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NEET2004PhysicsAtomic PhysicsActual

Energy E of a hydrogen atom with principal quantum number n is given by E= -13.6 n^2 eV . The energy of a photon ejected when the electron jumps for n=3 state n=2 state of hydrogen is approximately

Options

  1. A1.5 eV
  2. B0.85 eV
  3. C3.4 eV
  4. D1.9 eV

Correct answer

D. 1.9 eV

Step-by-step solution

Energy of photon =E₂-E₂ gathered = 13.6 9 - ( -13.6 9 )= 5 30 13.6 =1.9 eV gathered

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