = -N dΦ/dt negative sign, conservation of energy. Magnet moved towards coil south pole first, approaching south pole increasing flux into coil with south polarity, coil face nearest magnet becomes south pole to repel, opposing approach, so face becomes south pole, repelling magnet. Mutual inductance depends on the geometry (e.g., coil sizes, separation, orientation) and the medium’s permeability, affecting how much flux from one coil links with the other. Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B A

Ref: NCERT > Physics Book > Electromagnetic Induction > Lenz's Law, Eddy Currents and Applications

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Lenz's Law, Eddy Currents and Applications

A conducting loop is placed in a uniform magnetic field with its plane parallel to the field lines. Why is no emf induce

**Solenoid second coil** experiences emf only when current in solenoid changes because flux linkage changes only then, steady current gives constant Φ, dΦ/dt=0, no emf, when current changes, dΦ/dt ≠0, emf induced, illustrating Faraday's law requirement of changing flux. When the plane is parallel to the field, the flux through the loop is zero ( Φ = B A cos 90° = 0 ), so changing the field strength does not alter the flux, resulting in no emf. Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B A ω sinωt, L = μ₀

Ref: NCERT > Physics Book > Electromagnetic Induction > Lenz's Law, Eddy Currents and Applications

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Energy Stored in Inductor and Magnetic Energy

A solenoid’s current is switched off suddenly. What causes sensitive instruments nearby to be at risk of damage?

**Energy stored in inductor** U =½ L I² (J), L inductance (H), I current (A), resides in magnetic field, energy density u = B²/(2μ₀) (J/m³), B field (T), μ₀=4π×10⁻⁷ H/m. For solenoid B=μ₀ n I, volume V= A l, U = (B²/2μ₀) V =½ (μ₀ n² A l) I² =½ L I², consistent. The sudden drop in current induces a high emf due to self-inductance, which can generate significant currents or voltages in nearby circuits, risking damage. Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B A ω sinωt, L = μ₀ N²A/l,

Ref: NCERT > Physics Book > Electromagnetic Induction > Energy Stored in Inductor and Magnetic Energy

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