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An electron of a hydrogen like atom, having Z = 4 , jumps from 4 t h energy state to 2 n d energy state, The energy released in this process, will be: (Given R c h = 13 . 6 e V ) Where R = Rydberg constant c = Speed of light in vacuum h = Planck's constant

Options

  1. A13 . 6   e V
  2. B10 . 5   e V
  3. C3 . 4   e V
  4. D40 . 8   e V

Correct answer

D. 40 . 8   e V

Step-by-step solution

According to Bohr's atomic theory, whenever an electron jumps from one orbit to another, there is always an emission or absorption energy in terms of emission or absorption of photons. The formula to calculate the energy ∆ E of the emitted or absorbed photon when an electron jumps from a state with quantum number n 1 to another with quantum number n 2 is given by ∆ E =   - R c h Z 2 1 n i 2 - 1 n f 2 . . . . . . . . . . . . . . . . . . . . ( 1 ) where, R is Rydberg's constant, h is Planck&#

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