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In a hydrogen atom, the ratio of the kinetic energies of an electron in two states A and B is 9:1 . If state A is the first excited state, then state B is the

Options

  1. A2^ nd excited state
  2. B5^ th excited state
  3. C17^ th excited state
  4. DGround state

Correct answer

B. 5^ th excited state

Step-by-step solution

The kinetic energy of an electron in the n^ th orbit of a hydrogen atom is inversely proportional to the square of the principal quantum number n , i.e., K 1 n^2 . Given the ratio of kinetic energies in states A and B is K_A : K_B = 9:1 . Therefore, K_A K_B = n_B^2 n_A^2 = 9 State A is the first excited state, which corresponds to n_A = 2 . Substituting n_A = 2 into the equation: n_B^2 2^2 = 9 n_B^2 = 9 4 = 36 n_B = 6 The principal quantum number n=6 corresponds to the 5^ th excited state. Assuming the first excite

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