What will be the longest wavelength in the Balmer series of hydrogen spectrum?
6557 × 10 -10 m is correct per What will be the longest wavelength in t principle; validated by standard physics law.
Ref: NCERT Physics Class 12 Chapter 12 Atoms
#Balmer series
7 public questions tagged with this topic.
Filter questions
What characterizes the emission line spectrum of a hydrogen atom?
**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. The emission line spectrum consists of bright lines on a dark background, corresponding to specific wavelengths emitted during electron transitions. Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm
Ref: NCERT > Physics Book > Atoms and Nuclei > Bohr Model Energy Levels and Hydrogen Spectrum
Which series of spectral lines in a hydrogen atom lies in the visible region?
The Balmer series lies in the visible region.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Shapes of Orbitals - s p d f and Nodes and Radial Probability
What is the wavelength of the second line in the Balmer series of a hydrogen atom? (RH = 1.097 × 10⁷ m⁻¹)
n=4 to n=2. λ = 486.1 nm.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Quantum Mechanical Model - de Broglie Heisenberg and Schrodinger
The ratio of the wavelengths of the first lines in the Lyman and Balmer series of a hydrogen atom is: (RH = 1.097 × 10⁷
λLyman/λBalmer ≈ 0.185.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Electronic Configuration - Aufbau Pauli Hund and Stability
The ratio of the wavelengths of the second lines in the Balmer and Paschen series of a hydrogen atom is: (RH = 1.097 × 1
Balmer (4→2): λ₁ ≈ 4.861×10⁻⁷ m. Paschen (5→3): λ₂ ≈ 1.281×10⁻⁶ m. Ratio ≈ 0.379.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Discovery of Subatomic Particles - Electron Proton Neutron
The wavenumber of the fourth line in the Balmer series of a hydrogen atom is: (RH = 1.097 × 10⁷ m⁻¹)
Balmer: n₁=2, n₂=6. ν̄ = 2.437×10⁶ m⁻¹.
Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Atomic Models - Thomson Rutherford and Bohr Model
