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JEE Main20208 Jan 2020Evening ShiftPhysicsAtomic PhysicsActual

The first member of the Balmer series of hydrogen atom has a wavelength of 6561 A ∘ . The wavelength of the second member of the Balmer series (in nm) is_____________

Correct answer

0

Step-by-step solution

1 λ = R Z 2 1 n 1 2 - 1 n 2 2 1 λ 1 = R 1 2 1 2 2 - 1 4 2 = 3 R 16 λ 2 λ 1 = 20 27 λ 2 = 20 27 × 6561   A ∘ = 4860   A ∘ = 486   nm

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