AP EAMCET · PHYSICS · Atomic Physics
The difference between the frequencies of second and first Paschen lines of hydrogen atom is
(R - Rydberg constant and c - speed of light in vacuum)
- A \(\frac{9 \mathrm{Rc}}{16}\)
- B \(\frac{16 R c}{25}\)
- C \(\frac{9 R c}{400}\)
- D \(\frac{3 R c}{200}\)
Answer & Solution
Correct Answer
(C) \(\frac{9 R c}{400}\)
Step-by-step Solution
Detailed explanation
\( \nu = R c \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \) \( \nu_1 = R c \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = R c \left( \frac{1}{9} - \frac{1}{16} \right) \)…
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