JEE Main 2020 · 9 January · evening · Single correct · +4 / −1
The energy required to ionise a hydrogen like ion in its ground state is 9 Rydbergs. What is the wavelength of the radiation emitted when the electron in this ion jumps from the second excited state to the ground state ?
- (a) 35.8 nm
- (b) 11.4 nm
- (c) 8.6 nm
- (d) 24.2 nm
Answer and solution
Answer (b) 11.4 nm
Solution
Physics sir · 4 steps
The ionisation energy tells you Z². After that it is the Rydberg formula from n = 3 down to n = 1.
- Find Z. For a hydrogen-like ion, the ground-state ionisation energy is Z² Rydbergs. Here Z² = 9, so Z = 3, which is Li²⁺.
- Pick the levels. The ground state is n = 1, so the second excited state is n = 3, not n = 2. The jump is from n = 3 to n = 1.
1/λ = RZ²(1/1²-1/3²) = 9R × 8/9 = 8R
- Put in numbers. With R = 1.097 × 10⁷ m⁻¹, λ = 1/(8 × 1.097 × 10⁷) = 1.14 × 10⁻⁸ m. As a check, the photon energy is 8 × 13.6 = 108.8 eV, and 1240/108.8 = 11.4 nm.
- So. The emitted wavelength is about 11.4 nm, which is option (b).
The idea to keep. Ionisation energy from the ground state equals Z² Rydbergs, so it hands you Z. Also count carefully: the first excited state is n = 2 and the second excited state is n = 3.
Answer from the NTA final answer key. Solution by PrepWiser.