1.

Calculate the wavelength of minimum energy transition in the Balmer series of hydrogen spectra​

Answer»

λ = 1.46236× 10⁷ mExplanation:n₁ = 1n₂ = 2 (BALMER series)Rh - constant = 1.09677 × 10⁷ m⁻¹1 / λ = 1 / Rh [ 1/n₁² - 1/n₂²]        = 1 /  1.09677 × 10⁷ [ 1/1² - 1/2²]        = 1 /  1.09677 × 10⁷ [ 1/1 - 1/4]        = 1 /  1.09677 × 10⁷ [ 4 - 1 / 4]        = 1 /  1.09677 × 10⁷ [ 3 / 4]λ = 1.09677 × 10⁷ × 4 / 3λ = 1.46236× 10⁷ m



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