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PhysicsNEET 2024Hydrogen spectrum and energy levels

Atoms

Match List I with List II.

List I (Spectral lines of hydrogen for transitions from) List II (Wavelength, nm)
A. n2=3n_2 = 3 to n1=2n_1 = 2 I. 410.2
B. n2=4n_2 = 4 to n1=2n_1 = 2 II. 434.1
C. n2=5n_2 = 5 to n1=2n_1 = 2 III. 656.3
D. n2=6n_2 = 6 to n1=2n_1 = 2 IV. 486.1

Choose the correct answer from the options given below:

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

Solution

Balmer wavelengths follow 1λ∝(14−1n2)\frac{1}{\lambda} \propto \left(\frac14 - \frac{1}{n^2}\right): the smallest jump (n=3→2n=3\to2) gives the longest wavelength, 656.3 nm (H−α)(H-\alpha), then 486.1, 434.1 and 410.2 nm as nn rises. Hence A-III, B-IV, C-II, D-I.

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