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JEE Main20199 Apr 2019Morning ShiftPhysicsAtomic PhysicsActual

Taking the wavelength of first Balmer line in hydrogen spectrum ( n = 3 to n = 2 ) as 660 n m , the wavelength of the 2 n d Balmer line ( n = 4 to n = 2 ) will be :

Options

  1. A889.2 n m
  2. B488.9 n m
  3. C388.9 n m
  4. D642.7   nm

Correct answer

B. 488.9 n m

Step-by-step solution

From Rydberg’s equation, 1 λ = R 1 n 2 - 1 n 2 2 1 660 × 10 - 9 = R 1 2 2 - 1 3 2 1 λ = R 1 2 2 - 1 4 2 λ 600 × 10 - 9 = 5 9 × 4 × 16 × 4 12 λ = 488.9 n m

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