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#quantum levels

6 public questions tagged with this topic.

A hydrogen atom in the \( n = 3 \) state emits a photon and returns to the ground state. What is the maximum energy of t

**Bohr's quantization** angular momentum L = m v r = n h/2π, for n=2 L=2h/2π= h/π=2.11×10⁻³⁴ J·s, for n=5 L=5h/2π, de Broglie λ = h/p, p= m v, for first orbit v=2.2×10⁶ m/s, λ= h/(m v)=6.6×10⁻³⁴/(9.1×10⁻³¹×2.2×10⁶)=3.3×10⁻¹⁰ m, circumference 2πr=3.33×10⁻¹⁰ m, one wavelength fits for n=1. E₃ = -1.51 eV , E₁ = -13.6 eV . Δ E = 12.09 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 12.09 eV, consistent with Bohr model and nuclear binding energy systematics.

Ref: NCERT > Physics Book > Atoms and Nuclei > De Broglie Hypothesis and Quantization in Bohr Model

What is the speed of an electron in the \( n = 4 \) orbit of a hydrogen atom if its speed in \( n = 1 \) is \( 2.2 \time

**Bohr energy levels** E_n = -13.6/n² eV for hydrogen, negative indicating bound state, total energy = -13.6 eV ground state n=1, -3.4 eV n=2, -1.51 eV n=3, etc., photon energy for transition n_i → n_f is ΔE =13.6(1/n_f² -1/n_i²) eV, wavelength λ = hc/ΔE, h=6.6×10⁻³⁴ J·s, c=3×10⁸ m/s. Emission line spectrum characterized by discrete wavelengths because energy levels discrete. v_n = (v₁/n) . For n = 4 : v₄ = (2.2 × 10⁶/4) = 5.5 × 10⁵ m/s . 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

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

In Bohr’s model, what happens to the energy required to ionize a hydrogen atom as the electron’s orbit number increases?

**Bohr energy levels** E_n = -13.6/n² eV for hydrogen, negative indicating bound state, total energy = -13.6 eV ground state n=1, -3.4 eV n=2, -1.51 eV n=3, etc., photon energy for transition n_i → n_f is ΔE =13.6(1/n_f² -1/n_i²) eV, wavelength λ = hc/ΔE, h=6.6×10⁻³⁴ J·s, c=3×10⁸ m/s. Emission line spectrum characterized by discrete wavelengths because energy levels discrete. As n increases, the energy becomes less negative (closer to zero), so less energy is required to ionize the atom from higher orbits. 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 > Bohr Model Energy Levels and Hydrogen Spectrum

The radius of the first orbit in a hydrogen atom is \( 5.3 \times 10^{-11} \, \text{m} \). What is the radius of the sec

**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. Radius r_n = n² r₁ , where r₁ = 5.3 × 10⁻¹¹ m . For n = 2 : r₂ = 2² × 5.3 × 10⁻¹¹ = 4 × 5.3 × 10⁻¹¹ = 2.12 × 10⁻¹⁰ m . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀

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

What is the energy difference between the \( n = 5 \) and \( n = 3 \) states in a hydrogen atom? (Use \( E_n = -\frac{13

**Bohr energy levels** E_n = -13.6/n² eV for hydrogen, negative indicating bound state, total energy = -13.6 eV ground state n=1, -3.4 eV n=2, -1.51 eV n=3, etc., photon energy for transition n_i → n_f is ΔE =13.6(1/n_f² -1/n_i²) eV, wavelength λ = hc/ΔE, h=6.6×10⁻³⁴ J·s, c=3×10⁸ m/s. Emission line spectrum characterized by discrete wavelengths because energy levels discrete. E₅ = -0.544 eV , E₃ = -1.51 eV . Δ E = -0.544 - (-1.51) = 0.966 eV ≈ 0.97 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

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

Which electromagnetic waves are emitted by electrons transitioning between energy levels in atoms?

**IR effectiveness in remote sensing** because IR scattered less than visible by atmosphere, can detect heat signatures, used in satellite imaging for vegetation, moisture, military night vision, illustrating application based on thermal emission and low scattering. The document states that visible light is emitted when electrons in atoms move from one energy level to a lower energy level. Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields Visible light, illustrating EM wave transverse nature and Maxwell's displacement current concept.

Ref: NCERT > Physics Book > Electromagnetic Waves > Infrared, Visible Light and Applications