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#orbital radius

11 public questions tagged with this topic.

Hydrogen Atom Properties - Radius, Speed and Energy

The radius of the third orbit in a hydrogen atom is how many times the radius of the first orbit if \( r_1 = 5.3 \times

**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. r_n = n² r₁ . For n = 3 : r₃ = 3² × 5.3 × 10⁻¹¹ = 9 × 5.3 × 10⁻¹¹ = 4.77 × 10⁻¹⁰ m . Ratio = (r₃/r₁) = 9 . 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, consistent with Bohr model and nuclear binding energy systematics.

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

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Atomic Models - Rutherford, Thomson and Bohr

What is the ratio of the radius of the \( n = 4 \) orbit to the \( n = 2 \) orbit in a hydrogen atom?

**Bohr's stationary orbits** defined by angular momentum quantization L = n ħ, ħ = h/2π, n=1 ground state, electron in these orbits does not radiate despite acceleration, contrary to classical EM theory which predicts atom collapse due to energy loss via radiation, Bohr postulates to explain observed stability and discrete spectra. r_n = n² r₁ . r₄ = 16 r₁ , r₂ = 4 r₁ . Ratio = (r₄/r₂) = (16/4) = 4 . 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 4, consistent

Ref: NCERT > Physics Book > Atoms and Nuclei > Atomic Models - Rutherford, Thomson and Bohr

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Universal Law of Gravitation

A satellite orbits a planet at 8×107m from its center with a period of 20 hours. What is the planet’s mass? (G\=6.67×10−

M = 4π2r3GT2. T = 20×3600 = 72000s, T2 = 5.184×109s2. r3 = (8×107)3 = 5.12×1023m3. M = 4×(3.14)2×5.12×10236.67×10−11×5.184×109. M = 2.019×10253.458×10−1≈5.84×1025kg. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 5.8 × 10²⁵ kg. 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.

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Gravitational Effects and Applications

A planet orbits the Sun with a period of 10 years. If Earth’s orbital radius is 1.5×1011m, what is its semi-major axis?

Kepler’s third law: Tp2TE2 = ap3aE3. TE = 1year, Tp = 10years, aE = 1.5×1011m. 10212 = ap3(1.5×1011)3. 100 = ap33.375×1033. ap3 = 100×3.375×1033 = 3.375×1035. ap = (3.375×1035)1/3≈6.96×1011m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 7.0 × 10¹¹ m. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

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

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Gravitational Effects and Applications

A planet has a moon with an orbital radius of 5×108m and a period of 10 days. What is the planet’s mass? (G\=6.67×10−11N

T2 = 4π2GMR3. M = 4π2R3GT2. T = 10×86400 = 8.64×105s. T2 = (8.64×105)2 = 7.465×1011s2. R3 = (5×108)3 = 1.25×1025m3. M = 4×(3.14)2×1.25×10256.67×10−11×7.465×1011. M = 4.93×10264.979×101≈9.9×1024kg. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 9.9 × 10²⁴ kg. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

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

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Orbital Velocity and Time Period

A 1000kg satellite orbits Earth at 16RE from the center. What is its total energy? (ME\=6×1024kg,RE\=6.4×106m,G\=6.67×10

E = −GMEm2r. r = 16RE = 1.024×108m. E = −6.67×10−11×6×1024×10002×1.024×108. E = −4.002×10172.048×108≈−1.95×109J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to -1.9 × 10⁹ J. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.

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

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Orbital Velocity and Time Period

How much energy is required to move a 300kg satellite from 7RE to 14RE from Earth’s center? (ME\=6×1024kg,RE\=6.4×106m,G

ΔE = −GMEm(1r2−1r1). r1 = 4.48×107m, r2 = 8.96×107m. ΔE = −6.67×10−11×6×1024×300(18.96×107−14.48×107). ΔE = −1.201×1017(−1.116×10−8)≈1.34×109J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.3 × 10⁹ J. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.

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

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Orbital Velocity and Time Period

What is the kinetic energy of a 200kg satellite at 4RE from Earth’s center? (ME\=6×1024kg,RE\=6.4×106m,G\=6.67×10−11N m2

K = GMEm2r. r = 4RE = 2.56×107m. K = 6.67×10−11×6×1024×2002×2.56×107. K = 8.004×10165.12×107≈1.56×109J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.6 × 10⁹ J. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

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

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Orbital Velocity and Time Period

A moon orbits a planet with period 8 days and radius 3×108m. What is the planet’s mass? (G\=6.67×10−11N m2/kg2,1day\=864

M = 4π2r3GT2. T = 8×86400 = 6.912×105s. T2 = 4.776×1011s2. r3 = (3×108)3 = 2.7×1025m3. M = 4×(3.14)2×2.7×10256.67×10−11×4.776×1011. M = 1.065×10263.185×101≈3.34×1024kg. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 3.4 × 10²⁴ kg. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

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

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