Concept: The longest wavelength corresponds to the minimum energy transition in the Paschen series, which is from \( n = 4 \) to \( p = 3 \).
Formula:$$ \frac{1}{\lambda} = R_H \left( \frac{1}{3^2} - \frac{1}{4^2} \right) $$
Solution:- Using Rydberg's constant \( R_H \approx 1.097 \times 10^7 \text{ m}^{-1} \).
- \( \frac{1}{\lambda} = R_H \left( \frac{1}{9} - \frac{1}{16} \right) = R_H \left( \frac{16 - 9}{144} \right) = R_H \left( \frac{7}{144} \right) \).
- \( \lambda = \frac{144}{7 \times 1.097 \times 10^7} \approx \frac{144}{7.679 \times 10^7} \approx 1.875 \times 10^{-6} \text{ m} \).
- In standard decimal form, \( 1.87 \times 10^{-6} \text{ m} \) is written as \( 0.00000187 \text{ m} \).
Why other options are incorrect:Option A describes an absurdly massive distance. Options B and D are wave numbers (\( 1/\lambda \)) rather than actual wavelengths, as indicated by the "/ m" (per meter) units.
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