The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is equal to h²/xma₀². The value of 10x is ___________. (a₀ is radius of Bohr's orbit) (Nearest integer) [Given : π = 3.14]

Solution

Chemistry ma’am · 4 steps

Get the speed from Bohr's angular momentum rule, then put it into 1/2mv² for n = 2.

1. Second orbit radius. In hydrogen, rₙ = n² a₀. So the second orbit has r₂ = 4a₀.
2. Speed from quantisation. Bohr's rule says mvr = nh/2π. With n = 2 and r = 4a₀, this gives v = 2h/(2π m (4a₀)) = h/(4π m a₀).

   KE = 1/2mv² = 1/2m(h/(4π m a₀))² = h²/(32π² m a₀²)

3. Compare the forms. Matching with h²/(x m a₀²) gives x = 32π² = 32 × 9.8596 = 315.51.
4. So. 10x = 3155.1, which rounds to 3155.

The common slip: 789. Taking r₂ = 2a₀ instead of 4a₀ gives x = 8π², so 10x comes out as 789, not 3155.

The idea to keep. For any Bohr orbit, find rₙ = n² a₀ first, then get v from mvr = nh/2π. Both n values matter, the one in the radius and the one in the angular momentum.

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