Concept: The longest wavelength in any spectral series corresponds to the minimum energy transition. For the Balmer series, this occurs when an electron transitions from \( n = 3 \) to \( p = 2 \).
Formula:$$ \frac{1}{\lambda} = R_H \left( \frac{1}{p^2} - \frac{1}{n^2} \right) $$
Solution:- Substitute \( p = 2 \) and \( n = 3 \) into the Rydberg equation:
- \( \frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \)
- \( \frac{1}{\lambda} = R_H \left( \frac{1}{4} - \frac{1}{9} \right) = R_H \left( \frac{9 - 4}{36} \right) = R_H \left( \frac{5}{36} \right) \)
- Invert the result to solve for wavelength: \( \lambda = \frac{36}{5R_H} \).
Why other options are incorrect:The other options result from incorrect fractional subtractions or choosing the wrong series transitions.
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