NEETChemistryStructure of Atom
Match List-I with List-II: List-I (Species and Orbit) List-II (de-Broglie wavelength) (A) H atom, n=3 (I) 2 a₀ (B) He ^+ ion, n=4 (II) a₀ (C) Li ²⁺ ion, n=3 (III) 6 a₀ (D) Be ³⁺ ion, n=2 (IV) 4 a₀ Choose the correct answer from the options given below:
Options
- A(A) - (I), (B) - (II), (C) - (III), (D) - (IV)
- B(A) - (III), (B) - (I), (C) - (IV), (D) - (II)
- C(A) - (III), (B) - (IV), (C) - (I), (D) - (II)
- D(A) - (IV), (B) - (III), (C) - (II), (D) - (I)
Correct answer
C. (A) - (III), (B) - (IV), (C) - (I), (D) - (II)
Step-by-step solution
According to Bohr's quantization condition, mvr = nh 2 . The de-Broglie wavelength is = h mv . Therefore, 2 r = n = 2 r n . The radius of the n^ th Bohr orbit is given by r = a₀ n^2 Z . Substituting r into the wavelength equation, we get: = 2 n ( a₀ n^2 Z ) = 2 a₀ n Z Applying this to each case in List-I: (A) H atom ( Z=1, n=3 ): = 2 a₀ (3) 1 = 6 a₀ (Matches III) (B) He ^+ ion ( Z=2, n=4 ): = 2 a₀ (4) 2 = 4 a₀ (Matches IV) (C) Li ²⁺ ion ( Z=3, n=3 ): = 2 a₀ (3) 3 = 2 a₀ (Matches I) (D) Be ³⁺ ion ( Z=4, n=2 ): = 2 a