calculate the wavelength of minimum energy transition in the Balmer series of hydrogen spectra
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Answer:
λ = 1.46236× 10⁷ m
Explanation:
n₁ = 1
n₂ = 2 (balmer series)
Rh - constant = 1.09677 × 10⁷ m⁻¹
1 / 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 / 3
λ = 1.46236× 10⁷ m
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