COMEDK20269 May 2026Evening ShiftChemistryStructure of AtomActual
The frequency of photon which is emitted during a transition of electron of He ^+ ion from fifth energy level to third energy level will be:
Options
- A9.39 10¹⁴ s⁻¹
- B1.34 10⁻¹⁴ s⁻¹
- C2.34 10¹⁴ s⁻¹
- D8.29 10⁻¹⁴ s⁻¹
Correct answer
A. 9.39 10¹⁴ s⁻¹
Step-by-step solution
Using the Rydberg formula for the frequency of the emitted photon: = R c Z^2 ( 1 n₁^2 - 1 n₂^2 ) For the He ^+ ion, Z = 2 . The transition is from n₂ = 5 to n₁ = 3 . = R c (2)^2 ( 1 3^2 - 1 5^2 ) = R c 4 ( 1 9 - 1 25 ) = R c 4 16 225 = R c 64 225 Substituting the values R 1.097 10^7 m ⁻¹ and c = 3 10^8 m/s : = (1.097 10^7) (3 10^8) 64 225 = 3.291 10¹⁵ 0.2844 9.36 10¹⁴ s ⁻¹ This value is closest to 9.39 10¹⁴ s ⁻¹ . Answer: 9.39 10¹⁴ s⁻¹