Q28.An excited He+ ion emits two photons in succession, with wavelengths 108.5 nm and 30.4 nm in making a transition to the ground state. The quantum number n, corresponding to its initial excited state is 1240 eV (for a photon of wavelength λ, energy E = ) λ ( in nm ) (1) n = 6 (2) n = 5 (3) n = 7 (4) n = 4
What This Question Tests
This problem involves calculating the total energy emitted by an He+ ion during two successive transitions to the ground state, and then using the Bohr model to find the initial excited state quantum number.
Concepts Tested
Formulas Used
E = hc/λ = 1240 eV-nm/λ(nm)
E_n = -13.6 Z²/n² eV
ΔE = E_initial - E_final
📚 NCERT Sections This Tests
12.5 — A Hydrogen Atom Initially In The Ground Level Absorbs A Photon,
Physics Class 12 · Chapter 12
12.5 A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of photon.
12.8 — A 12.5 Ev Electron Beam Is Used To Bombard Gaseous Hydrogen At
Physics Class 12 · Chapter 12
12.8 A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?
12.3 — A Difference Of 2.3 Ev Separates Two Energy Levels In An Atom. What
Physics Class 12 · Chapter 12
12.3 A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom make a transition from the upper level to the lower level?
📋 Question Details
- Chapter
- Atoms
- Topic
- Atomic Spectra and Bohr Model for Hydrogen-like Atoms
- Year
- 2019
- Shift
- 12 Apr Shift 1
- Q Number
- Q28
- Type
- Multi concept
- NCERT Ref
- Class 12 Physics Ch 12: Atoms
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