Hydrogen atoms in the ground state are excited using monochromatic radiation of wavelength 969.69 angstroms. After absorbing the radiation, the atoms jump to an excited state. In which principal quantum number n does the excited electron reside? (Given: Rydberg constant R = 1.1 * 10⁷ m⁻¹)

Question

Accepted Answer

n = 4. Wavelength lambda = 969.69 angstroms = 969.69 * 10⁻¹⁰ m. 1/lambda = 1/(969.69 * 10⁻¹⁰) ≈ 1.031 * 10⁷ m⁻¹. Using Rydberg formula: 1/lambda = R(1/1 - 1/n²). So 1 - 1/n² = (1.031 * 10⁷)/(1.1 * 10⁷) = 0.9375 = 15/16. Therefore 1/n² = 1/16, giving n = 4.

Hydrogen atoms in the ground state are excited using monochromatic radiation of wavelength 969.69 angstroms. After absorbing the radiation, the atoms jump to an excited state. In which principal quantum number n does the excited electron reside? (Given: Rydberg constant R = 1.1 * 10⁷ m⁻¹)

Solution

Related JEE Advanced Physics questions