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NEET
Exam
NEET
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Physics
Chapter
Atomic Physics

NEET · Physics · Atomic Physics

Match List I with List II. List I List II (A) 2^ nd orbit of H (I)...

Match List I with List II.

List IList II
(A) 2nd2^{\text{nd}} orbit of H(I) 1a01 a_0
(B) 3rd3^{\text{rd}} orbit of He+^+(II) 3a03 a_0
(C) 3rd3^{\text{rd}} orbit of Li2+^{2+}(III) 4a04 a_0
(D) 2nd2^{\text{nd}} orbit of Be3+^{3+}(IV) 4.5a04.5 a_0

Choose the correct answer from the options given below:

Options

  1. A(A)-(I), (B)-(II), (C)-(III), (D)-(IV)
  2. B(A)-(III), (B)-(IV), (C)-(II), (D)-(I)
  3. C(A)-(III), (B)-(II), (C)-(IV), (D)-(I)
  4. D(A)-(IV), (B)-(III), (C)-(I), (D)-(II)

Correct answer

B. (A)-(III), (B)-(IV), (C)-(II), (D)-(I)

Step-by-step solution

The radius of the nthn^{\text{th}} Bohr orbit of a hydrogen-like species is given by:
rn=a0n2Zr_n = a_0 \dfrac{n^2}{Z}

(A) For the 2nd2^{\text{nd}} orbit of H (n=2,Z=1n = 2, Z = 1):
r=a0221=4a0r = a_0 \dfrac{2^2}{1} = 4 a_0

(B) For the 3rd3^{\text{rd}} orbit of He+^+ (n=3,Z=2n = 3, Z = 2):
r=a0322=4.5a0r = a_0 \dfrac{3^2}{2} = 4.5 a_0

(C) For the 3rd3^{\text{rd}} orbit of Li2+^{2+} (n=3,Z=3n = 3, Z = 3):
r=a0323=3a0r = a_0 \dfrac{3^2}{3} = 3 a_0

(D) For the 2nd2^{\text{nd}} orbit of Be3+^{3+} (n=2,Z=4n = 2, Z = 4):
r=a0224=1a0r = a_0 \dfrac{2^2}{4} = 1 a_0

Therefore, the correct matching is (A)-(III), (B)-(IV), (C)-(II), (D)-(I).

Answer: (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
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