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#joules

15 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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Nuclear Reactions - Fission, Fusion and Energy Release

What is the energy equivalent of \( 0.005 \, \text{kg} \) of matter in Joules? (Given \( c = 3 \times 10^8 \, \text{m/s}

**Nuclear fusion** source of energy in Sun, proton-proton cycle 4p→He+2e⁺+2ν+26.7 MeV, high temperature ~10⁷ K needed to give kinetic energy to overcome repulsion, thermal motion at high T allows tunneling, energy release because He BE/A higher than H. Fission releases energy because heavy nucleus BE/A ~7.6 MeV splits to intermediate ~8.5 MeV. E = m c² . m = 0.005 kg , c² = 9 × 10¹⁶ m²/s² . E = 0.005 × 9 × 10¹⁶ = 4.5 × 10¹⁴ 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

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

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Nuclear Size, Density and Structure

What is the energy equivalent of \( 0.001 \, \text{kg} \) of matter in Joules? (Given \( c = 3 \times 10^8 \, \text{m/s}

**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 = m c² . m = 0.001 kg , c² = (3 × 10⁸)² = 9 × 10¹⁶ m²/s² . E = 0.001 × 9 × 10¹⁶ = 9 × 10¹³ 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 9 × 10¹³ J, 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

What is the energy equivalent of \( 0.05 \, \text{g} \) of matter in Joules? (Given \( c = 3 \times 10^8 \, \text{m/s} \

**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 = m c² . m = 0.05 × 10⁻³ kg = 5 × 10⁻⁵ kg , c² = 9 × 10¹⁶ m²/s² . E = 5 × 10⁻⁵ × 9 × 10¹⁶ = 4.5 × 10¹² J . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R =

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

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Hydrogen Atom Properties - Radius, Speed and Energy

What is the energy equivalent of \( 0.15 \, \text{g} \) of matter in Joules? (Given \( c = 3 \times 10^8 \, \text{m/s} \

**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 = 0.15 × 10⁻³ kg = 1.5 × 10⁻⁴ kg , c² = 9 × 10¹⁶ m²/s² . E = 1.5 × 10⁻⁴ × 9 × 10¹⁶ = 1.35 × 10¹³ 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 1.35 × 10¹³ J,

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 =

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

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Mass Defect, Binding Energy and Binding Energy per Nucleon

What is the energy equivalent of \( 0.002 \, \text{kg} \) of matter in Joules? (Given \( c = 3 \times 10^8 \, \text{m/s}

**Energy equivalent** E= m c², 0.005 kg matter E=0.005×9×10¹⁶=4.5×10¹⁴ J, 0.01 kg 9×10¹⁴ J, mass defect 0.1 u => BE=0.1×931.5=93.15 MeV, mass defect from BE 149.04 MeV => Δm=149.04/931.5=0.16 u, BE per nucleon 8.5 MeV A=20 total BE=170 MeV. E = m c² . m = 0.002 kg , c² = 9 × 10¹⁶ m²/s² . E = 0.002 × 9 × 10¹⁶ = 1.8 × 10¹⁴ 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 1.8 × 10¹⁴ J, consistent with Bohr

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

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Mass Defect, Binding Energy and Binding Energy per Nucleon

What is the energy equivalent of \( 0.5 \, \text{g} \) 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 = 0.5 × 10⁻³ kg , c² = 9 × 10¹⁶ m²/s² . E = 0.5 × 10⁻³ × 9 × 10¹⁶ = 4.5 × 10¹³ J . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀

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

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Mass Defect, Binding Energy and Binding Energy per Nucleon

What is the energy equivalent of \( 1 \, \text{kg} \) of matter in Joules? (Given \( c = 3 \times 10^8 \, \text{m/s} \))

**Energy equivalent** E= m c², 0.005 kg matter E=0.005×9×10¹⁶=4.5×10¹⁴ J, 0.01 kg 9×10¹⁴ J, mass defect 0.1 u => BE=0.1×931.5=93.15 MeV, mass defect from BE 149.04 MeV => Δm=149.04/931.5=0.16 u, BE per nucleon 8.5 MeV A=20 total BE=170 MeV. E = m c² . m = 1 kg , c² = 9 × 10¹⁶ m²/s² . E = 1 × 9 × 10¹⁶ = 9 × 10¹⁶ 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 9 × 10¹⁶ J, consistent with Bohr

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

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Work Heat Distinction and Energy Transfer Modes

How many joules are equivalent to 350 cal of heat? (1 cal = 4.186 J )

**Energy transfer** first law ΔU = Q - W, W includes P-V work, shaft work, electrical work, Q includes conduction Fourier law, convection, radiation Stefan-Boltzmann, distinction important because work is controllable, heat spontaneous from hot to cold, entropy associated with heat not work, explaining why heat engine efficiency

Ref: NCERT > Physics Book > Thermodynamics > Work Heat Distinction and Energy Transfer Modes

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Isochoric Processes and Pressure-Temperature

How many calories are equivalent to 4186 J of heat? (1 cal = 4.186 J )

**Isochoric process** constant volume ΔV=0, work W=0, first law ΔU = Q, all heat goes to internal energy, P/T = constant from ideal gas law P V = n R T at constant V, pressure proportional to temperature, P₁/T₁ = P₂/T₂, e.g., heating gas in rigid container pressure rises proportionally to T. Heat in cal = Heat in J4.186 . (4186)/(4.186) = 1000 cal . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic P V^γ = const and η = 1 - T_c/T_h, evaluation yields 1000 cal,

Ref: NCERT > Physics Book > Thermodynamics > Isochoric Processes and Pressure-Temperature

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