AP EAMCET · PHYSICS · Atomic Physics
The ratio of the shortest wavelengths of Bracket and Balmer series of hydrogen atom is
- A \(2: 1\)
- B \(3: 2\)
- C \(4: 1\)
- D \(6: 5\)
Answer & Solution
Correct Answer
(C) \(4: 1\)
Step-by-step Solution
Detailed explanation
\( \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \) For shortest wavelength, \( n_2 = \infty \). Thus, \( \lambda = \frac{n_1^2}{R_H} \). Balmer series (\( n_1 = 2 \)): \( \lambda_B = \frac{2^2}{R_H} = \frac{4}{R_H} \) Brackett series (\( n_1 = 4 \)):…
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