The time period of revolution of electron in its ground state orbit in a hydrogen atom is 1.6 × 10⁻¹⁶ s. The frequency of revolution of the electron in its first excited state (in s⁻¹) is :

Solution

Physics sir · 4 steps

Bohr's period grows as n³, so going from n = 1 to n = 2 makes one revolution 8 times longer.

1. Period and n. In Bohr's model the orbit radius grows as r ∝ n² and the speed falls as v ∝ 1/n. The time for one revolution is the distance around the orbit divided by the speed.

   T = (2π r)/v ∝ n²/(1/n) = n³

2. First excited state. The first excited state is n = 2, so T₂ = 2³ × T₁ = 8 × 1.6 × 10⁻¹⁶ = 1.28 × 10⁻¹⁵ s.
3. Frequency. f₂ = 1/T₂ = 1/(1.28 × 10⁻¹⁵) ≈ 7.8 × 10¹⁴ s⁻¹.
4. So. The frequency of revolution in the first excited state is 7.8 × 10¹⁴ s⁻¹, which is option (c).

The idea to keep. In a hydrogen-like atom, r ∝ n², v ∝ 1/n and T ∝ n³. Build the period from 2π r/v and you never have to memorise the power.

Answer from the NTA final answer key. Solution by PrepWiser.