Skip to content
Login Register

#total energy

13 public questions tagged with this topic.

Hydrogen Atom Properties - Radius, Speed and Energy

An electron in a hydrogen atom has a total energy of -3.4 eV. What is its kinetic energy?

**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 = -K , K = -E = -(-3.4) = 3.4 eV . 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 3.4 eV, consistent with Bohr model and nuclear binding energy systematics.

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

View Details
Hydrogen Atom Properties - Radius, Speed and Energy

In a hydrogen atom, the total energy of an electron in the \( n = 3 \) state is -1.51 eV. What is its potential energy?

**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 = K + U , K = -E = 1.51 eV . U = -2K = -2 × 1.51 = -3.02 eV . 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 -3.02

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

View Details
Hydrogen Atom Properties - Radius, Speed and Energy

What is the total energy of an electron in the \( n = 5 \) state of a hydrogen atom? (Use \( E_n = -\frac{13.6}{n^2} \,

**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. E₅ = -(13.6/5²) = -(13.6/25) = -0.544 eV . 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 -0.544 eV, consistent with Bohr model and nuclear binding energy systematics.

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

View Details
Hydrogen Atom Properties - Radius, Speed and Energy

What is the significance of the negative total energy of an electron in a hydrogen atom according to Bohr’s model?

**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 negative total energy indicates that the electron is bound to the nucleus, requiring energy to be supplied to free it. 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 Electron is bound to the nucleus, consistent with Bohr model and

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

View Details
Spring-Mass System and Combination of Springs

A spring-mass system has \( m = 0.8 \, \text{kg}, k = 320 \, \text{N/m} \). If displaced by \( 5 \, \text{cm} \), what i

**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 = 0.05 m, k = 320 N/m . E = 0.5 × 320 × (0.05)² = 0.5 × 320 × 0.0025 = 0.4 J . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.4 J follows, reflecting SHM dependence on amplitude A, ω and system parameters.

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

View Details
Spring-Mass System and Combination of Springs

A spring-mass system has \( m = 0.4 \, \text{kg}, k = 160 \, \text{N/m} \). If displaced by \( 6 \, \text{cm} \), 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 = 0.06 m, k = 160 N/m . E = 0.5 × 160 × (0.06)² = 0.5 × 160 × 0.0036 = 0.288 J . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.288 J

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

View Details
Spring-Mass System and Combination of Springs

A spring-mass system has \( m = 0.9 \, \text{kg}, k = 360 \, \text{N/m} \). If displaced by \( 4 \, \text{cm} \), what i

**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 = 0.04 m, k = 360 N/m . E = 0.5 × 360 × (0.04)² = 0.5 × 360 × 0.0016 = 0.288 J . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.288 J follows, reflecting SHM dependence on amplitude A, ω and system parameters.

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

View Details
Spring-Mass System and Combination of Springs

A spring-mass system has \( m = 0.5 \, \text{kg}, k = 200 \, \text{N/m} \). If the amplitude is \( 4 \, \text{cm} \), wh

**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 = 0.04 m, k = 200 N/m . E = 0.5 × 200 × (0.04)² = 0.5 × 200 × 0.0016 = 0.16 J . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.16 J follows, reflecting SHM dependence on amplitude A, ω and system parameters.

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

View Details
Energy in SHM - Kinetic, Potential and Total

A mass of \( 3 \, \text{kg} \) is attached to a spring with \( k = 300 \, \text{N/m} \). If displaced by \( 15 \, \text{

**Conservation of mechanical energy** in undamped SHM implies total energy proportional to amplitude squared A² and spring constant k. E = ½ k A² allows amplitude determination from known E and k via A = √(2E/k), with k = m ω² linking dynamical and energetic descriptions for spring-mass system. Total energy: E = (1/2) k A² . A = 0.15 m, k = 300 N/m . E = (1/2) × 300 × (0.15)² = 0.5 × 300 × 0.0225 = 3.375 J . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A²,

Ref: NCERT > Physics Book > Oscillations > Energy in SHM - Kinetic, Potential and Total

View Details
Escape Velocity and Bound Systems

A 700kg satellite orbits Earth at 10RE from the center. What is its total energy? (ME\=6×1024kg,RE\=6.4×106m,G\=6.67×10−

E = −GMEm2r. r = 10RE = 6.4×107m. E = −6.67×10−11×6×1024×7002×6.4×107. E = −2.801×10171.28×108≈−2.19×109J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to -2.2 × 10⁹ J. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: Halliday & Resnick, Fundamentals of Physics, 11th ed., Chapter 13: Gravitation.

View Details