#energy conversion

The minimum wavelength of X-rays from a \( 25 \, \text{kV} \) tube is: (Take \( h = 6.63 \times 10^{-34} \, \text{J s} \

**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. E = e V = 1.6 × 10⁻¹⁹ × 25 × 10³ = 4.0 × 10⁻¹⁵ J . λₘiₙ = (h c/E) = (6.63 × 10⁻³⁴ × 3 × 10⁸/4.0 × 10⁻¹⁵) = 4.9725 × 10⁻¹¹ m ≈ 0.0497 nm . Applying E = h f = h c/λ,

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

A photon of wavelength \( 300 \, \text{nm} \) is incident on a metal. What is its energy in eV? (Take \( h c = 1240 \, \

**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. Energy E = (h c/λ) . E = (1240/300) ≈ 4.13 eV . 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 4.13 eV 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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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

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

A photon’s energy is \( 3.0 \, \text{eV} \). What is its momentum? (Take \( c = 3 \times 10^8 \, \text{m/s} \), \( 1 \,

**Photon energy** E = h f = h c/λ, momentum p = h/λ = E/c, power P = N E/t, N number of photons per second, wavelength λ = h c/E, for photon momentum 1.5×10⁻²⁷ kg·m/s frequency f = p c/h? Actually E = p c, f = E/h = p c/h =1.5×10⁻²⁷×3×10⁸/6.63×10⁻³⁴=6.79×10¹⁴ Hz, energy E= h f =4.5×10⁻¹⁹ J, for 5.0×10⁻¹⁹ J frequency f=E/h=7.54×10¹⁴ Hz. E = 3.0 × 1.6 × 10⁻¹⁹ = 4.8 × 10⁻¹⁹ J . For a photon, p = (E/c) = (4.8 × 10⁻¹⁹/3 × 10⁸) = 1.6 × 10⁻²⁷ kg m/s . Applying E = h f = h

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

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Radioactive Decay, Nuclear Forces and Stability

How much energy is equivalent to a mass defect of \( 0.1 \, \text{u} \)? (Given \( 1 \, \text{u} = 931.5 \, \text{MeV/c}

**Stability** belt of stability N≈Z for light, N>Z for heavy due to Coulomb, beyond leads to alpha decay, fission, stability requires balance, nuclear force saturated explains constant density and BE/A. Energy = Δ M · c² . Δ M = 0.1 u . Energy = 0.1 × 931.5 = 93.15 MeV . 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 93.15 MeV, consistent with Bohr model and nuclear binding energy systematics.

Ref: NCERT > Physics Book > Atoms and Nuclei > Radioactive Decay, Nuclear Forces and Stability

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Nuclear Size, Density and Structure

What is the binding energy of a nucleus with mass defect \( 0.12 \, \text{u} \)? (Given \( 1 \, \text{u} = 931.5 \, \tex

**Nuclear size** order 10⁻¹⁵ m femtometer, atomic size 10⁻¹⁰ m, ratio 10⁵, nucleus contains protons and neutrons bound by strong force, density independent of A indicates incompressibility, R₀ determined from electron scattering experiments. E_b = Δ M · c² . Δ M = 0.12 u . E_b = 0.12 × 931.5 = 111.78 MeV . 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 111.78 MeV, consistent with Bohr model and nuclear binding energy systematics.

Ref: NCERT > Physics Book > Atoms and Nuclei > Nuclear Size, Density and Structure

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Hydrogen Atom Properties - Radius, Speed and Energy

The mass of a proton is \( 1.67262 \times 10^{-27} \, \text{kg} \). What is its energy equivalent in Joules? (Given \( c

**Angular momentum** L_n = n h/2π, h=6.6×10⁻³⁴ J·s, L₂=2×6.6×10⁻³⁴/2π=2.11×10⁻³⁴ J·s, speed v_n = e²/(2 ε₀ h) ×1/n ≈2.2×10⁶/n m/s, kinetic energy ½ m v² =13.6/n² eV, illustrating Bohr model predictions for hydrogen-like atoms. E = m c² . m = 1.67262 × 10⁻²⁷ kg , c² = (3 × 10⁸)² = 9 × 10¹⁶ m²/s² . E = 1.67262 × 10⁻²⁷ × 9 × 10¹⁶ ≈ 1.505 × 10⁻¹⁰ J . 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 1.505 × 10⁻¹⁰ J, consistent

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

What is the energy equivalent of \( 0.1 \, \text{g} \) of matter in Joules? (Given \( c = 3 \times 10^8 \, \text{m/s} \)

**Mass defect** Δm = Z m_p + N m_n - M_nucleus, binding energy BE = Δm c², 1 u =931.5 MeV/c², BE per nucleon = BE/A, measures stability, peak ~8.8 MeV at Fe-56, for A=36 BE=288 MeV BE/A=8 MeV, for A=12 BE=96 MeV BE/A=8 MeV, for A=16 BE=127.5 MeV BE/A≈7.97 MeV, higher BE/A more stable. E = m c² . m = 0.1 × 10⁻³ kg = 10⁻⁴ kg , c² = 9 × 10¹⁶ m²/s² . E = 10⁻⁴ × 9 × 10¹⁶ = 9 × 10¹² J . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R =