Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.

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Payal Gupta | Contributor-Level 10

3 months ago

According to Balmer formula? = 1 / λ = R_H [1/n₁² – 1/n₂²]

For the longest wavelength transition in the Balmer series of atomic hydrogen, wave number must be least. This is possible in case n₂– n₁ = minimum; i.e. n₁= 2 and n₂= 3. Substituting the values:

? = 1 / λ = (1.097 x 10⁷ m^-1) [1/2² – 1/3²] = (1.097 x 10⁷ m^-1) [5/36] = 1.523 x 10⁶ m^-1

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