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NEET PHYSICSHard

The photon radiated from hydrogen corresponding to the second line of Lyman series is absorbed by a hydrogen-like atom X in the second excited state. As a result the hydrogen-like atom X makes a transition to nth orbit. Then:

A

X = He⁺, n = 4

B

X = Li⁺⁺, n = 6

C

X = He⁺, n = 9

D

X = Li⁺⁺, n = 9

Step-by-Step Solution

  1. Calculate energy of the photon from Hydrogen: The second line of the Lyman series corresponds to the transition from n=3 to n=1. E = 13.6 × (1/1² - 1/3²) = 13.6 × (8/9) eV.
  2. Analyze Atom X: It absorbs this photon starting from the second excited state (n=3). The energy difference for transition n=3 to n in atom X (atomic number Z) is: \Delta E = 13.6 × Z² × (1/3² - 1/n²).
  3. Equate Energies: 13.6 × (8/9) = 13.6 × Z² × (1/9 - 1/n²) 8/9 = Z² × (1/9 - 1/n²)
  4. Test Options:
  • For He⁺ (Z=2): 8/9 = 4(1/9 - 1/n²) ⇒ 2/9 = 1/9 - 1/n² ⇒ 1/n² = -1/9 (Not possible).
  • For Li⁺⁺ (Z=3): 8/9 = 9(1/9 - 1/n²) ⇒ 8/81 = 1/9 - 1/n² ⇒ 1/n² = 9/81 - 8/81 = 1/81 ⇒ n = 9. Therefore, X is Li⁺⁺ and the final orbit is n=9.
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