Concept: The shortest wavelength in the Lyman series corresponds to the transition with the maximum energy. This happens when an electron falls from infinity (\( n = \infty \)) down to the ground state (\( p = 1 \)).
Formula:$$ \frac{1}{\lambda} = R_H \left( \frac{1}{p^2} - \frac{1}{n^2} \right) $$
Solution:- Set \( p = 1 \) (Lyman series) and \( n = \infty \) (shortest wavelength limit).
- Substitute into the Rydberg formula: \( \frac{1}{\lambda} = R_H \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) \).
- Since \( \frac{1}{\infty^2} = 0 \), the equation simplifies strictly to \( \frac{1}{\lambda} = R_H(1 - 0) = R_H \).
- Inverting both sides gives \( \lambda = \frac{1}{R_H} \).
Why other options are incorrect:Option A is the wave number (\( 1/\lambda \)), not the wavelength. Options C and D represent calculations for entirely different transitions or incorrect algebraic inversions.
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