JEE Main 2022 · 27 June · evening · Single correct · +4 / −1
Given below are two statements :
Statement I : In hydrogen atom, the frequency of radiation emitted when an electron jumps from lower energy orbit (E₁) to higher energy orbit (E₂), is given as hf = E₁ - E₂
Statement II : The jumping of electron from higher energy orbit (E₂) to lower energy orbit (E₁) is associated with frequency of radiation given as f = (E₂ - E₁)/h
This condition is Bohr's frequency condition.
In the light of the above statements, choose the correct answer from the options given below :
- (a) Both Statement I and Statement II are true.
- (b) Both Statement I and Statement II are false.
- (c) Statement I is correct but Statement II is false.
- (d) Statement I is incorrect but Statement II is true.
Answer and solution
Answer (d) Statement I is incorrect but Statement II is true.
Solution
Physics sir · 4 steps
Going up an orbit costs energy, so the atom absorbs a photon there; it emits only on the way down.
- Bohr's frequency condition. When an electron moves between two orbits, the photon carries exactly the energy difference between them. Photon energy hf is always positive, so it equals the higher energy minus the lower energy.
hf = E₂ - E₁ (E₂ > E₁)
- Check Statement I. A jump from E₁ up to E₂ needs energy, so the atom absorbs radiation and does not emit it. Also, E₁ - E₂ is negative, and a photon cannot have negative energy. So Statement I is false.
- Check Statement II. A jump from E₂ down to E₁ releases the extra energy as a photon with f = (E₂ - E₁)/h. This is exactly Bohr's frequency condition, so Statement II is true.
- So. Statement I is incorrect and Statement II is true, which is option (d).
The idea to keep. Down an orbit means the atom emits, up an orbit means it absorbs. Either way hf equals the higher energy minus the lower energy, so the frequency always comes out positive.
Answer from the NTA final answer key. Solution by PrepWiser.