For the complete Lyman series of hydrogen, let λₘₐₓ be its maximum wavelength and λₘᵢₙ its series-limit minimum wavelength. If λₘₐₓ ÷ λₘᵢₙ = N ÷ 3, find N.
Physics · Atomic Structure and X-rays · atomic structure and bohr atomic model
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Question

For the complete Lyman series of hydrogen, let λₘₐₓ be its maximum wavelength and λₘᵢₙ its series-limit minimum wavelength. If λₘₐₓ ÷ λₘᵢₙ = N ÷ 3, find N.

G1 Physics · Accepted Answer

For the Lyman series, all transitions end at n = 1. Maximum wavelength corresponds to the smallest Lyman energy difference, from the second level to the ground state: 1 ÷ λₘₐₓ = R[1−1 ÷ 2²] = 3R ÷ 4. Therefore, λₘₐₓ = 4 ÷ (3R). Minimum wavelength is the series limit, with the initial level approaching infinity: 1 ÷ λₘᵢₙ = R. Thus, λₘᵢₙ = 1 ÷ R. Hence, λₘₐₓ ÷ λₘᵢₙ = 4 ÷ 3. Comparing with N ÷ 3 gives N = 4.