Exams › JEE Main › Chemistry
For any given series of spectral lines of atomic hydrogen, let Δν̄ = ν̄_max − ν̄_min be the difference in maximum and minimum frequencies in cm⁻¹. The ratio Δν̄_Lyman / Δν̄_Balmer is:
- 4: 1
- 9: 4
- 5: 4
- 27: 5
Correct answer: 9: 4
Solution
The ratio Δν̄_Lyman / Δν̄_Balmer is derived from the differences in energy levels involved in the transitions for each series. The Lyman series transitions occur from higher energy levels to n=1, while the Balmer series transitions go to n=2, resulting in a greater frequency range for the Lyman series, specifically leading to a ratio of 9:4 when calculated.
Related JEE Main Chemistry questions
- At what wavenumber does the first line of the hydrogen Balmer series occur in the atomic spectrum?
- In the hydrogen emission spectrum, a spectral series has its limit at 12186.3 cm⁻¹. This series is identified as
- Two rapidly moving particles X and Y have de Broglie wavelengths of 1 nm and 4 nm, respectively. If the mass of X is nine times the mass of Y, what is the ratio of their kinetic energies, X: Y?
- Given that Planck’s constant is 6.63 × 10⁻³⁴ J s and the speed of light is 3.0 × 10⁸ m s⁻¹, which of the following is nearest to the wavelength, in nanometres, of light having frequency 8 × 10¹⁵ s⁻¹?
- A lithium ion and a proton are each accelerated through the same potential difference. If their de Broglie wavelengths are λ₁ and λ₂, respectively, and m_(Li) = 9mₚ, what is the ratio λ₁: λ₂?
- The ionisation energy of He⁺ is 19.6 × 10⁻¹⁸ J atom⁻¹. What is the energy of the first stationary state (n = 1) of Li²⁺?
⚔️ Practice JEE Main Chemistry free + battle 1v1 →