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AIEEE 2012

JEE Main 2012 Chemistry question on Atomic Structure

The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [a0a_0 is Bohr radius]
  1. A.
    h24π2ma02\dfrac{h^2}{4\pi^2 m a_0^2}
  2. B.
    h216π2ma02\dfrac{h^2}{16\pi^2 m a_0^2}
  3. C.
    h232π2ma02\dfrac{h^2}{32\pi^2 m a_0^2}
  4. D.
    h264π2ma02\dfrac{h^2}{64\pi^2 m a_0^2}
Show answer and solution

Correct option: C

Solution

For the nth Bohr orbit, the radius is rn = n²a₀ (for hydrogen, Z=1), so for n=2, r₂ = 4a₀.

The kinetic energy equals the magnitude of total energy: KE = Ze²/(2rn) = e²/(8a₀) for n=2. To express this in terms of h, use the Bohr quantization condition: mvr = nh/2π, so mv = nh/(2πr). Then KE = (1/2)mv² = (mv)²/(2m) = n²h²/(8π²mrn²).

Substituting n=2 and r₂ = 4a₀:

KE = (4h²)/(8π²m·16a₀²) = 4h²/(128π²ma₀²) = h²/(32π²ma₀²).

This matches option 3.

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